Ionic Equilibria in Aqueous Solutions

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					            Ionic Equilibria in Aqueous Solutions
           Equilibria of Acid-Base Buffer Systems
               The Common Ion Effect
               The Henderson-Hasselbalch Equation
               Buffer Capacity and Range
               Preparing a buffer
           Acid-Base Titration Curves
               Acid-Base Indicators
               Strong-Acid-Strong Base Titrations
               Weak Acid-Strong Base Titrations
               Weak Base-Strong Acid Titrations
               Polyprotic Acid Titrations
               Amino Acids as Polyprotic Acids
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            Ionic Equilibria in Aqueous Solutions
           Equilibria of Slightly Soluble Ionic Compounds
               The Solubility-Product Constant
               Calculations involving Ksp
               The Effect of a Common Ion
               The Effect of pH
               Qsp vs. Ksp
           Equilibria involving Complex Ions
               Formation of Complex Ions
               Complex Ions and Solubility
               Amphoteric Hydroxides




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            Ionic Equilibria in Aqueous Solutions
           The simplest acid-base equilibria are those in which a
            single acid or base solute reacts with water.

           In this chapter, we will look at solutions of weak acids
            and bases through acid/base ionization, the reactions of
            salts with water, and titration curves. All of these
            processes involve equilibrium theory.




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            Ionic Equilibria in Aqueous Solutions
           Equilibria of Acid-Base Buffer Systems
               Buffer
                   An Acid-Base Buffer is a species added to an
                    solution to minimize the impact on pH from the
                    addition of [H3O+] or [OH-] ions
                   Small amounts of acid or base added to an
                    unbuffered solution can change the pH by several
                    units.
                   Since pH is a logarithmic term, the change in
                    [H3O+] or [OH-] can be several orders of magnitude




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            Ionic Equilibria in Aqueous Solutions
           The Common Ion Effect
             Buffers work through a phenomenon known as the:

                         Common Ion Effect
               The common ion effect occurs when a given ion is
                added to an “equilibrium mixture of a weak acid or
                base that already contains that ion
               The additional “common ion” shifts the equilibrium
                away from its formation to more of the undissociated
                form; i.e., the acid or base dissociation decreases.
               A buffer must contain an “acidic” component that can
                react with the added OH- ion, and a “basic”
                component than can react with the added [H3O+]
               The buffer components cannot be just any acid or
                base
               The components of a buffer are usually the conjugate
                acid-base pair of the weak acid (or base) being
                buffered
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            Ionic Equilibria in Aqueous Solutions
               Ex: Acetic Acid & Sodium Acetate
             CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
                  Acid           Base     Conjugate Base   Conjugate Acid
               Acetic acid is a weak acid, slightly dissociated
               Now add Sodium Acetate (CH3COONa), a strong
                electrolyte (acetate ion is the conjugate base)
            CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq, added)
               The added acetate shifts the equilibrium to the left
                forming more undissociated acetic acid
               This lowers the extent of acid dissociation, which
                lowers the acidity by reducing the [H3O+]
               Similarly, if acetic acid is added to a solution of
                Sodium Acetate, the acetate ions already present act
                to suppress the dissociation of the acid
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       Ionic Equilibria in Aqueous Solutions




     When a small amount of [H3O+] is add to acetic acid/acetate buffer, that
     same amount of acetate ion (CH3COO-) combines with it, increasing the
     concentration of acetic acid (CH3COOH). The change in the [HA]/[A-]
     ratio is small; the added [H3O+] is effectively tied up; thus, the change
     in
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            Ionic Equilibria in Aqueous Solutions
           Essential Features of a Buffer
             A buffer consists of high concentrations of the
              undissociated acidic component [HA] and the
              conjugate base) [A-] component
                                            +         -
             When small amounts of [H3O ] or [OH ] are added to
              the buffer, they cause small amounts of one buffer
              component to convert to the other component
              Ex. When a small amount of H3O+ is added to an
              acetate buffer, that same amount of CH3COO-
              combines with it, increasing the amount of
              undissociated CH3COOH- tying up potential [H3O+]
              Similarly, a small amount of OH- added combines with
              undissociated CH3COOH to form CH3COO- tying up
              potential [OH-]
              In both cases, the relative changes in the amount of
              buffer components is small, but the added [H3O+] or
              [OH-] ions are tied up as undissociated components;
              thus little impact on pH
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            Ionic Equilibria in Aqueous Solutions
           The equilibrium perspective
                            [H + ][A- ] [H 3O+ ][CH 3COO- ]
                       Ka =            =
                              HA            [CH 3COOH]
                         +        [CH 3COOH]
                   [H 3O ] = Ka ×
                                   [CH 3COO - ]
           The [H3O+] (pH) of the solution depends directly on the
            buffer-component concentration ratio
           If the ratio [HA]/[A-] goes up, [H3O+] goes up (pH down)
           If the ratio [HA]/[A-] goes down, [H3O+] goes down
            ([OH-] goes up) and pH increases
           When a small amount of a strong acid is added, the
            increased amount of [H3O+] reacts with a stoichiometric
            amount of acetate ion from the buffer to form more
            undissociated acetic acid
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                                Practice Problem
     What is the pH of a buffer solution consisting of 0.50 M
     CH3COOH (HAc) and 0.50 M CH3COONa (NaAc)?
                                Ka (CH3COOH) = 1.8 x 10-5

            x = [CH 3COOH]dissoc = [H 3O + ]
            [H 3O + ] from autoionization of H 2O is neglible and is neglected

              Concentration       HAc(aq)    +   H2O(l)      Ac-(aq)    +   H3O+
                 Initial Ci        0.50            -           0.50           0
                                   -X             -            +X            +X
               Equilibrium Cf     0.50 - X         -          0.50 + X        X


      If x is small, then:
                              [CH3COOH] = 0.50 - x  0.50
                              [CH 3COO - ] = 0.50 + x  0.50                        Con’t
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            Practice Problem (Con’t)
                      [H 3O + ][CH 3COO- ]
                 Ka =
                          [CH 3COOH]
                         +     [CH 3COOH]
             X = [H 3O ] = Ka×
                                [CH 3COO -]
                                 0.5
              x = 1.8 x 10 -5×       = 1.8×10 -5M
                                 0.5
             pH = -log[H 3O +] = -log(1.8 x 10 -5
                                                )
                             pH = 4.74




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                   Practice Problem (Con’t)
     Calculate the pH of the previous buffer solution after
     adding 0.020 mol of solid NaOH to 1 L of the solution
     Set up reaction table for the stoichiometry of NaOH (strong
     base) to Acetic acid (weak acid)
                            0.020mol
          Molarity NaOH =             = 0.020 mol / L (M)
                                1L
            Concentration    HAc(aq)    +   OH-(aq)      Ac-(aq)    +    H2O(l)
               Initial Ci     0.50           0.020         0.50             -
                            - 0.020        -0.020        + 0.020           -
            Equilibrium Cf     -.48           0.0          0.52             -

     Set up reaction table for acid dissociation using new
     concentrations
            Concentration    HAc(aq)    +   H2O          Ac-(aq)    +   H3O+(aq)
               Initial Ci     0.48            -            0.50             0
                              -x             -           + 0.020          +X
            Equilibrium Cf   0.48 - X         -           0.52 + X          X       Con’t
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                Practice Problem (Con’t)
     Assuming x is small:
     [CH 3COOH] = 0.48 M - x = 0.48 M and [CH 3COO - ] = 0.52 M + x = 0.52 M

                                   [CH 3COOH]                   0.48
            x = [H 3O + ] = Ka ×                = (1.8×10-5 ) ×
                                    [CH 3OO - ]                 0.52

                           [H 3O + ] = 1.7 × 10-5 M
                    pH = -log[H 3O + ] = -log(1.7 ×10-5 )
                    pH = 4.77

    The addition of the strong base increased the concentration
    of the basic buffer [CH3COO-] component (0.52 M) at the
    expense of the acidic buffer component [CH3COOH] (0.48 M)
    The concentration of the [H3O+] (x) also increased, but only
    slightly, thus, the pH changed only slightly from 4.74 to 4.77
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            Ionic Equilibria in Aqueous Solutions
           The Henderson-Hasselbalch Equation
               For any weak acid, HA, the dissociation equation and
                Ka expression are:
                         HA + H 2O           H 3O + + A -
                                      [H 3O + ][A - ]
                                 Ka =
                                          [HA]
               The key variable that determines the [H3O+] is the
                ratio of acid species to base species
                             +         [Ha] (acid species)
                        [H 3O ] = ka ×
                                       [A - ] (base species)
                             +                  [HA]
                    -log[H 3O ] = - logKa - log -
                                                 [A ]
                                              [A - ]     Note change in sign of “log”
                             pH = pKa + log              term and inversion of acid-
                                             [HA]        base terms
                                             [base]
                             pH = pKa + log
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                                             [acid]                                     14
                       Practice Problem
     What is the pH of a buffer formed by adding 0.15 mol of
     acetic acid and 0.25 mol of sodium acetate to 1.5 L of
     water?
     Ans:   Both the acid and salt forms are added to the
            solution at somewhat equal concentrations
            Neither form has much of a tendency to react
            Dissociation of either form is opposed in bi-directional
            process
            The final concentrations will approximately equal the
            initial concentrations
                     [HA] = 0.15 mol / 1.5L = 0.10 M
                  [A - ] = 0.25 mol / 1.5L = 0.167 M
                   Ka for acetic acid = 1.7 x 10-5
                   pKa = -log(Ka) = -log(1.7 x 10-5 ) = 4.77      Con’t
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                         Practice Problem
     From the Henderson-Hasselbalch equation

                                           [A - ] 
                           pH = Pka + log         
                                           [HA] 

                             0.167M 
            pH = 4.77 + log          = 4.77 + log(1.67) = 4.77 + 0.22
                             0.10M 
            pH = 4.99




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            Ionic Equilibria in Aqueous Solutions
           Buffer Capacity
               A buffer resists a pH change as long as the
                concentration of buffer components are large
                compared with the amount of strong acid or base
                added
               Buffer Capacity is a measure of the ability to resist pH
                change
               Buffer capacity depends on both the absolute and
                relative component concentrations
                   Absolute - The more concentrated the components
                    of a buffer, the greater the buffer capacity
                   Relative – For a given addition of acid or base, the
                    buffer-component concentration ratio changes less
                    when the concentrations are similar than when they
                    are different
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            Ionic Equilibria in Aqueous Solutions
           Ex. Consider 2 solutions
               Solution 1 – Equal volumes of 1.0M HAc and 1.0 M Ac-
               Solution 2 – Equal volumes of 0.1M Hac and 0.1M Ac-
               Same pH (4.74) but 1.0 M buffer has much larger
                buffer capacity




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            Ionic Equilibria in Aqueous Solutions
     Ex. Buffer #1 [HA] = [A-] = 1.000M
           Add 0.010 mol of OH- to 1.00 L of buffer
           [HA] changes to 0.990 M    [A-] changes to 1.010 M
                [A - ]init 1.000M                      [A - ]final   1.010M
                          =       = 1.000                          =        = 1.020
               [HA]init 1.000M                        [HA]final 0.990M
                                              1.020 - 1.000
                         Percent change =                   ×100 = 2%
                                                 1.000
             Buffer #2 [HA] = 0.250 M     [A-] = 1.75 M
               Add 0.010 mol of OH- to 1.00 L of buffer
                [HA] changes to 0.240 M      [A-] changes to 1.760 M
                [A - ]init   1.750M                    [A - ]final   1.760M
                           =        = 7.000                        =        = 7.330
               [HA]init 0.250M                        [HA]final 0.240M
                                       7.330 - 7.000
                        Percent change =             ×100 = 4.7%
                                          7.000
            Buffer-component concentration ratio much larger when initial
4/30/2010   concentrations of the components are very different                       19
            Ionic Equilibria in Aqueous Solutions
           A buffer has the highest capacity when the component
            concentrations are equal - [A-]/[HA] = 1
                               [A - ] 
               pH = pKa + log          = pKa + log1 = pKa + 0 = pKa
                               [HA] 
           Corollary: A buffer whose pH is equal to or near the pKa
            of its acid component has the highest buffer capacity




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            Ionic Equilibria in Aqueous Solutions
           Buffer Range
               pH range over which the buffer acts effectively is
                related to the relative component concentrations
               The further the component concentration ratio is from
                1, the less effective the buffering action
               If the [A-]/[HA] ratio is greater than 10 or less than
                0.1 (or one component concentration is more than 10
                times the other) the buffering action is poor
                                10                           1 
                pH = pKa + log   = pKa + 1   pH = pKa + log   = pKa - 1
                                1                            10 
               Buffers have a usable range within ± 1 pH unit of the
                pKa of the acid components



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            Ionic Equilibria in Aqueous Solutions
           Equilibria of Acid-Base Systems
             Preparing a Buffer
               Choose the Conjugate Acid-Base pair
                   Driven by pH
                   Ratio of component concentrations close to 1
                    and pH ≈ pKa
               Ex. Assume you need a biochemical buffer whose
                 pH is 3.9
                   pKa of acid component should be close to 3.9
                    Ka = 10-3.9 = 1.3 x 10-4
                   From a table select possibilities
                     Lactic acid     (pKa = 3.86)
                     Glycolic acid (pKa = 3.83)
                     Formic Acid     (pKa = 3.74)
                                                                   Con’t
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            Ionic Equilibria in Aqueous Solutions
       Preparing a Buffer (Con’t)
           To avoid common biological species, select Formic
            Acid
                Buffer components
                  Formic Acid – HCOOH
                  Formate Ion – HCOO
                                         -
                  Obtain soluble Formate salt – HCOONa
           Calculate Ratio of Buffer Component Concentrations
            ([A-]/[HA]) that gives desired pH
                                  [A - ]                           [HCOO- ] 
                  pH = Pka + log                 3.9 = 3.74 + log           
                                  [HA]                             [HCOOH] 
                 [HCOO- ]                               [HCOO- ]       0.16
            log            = 3.9 - 3.74 = 0.16    so              = 10      = 1.4
                 [HCOOH]                                [HCOOH] 

                1.4 mol of HCOONa is needed for every mole HCOOH
                                                                                    Con’t
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            Ionic Equilibria in Aqueous Solutions
               Preparing a Buffer (Con’t)
                   Determine the Buffer Concentrations
                       The higher the concentration of the components,
                        the higher the buffer capacity
                       Assume 1 Liter of buffer is required and you
                        have a stock of 0.40 M Formic Acid (HCOOH)
                       Compute moles and then grams of Sodium
                        Formate (CHOONa) needed to produce 1.4/1.0
                        ratio
                                       0.40 mol HCOOH
            Moles of HCOOH = 1.0 L ×                  = 0.40 mol HCOOH
                                           1.0 L soln

                                                 1.40 mol HCOONa
            Moles of HCOONa = 0.40 mol HCOOH ×                   = 0.56 mol HCOONa
                                                  1.0 mol HCOOH

                                                 68.01 g HCOONa
            Mass of HCOONa = 0.56 mol HCOONa ×                  = 38 g HCOONa Con’t
                                                  1 mol HCOONa
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            Ionic Equilibria in Aqueous Solutions
               Preparing a Buffer (Con’t)
                   Mix the solution and adjust the pH
                       The prepared solution may not be an ideal
                        solution (see Chapter 13, section 6) – The
                        desired pH (3.9) may not exactly match the
                        actual value of the buffer solution
                       The pH of the buffer solution can be adjusted by
                        a few tenths of a pH unit by adding strong acid
                        or strong base.




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                            Practice Problem
    Example of preparing a buffer solution without using the
    Henderson-Hasselbalch equation
    How many grams of Sodium Carbonate (Na2CO3) must be
    added to 1.5 L of a 0.20 M Sodium Bicarbonate (NaHCO3)
    solution to make a buffer with pH = 10.0?
                            Ka (HCO3-) = 4.7 x 10-11
             Conjugate Pair - HCO3- (acid) and CO3- (base)
             Convert pH to [H3O+] (H3O+] = 10-pH = 1.0 x 10-10 M)
             Use Ka to compute [CO32-]

                     HCO 3- (aq) + H 2O(l)      H 3O - (aq) + CO 3 2- (aq)
                                         [H 3O + ][CO 3 2- ]
                                    Ka =
                                             [HCO 3- ]
                             [HCO 3- ]            -11  (0.20) 
                     -
                [CO 3 ] = Ka           = (4.7  10 )           -10 
                                                                      = 0.094 M   Con’t
                                   +
                              [H 3O ]                  1.0  10 
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                   Practice Problem (Con’t)
        Compute the amount of CO3- needed for the given volume (1.5 L)
                                       0.094 mol CO 3 2-
        Moles CO 3 2-   = 1.5 L soln ×                   = 0.14 mol CO 3 2-
                                           1 L soln
      Compute the mass (g) of Na2CO3 needed
                                                 105.99 g Na 2CO 3
      Mass (g) of Na 2CO 3 = 0.14 mol Na 2CO 3 x
                                                  1mol Na 2CO 3
      Mass Na2CO3 = 15g

     Dissolve the 15g Na2CO3 in slightly less than 1.5 L of the NaHCO3,
     say 1.3 L.
     Add additional NaHCO3 to make 1.5 L; adjust pH to 10.0 with strong acid
     Check: For a useful buffer range, the concentration of the acidic component
     [HCO3-] must be within a factor of “10” of the concentration of the basic
     component [CO32-]
     (1.5 L) (0.20 M HCO3-) = 0.30 mol HCO3- vs. 0.14 mol CO32-

4/30/2010
             0.30 / 0.14 = 2.1 << 10; therefore buffer solution is reasonable      27
            Ionic Equilibria in Aqueous Solutions
           Acid-Base Titration Curves
               Acid-Base Indicators
                   Weak organic acid (HIn) that has a different color
                    than its conjugate base (In-)
                   The color change occurs over a relatively narrow
                    pH range
                   Only small amounts of the indicator are needed;
                    too little to affect the pH of the solution
                   The color range of typical indicators reflects a 102
                    - fold range in the [HIn]/In-] ratio
                   This corresponds to a pH range of 2 pH units




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                     Mixing Acids & Bases
           Acids and bases react through neutralization reactions
           The change in pH as an acid is mixed with a base is
            tracked with an Acid-Base titration curve
           Titration:
             Titration Curve: Plot of pH vs. the volume of the
               “Strong” acid or “Strong” base being added via buret
             Solution in buret is called the “titrant”
           Equivalence Point: Point in a titration curve where
            stoichiometric amounts of acid and base have been
            mixed (point of complete reaction)
           3 Important Cases:
             Strong Acid     +       Strong Base (& vice versa)
             Weak Acid       +       Strong Base
             Weak Base       +       Strong Acid
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                           Titration Curves
           Titration Curve: Plot of measured pH versus Volume of
            acid or base added during a “Neutralization” experiment
           All Titration Curves have a characteristic “sigmoid (S-
            shaped) profile
               Beginning of Curve:   pH changes slowly
               Middle of Curve:      pH changes very rapidly
               End of Curve:         pH changes very slowly again
           pH changes very rapidly in the titration as the
            equivalence point (point of complete reaction) is reached
            and right after the equivalence point




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                    Neutralization Reactions
           Acids and Bases react with each other to form salts (not
            always) and water (not always) through
                            Neutralization Reactions
           Neutralization reaction between a strong acid and strong base
                  HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)
                    acid           base           salt        water
           Neutralization of a strong acid and strong base reaction lies
            very far to the right (Kn is very large); reaction goes to
            completion; net ionic equation is
                          H3O+(aq) + OH-(aq)  2 H2O(l)
                          [H2O]2          1           1
                  Kn =              =              =    = 1×1014
                       [H3O+ ][OH- ] [H3O+ ][OH- ]   Kw

           H3O+ and OH- efficiently react with each other to form water


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                         Titration Curves
            Curve for Titration of a Strong Acid by a Strong Base




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                           Practice Problem
    Calculate pH after the addition of 10, 20, 30 mL NaOH
    during the titration of 20.0mL of 0.0010 M HCL(aq) with
    0.0010 M NaOH(aq)
                 Strong Acid + Strong Base Mixture
                   H 3O +(aq) +    OH - (aq)       2H 2O(l)

      Initial pH HCl soln: [HCl] = [H3O+] = 0.0010 M
              pH = - log[H 3O + ] = - log(0.0010) = 3.0
    Add 10 mL 0.0010 M NaOH
            1 mol H3O+ disappears for each mole OH- added
               moles = Molarity(mol / L) × Volume(L)
        Initial moles of H 3O + = 0.0010 mol / L x 0.020L = 0.00002 mol H 3O +
        -moles OH -added       = 0.0010 mol / L x 0.010L = 0.00001 mol OH -
       moles of H 3O + remaining                          = 0.00001 mol H 3O +   Con’t
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                    Practice Problem (Con’t)
       Calculate [H3O+] & pH, taking “Total Volume” into account
                            amount(mol) of H 3O +remaining
         [H 3O + ] =
                     original volume of acid + volume of added base
                        0.000010 mol H 3O +
            [H 3O + ] =                     = 0.000033M
                          0.020L + 0.010L
            pH = -log[H 3O + ] = -log(0.0000333) = 3.48
       pH at “Equivalence Point”
        Equal molar amounts of HCL & NaOH have been mixed
        20 mL x 0.0010 M HCl reacts with 20 mL x 0.0010 M NaOH
        At the equivalence point all H3O+ has been neutralized by
        the addition of all the NaOH, except for [H3O+] from
        autoionization of water, i.e., negligible
                         [H3O+] = pure water = 1 x 10-7 M           Con’t
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                        Practice Problem (Con’t)
           After the equivalence point
                   After the equivalence point, the pH calculations are
                    based on the moles of excess OH- present
                   A total of 30 mL of NaOH has been added
      Total moles of OH - added = 0.0010 mol / L x 0.030L = 0.00003 mol OH -
      -moles H 3O + consumed       = 0.0010 mol / L x 0.020L = 0.00002 mol H 3O +-
       moles of excess OH -                                  = 0.00001 mol OH -

                -         amount excess OH -              0.000010 moles OH -
       [OH ] =                                          =
               original volume acid + volume base added     0.020 L + 0.030 L
       [OH - ] = 0.00020M

       p[OH - ] = - log[OH - ] = - log(0.00020) = 3.70

       pH = pKw - pOH = 14.0 - 3.7 = 10.3
                                                                                     Con’t
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                    Neutralization Reactions
           Neutralization reactions between weak acids and strong
            bases (net ionic equation shown below)
    C6H5COOH(aq) + NaOH  C6H5COONa(aq) + H2O(l)                 KN=? benzoic
       acid        base    benzoate “salt” water
           Above equilibrium also lies very far to the right; the two
            equilibria below can be added together to provide the
            overall neutralization reaction shown above
        (1) C6H5COOH(aq) + H2O(l)  C6H5COO-(aq) + H3O+(aq)
                                    Ka = 1.7 x 10-5
        (2) H3O+(aq) + OH-(aq)  2 H2O(l)
                           1/Kw = (1 x 10-14)-1 = 1 x 1014
           The overall neutralization equilibrium constant (Kn) is the product
            of the two intermediate equilibrium constants (Ka & 1/Kw)
        KN = Ka X 1/Kw = (1.7 x 10-5)(1 x 1014) = 1.9 x 109 (very large!)
    
4/30/2010   Weak acids react completely with strong bases                     36
                         Titration Curves
            Curve for Titration of a Weak Acid by a Strong Base




4/30/2010                                                         37
        Weak Acid-Strong BaseTitration Curve
           Consider the reaction between Propanoic Acid (weak
            acid) and Sodium Hydroxide (NaOH) (strong base)
           Ka for CH3CH2COOH (HPr) - 1.3 x 10-5
           The Titration Curve (see previous slide)
               The bottom dotted curve corresponds to the strong
                acid-strong base titration
               The Curve consists of 4 regions, the first 3 of which
                differ from the strong acid case
                   The initial pH is “Higher”
                       The weak acid dissociates only slightly
                        producing less [H3O+] than with a strong acid
                   The gradually arising portion of the curve before
                    the steep rise to the equivalence point is called the
                    “buffer region”
4/30/2010                                                                   38
       Weak Acid-Strong Base Titration Curve
               As HPr reacts with the strong base, more and
                more of the conjugate base (Pr-) forms creating
                an (HPr/Pr-) buffer
               At the midpoint of the “Buffer” region, half the
                original HPr has reacted
                                              [Pr - ]
                      [HPr = [Pr - ] or               =1
                                             [HPr]
                                  [Pr - ] 
                  pH = Pka + log           = pKa + log(1) = pKa
                                  [HPr] 
                   pH at midpoint of “Buffer” region is common
                   method of estimating pKa of an “unknown”
                   acid




4/30/2010                                                           39
       Weak Acid-Strong Base Titration Curve
               The pH at the “equivalence point” is greater than
                7.00
                    The weak-acid anion (Pr-) acts as a weak base
                     to accept a proton from H2O forming OH-
                    The additional [OH-] raises the pH
               Beyond the equivalence point, the ph increases
                slowly as excess OH- is added




4/30/2010                                                            40
                       Practice Problem
     Calculate the pH during the titration of 40 ml of 0.1000 M
     of the weak acid, Propanoic acid (HPr); Ka = 1.3 x 10-5 after
     the addition of the following volumes of 0.1000 M NaOH
            0.00 mL      30.0 mL         40.0 mL          50.0 mL
     a. 0.0 mL NaOH added
           HPr(aq) + H 2O             H 3O + (aq) + Pr - (Aq)
               [HPr] = [HPr]init - [HPr]dissoc  [HPr]init
                          [H 3O+ ] = [Pr - ] = x
                    [H 3O + ][Pr - ]                 (X)(X)
               Ka =                  = 1.3  10-5 
                        [HPr]                       [HPr]init
               X2 = [H 3O+ ]2 = 1.3 × 10-5 × 0.1000 M
                         [H 3O+ ] = 1.1  10-3 M
                                                                    Con’t
4/30/2010
             pH = - log[H 3O ] = - log(1.1  10 ) = 2.96
                               +                    -3
                                                                        41
                     Practice Problem (Con’t)
     b. 30.00 mL 0.1000 M NaOH added
            Original moles of HPr = 0.0400L x 0.1000 M = 0.004000 mol HPr
            moles of NaOH added = 0.03000 L x 0.1000 M = 0.003000 mol OH -
               For each mol of NaOH that reacts, 1 mol Pr- forms
                Amount (mol)    HPr(aq)     +    OH-(aq)     →       Pr-      +   H2O(l)
                   Initial     0.004000          0.003000            0              -
                  Change       - 0.003000        -0.003000       + 0.003000         -
                   Final       0.001000             0             0.003000          -

               Last line shows new initial amounts of HPr & Pr- that
               react to attain a “new’ equilibrium
                                 [HPr]   0.001000 mol
               At equilibrium:         =              = 0.3333
                                       [Pr - ]      0.003000 mol
                                                       [H 3O+ ][Pr - ]
                                                  Ka =
                                                           [HPr]
                                            [HPr]
                      [H 3O + ] = Ka ×               = (1.3 x 10-5 ) (0.3333) = 4.3 x 10-6 M
                                             [Pr - ]
                                                                                               Con’t
4/30/2010
                               pH = -log[H 3O+ ] = -log(4.3  10-6 ) = 5.37                        42
                      Practice Problem (Con’t)
     c. 40.00 mL NaOH (0.00400 mol) added
         Reaction is at “equivalence point”
         All the HPr has reacted, leaving 0.004000 mol Pr-
         Calculate concentration of Pr
                                        -
                      -        0.004000 mol
                  [Pr ] =                        = 0.05000 M
                          0.04000 L + 0.004000 L
                                                       Kw   1.0  10-14
               Calculate Kb                      Kb =    =             = 7.7  10-10
                                                       Ka   1.3  10-5
                         -
             [HPr] = [OH ]                       Kb =
                                                       [HPr][OH - ] [OH - ]2
                                                                   =
                                                             -
                                                          [Pr ]      [Pr - ]

                  [OH - ] = K b x [Pr - ]         K w = [H 3O+ ][OH - ]

                                 Kw                    1.0  10-14
                [H 3O + ]                    =                              = 1.6  10-9 M
                              K b  [Pr - ]       (7.7  10-10 ) (0.05000)

                pH = - log[H 3O + ] = - log(1.6  10-9 ) = 8.80
                                                                                              Con’t
4/30/2010                                                                                         43
                      Practice Problem (Con’t)
     d. 50 mL 0.1000 M NaOH added
                                                   -
        Beyond equivalence point, just excess OH is being
         added
        Calculation of pH is same as for strong acid-strong
         base titration
                                           -
        Calculate the moles of excess OH from the initial
         concentration (0.1000 M) and the remaining volume
         (50.0 mL - 40.0 mL)
            Moles excess OH- = (0.1000 M)(0.05000 L - 0.04000 L) = 0.001000 mol
                     -      moles of excess OH -        0.001000 mol
                  [OH ] =                           =                    = 0.01111 M
                               total volume            0.04L + 0.05L
                             Kw                            moles of excess OH -
                [H 3O + ] =              where [OH - ] =
                            [OH - ]                             total volume
                             Kw       1.0  10-14
                      +
                 [H 3O ] =          =             = 9.0  10-13
                            [OH - ]    0.01111
                   pH = - log[H 3O + ] = - log(9.0  10-13 ) = 12.05
4/30/2010                                                                              44
            Weak Base-Strong Acid Titration
      Neutralization reactions between weak bases and strong
       acids (net ionic equation shown below)
              NH3(aq) + H3O+(aq)  NH4+(aq) + H2O(l)           KN=?
               base          acid           salt           water

      Above equilibrium also lies very far to the right; the two
       equilibria below can be added together to provide the
       overall neutralization reaction shown above
            NH3(aq) + H2O(aq)  NH4+ + OH-(aq)             Kb = 1.8 x 10-5

            H3O+(aq) + OH-(aq)  2 H2O(l)                1/Kw = 1.0 x 1014

            KN = Kb X 1/Kw = (1.8 x 10-5)(1 x 1014) = 1.8 x 109 (very large!)

      Weak bases react completely with strong acids


4/30/2010                                                                       45
            Weak Base-Strong Acid Titration
            Curve for Titration of a Weak Base by a Strong Acid




4/30/2010                                                         46
                     Polyprotic Acid Titrations
           Polyprotic Acids have more than one ionizable proton
           Except for Sulfuric Acid (H2SO4), the common polyprotic
            acids are all weak acids
           Successive Ka values for a polyprotic acid differ by
            several orders of magnitude
           The first H+ is lost much more easily than subsequent
            ones
           All “1st” protons (H+) are removed before any of the
            “2nd” protons
               H 2SO3 (aq) + H 2O(l)     HSO3 - (aq) + H 3O+ (aq)

                    K a1 = 1.4 x 10-2    and    pK a1 = 1.85

               HSO3 - (aq) + H 2O(l)     SO3 2- (aq) + H 3O + (aq)

                    K a2 = 6.5 x 10-8    and     pK a2 = 7.19
4/30/2010                                                             47
                  Polyprotic Acid Titrations
           Each mole of H+ is titrated separately
           In a diprotic acid two OH- ions are required to
            completely remove both H+ ions
           All H2SO3 molecules lose one H+ before any HSO3- ions
            lose a H+ to form SO3-
                                   
                H 2SO3  HSO3  SO3 2-
                         1 mol OH -   -   1 mol OH -




                                                       Titration curves looks like two
                                                       weak acid-strong base curves
                                                       joined end-to-end
                                                       HSO3- is the conjugate base
                                                       of H2SO3 (Kb = 7.1x10-13)
                                                       HSO3- is also an acid
                                                       (Ka=6.5x10-8) dissociating to
                                                       form its conjugate base SO32-



4/30/2010                                                                                48
      Equilibria - Slightly Soluble Ionic Compounds
           Equilibria of Slightly Soluble Ionic Compounds
               Slightly soluble compounds reach “equilibrium” with
                relatively “little” solute dissolved




           Slightly soluble compounds can produce complex
            mixtures of species
           Discussion here is to assume that the small amount of
            slightly soluble ionic compounds dissociate completely
4/30/2010                                                             49
      Equilibria - Slightly Soluble Ionic Compounds
           Equilibrium exists between solid solute and the aqueous
            ions
                    PbSO4(s) ⇄ Pb2+(aq)               + SO42-(aq)
           Set up “Reaction Quotient”
                                 [Pb 2+ ][SO 4 2- ]
                            Qc =
                                   [PbSO 4 ]
           Incorporate “constant” concentration of solid into “Qc”
                              Qsp = Qc[PbSO4]
                       Qc [PbSO 4 ] = Qsp = [Pb 2+ ][SO 4 2- ]
           When solid PbSO4 reaches equilibrium with Pb2+ and SO4-
            at saturation, the numerical value of Qsp attains a
            constant value called the solubility-product constant (Ksp)
            Qsp at saturation = K sp = solubility - product constant
4/30/2010                                                                 50
      Equilibria - Slightly Soluble Ionic Compounds
           The Solubility Product Constant (Ksp)
               Value of Ksp depends only on temperature, not
                individual ion concentrations
               Saturated solution of a slightly soluble ionic compound,
                MpXq, composed of ions Mn+ and Xz-, the equilibrium
                condition is:
                          Qsp = [M n+ ]p [X z- ]q = K sp
               Single-Step Process
                  Cu(OH)2 (s)      Cu 2+ (aq) + 2OH - (aq)
                             K sp = [Cu 2+ ][OH - ]2
                Multi-Step Process
                          MnS(s)          Mn2+ (aq) + S2- (aq)
                S 2- (aq) + H 2O(l)    HS- (aq) + OH - (aq)
                MnS(s) + H 2O(l)        Mn 2+ (aq) + HS- (aq) + OH- (aq)
4/30/2010                  K sp = [Mn 2+ ][HS - ][OH - ]                    51
      Equilibria - Slightly Soluble Ionic Compounds




4/30/2010                                             52
                               Practice Problem
     Determine Ksp from solubility
            The solubility of PbSO4 at 25oC is 4.25 x10-3 g /100 mL
            solution. Calculate the Ksp
                 PbSO4 (s) = Pb 2+ (aq) + SO4 2- (aq)
            Convert solubility to molar solubility
                                 0.00425 g PbSO 4   1000 ml    1 mol PbSO 4
            Molar Solubility =                    x         x
                                     100 mL           1L      303.3 g PbSO 4
                             = 1.40 x 10-4 M PbSO 4
            Some PbSO4 dissolves into Pb2+ & SO42-
                                  [Pb2+ ] = [SO4 2-]
                        [Pb 2+ ] = [SO4 2-] = 1.40 x 10-4 M
               K sp = [Pb 2+ ][SO 4 2- ] = (1.40 x 10-4 )2 = 1.96 x 10-8
4/30/2010                                                                      53
                                  Practice Problem
     Determine Solubility from Ksp
            Determine the solubility of Calcium Hydroxide, Ca(OH)2
            Ksp for Ca(OH)2 = 6.5 x 10-6
                 Ca(OH)2 (s)  Ca 2+ (aq) + 2OH - (aq)
            Set up reaction table (S = molar solubility)
                 Concentration (M)   Ca(OH)2(s)   ⇄    Ca2+(aq)   +    2OH-(aq)
                       Initial                           0               0
                      Change                            +S              +2S
                    Equilibrium                          S              2S


             K sp = [Ca 2+ ][OH - ]2 = (S)(2S)2 = (S)(4S 2 ) = 4S 3 = 6.5 x 10-6

                            2+               6.5x10-6
                      [Ca ] = S =        3            = 1.2 x 10-2 M
                                                4
                      [OH- ] = 2S = 2 x 1.2 x 10-2 = 4.2 x 10-2 M
4/30/2010                                                                          54
      Equilibria - Slightly Soluble Ionic Compounds
           The Effect of the Common Ion
               The presence of a “Common Ion” decreases the
                solubility of a slightly soluble ionic compound
                    PbCrO4 (s)     Pb 2+ (aq) + CrO4 2- (aq)
                   K sp = [Pb 2+ ][CrO 4 2- ] = 2.3 x 10-13
             Add some Na2CrO4, a soluble salt, to the saturated
              PbCrO4 solution
                   Concentration of CrO42- increases
                   Some of excess CrO42- combines with Pb2+ to form
                    PbCrO4(s)
                   Equilibrium shifts to the “Left”
                   This shift “reduces” the solubility of PbCrO4


4/30/2010                                                              55
                                   Practice Problem
     What is the solubility of Ca(OH)2 in 0.10 M Ca(NO3)2?
                     Ksp Ca(OH)2 = 6.5 x 10-6
     The addition of the common ion, Ca2+, should lower the
     solubility of Ca(OH)2
              Ca(OH)2 (s)  Ca 2+ (aq) + 2OH - (aq)
               Concentration (M)   Ca(OH)2(s)   ⇄     Ca2+(aq)        +     2OH-(aq)
                      Initial                          0.10                    0
                     Change                             +S                    +2S
                    Equilibrium                    0.10 + S = 0.1             2S

     Since Ksp is small, S << 0.10 then, 0.10 M + S  0.1 M
            K sp = [Ca 2+ ][OH - ]2 = 6.5 x 10-6  (0.1)(2S)2
                2 6.5 x 10-6                                              6.5 x 10-5
             4S =            = 6.5 x 10-5             S = [Ca ] =2+
                                                                                     = 4.03 x 10-3 M
                    0.10                                                      4
            4.0 x 10-3 M
                          x 100 = 4.0% < 5% Assumption : S << 0.10 is valid
              0.10 M
                        [Ca 2+ ]this prob vs. [Ca 2+ ]prev prob ( 0.0043 M vs. 0.012 M)
4/30/2010                                                                                              56
      Equilibria - Slightly Soluble Ionic Compounds
           The Effect of pH on Solubility
             The Hydronium Ion (H3O+) of a strong acid increases
              the solubility of a solution containing the anion of a
              weak acid (HA-)
             Adding some strong acid to a saturated solution of
              calcium carbonate (CaCO3) introduces large amount of
              H3O+ ion, which reacts immediately with the CaCO3 to
              form the weak acid HCO3-
                         CaCO3 (s)  Ca 2+ (aq) + CO 3 2- (aq)
                  CO3 2- (aq) + H 3O+  HCO 3 - (aq) + H 2O(l)
               Additional H3O+ reacts with the HCO3- to form
                carbonic acid, H2CO3, which immediately decomposes
                to H2O and CO2
                HCO3 - (aq) + H 3O+    H 2CO3 (aq) + H 2O(l)    CO2 (g) + 2H 2O(l)

4/30/2010                                                                              57
      Equilibria - Slightly Soluble Ionic Compounds
           The equilibrium shifts to the “Right” and more CaCO3
            dissolves – increased solubility
           The overall reaction is:
                                                 H3O+           H3O+
                CaCO 3 (s)    Ca   2+
                                                           
                                         + CO 3  HCO 3 
                                            2-              -



                                     H 2CO 3  CO 2 (g) + H 2O + Ca 2+

           Adding H3O+ to a saturated solution of a compound with
            a strong acid anion – AgCl
               Chloride ion, Cl-, is the conjugate base of a strong
                acid (HCl)
               It coexists with water, i.e. does not react with water
               There is “No” effect on the equilibrium


4/30/2010                                                                58
                            Practice Problem
     Write balanced equations to explain whether addition of
     H3O+ from a strong acid affects the solubility of the
     following ionic compounds
            Lead Bromide (PbBr2 )
                    PbBr2       Pb 2+ + 2Br -
                    No effect
                    Bromide (Br - ) is the anion of the strong acid HBr
                    Bromide ion does not react with water
            Copper(II) Hydroxide (Cu(OH)2
                    Cu(OH)2 ( s )  Cu 2+ (aq) + 2OH - (aq)
                    Increases Solubility
                    OH - is anion of H 2O, a very weak acid
                    It reacts with added H 3O + to form water (H 2O)
                    OH - (aq) + H 3O + (aq)  2H 2O
                        more Cu(OH)2 dissolves to replace OH -
4/30/2010                                                                 59
                         Practice Problem
     Iron(II) Sulfide (FeS)
            FeS(s) + H 2O(l)      Fe2+ (aq) + HS- (aq) + OH- (aq)
                  HS- (aq) + H 3O+ (aq)  H 2S(aq) + H 2O(l)
                         OH- (aq) + H 3O+  2H 2O(l)

            Additional H 3O + Increases Solubility
            The S 2- ion reacts immediately with H 2O to form HS - and OH -
            The added H 3O + reacts with both HS - and OH - , weak acid anions

    Calcium Fluoride (CaF2)
            CaF2 (s)     Ca2+ (aq) + 2F - (aq)
            F- is the conjugate base of a weak acid (HF)
            F - reacts with added H 3O+ to form the acid and water
                    F - (aq) + H 3O+ (aq)  HF(aq) + H 2O(l)
4/30/2010
            The solubility is increased as more CaF2 dissolves                   60
      Equilibria - Slightly Soluble Ionic Compounds
           Predicting Formation of a Precipitate
               Qsp = Ksp when solution is “saturated”
               Qsp > Ksp solution momentarily “supersaturated”
                   Additional solid precipitates until Qsp = Ksp again
               Qsp < Ksp solution is “unsaturated”
                   No precipitate forms at that temperature
                   More solid dissolves until Qsp = Ksp




4/30/2010                                                                 61
                           Practice Problem
     Does CaF2 precipitate when 0.100 L of 0.30 M Ca(NO3)2 is
     mixed with 0.200 L of 0.060 M NaF?
                       Ksp (CaF2) = 3.2 x 10-11
        Ions present: Ca2+ Na+ NO3-             F-
        All sodium & nitrate salts are soluble
           moles Ca 2+ = 0.30 M Ca 2+ x 0.100L = 0.030 mol Ca 2+
                  2+           0.030 mol Ca2+
              [Ca ]init     =                   = 0.10 M Ca 2+
                              0.100 L + 0.200 L
            moles F  = 0.60 M F  x 0.200L = 0.012 mol F 
                       -        0.012 mol F -
                 [F ]init   =                   = 0.040 M F -
                              0.100 L + 0.200 L
                  Qsp = [Ca 2+ ]init [F - ]init = (0.10)(0.040) 2 = 1.6  10-4
                                           2



                  Qsp > K sp (1.6  10-4 > 3.2  10 -11 )

4/30/2010         CaF2 will precipitate until Q sp = K sp                       62
      Equilibria - Slightly Soluble Ionic Compounds
           Selective Precipitation of Ions
               Separation of one ion in a solution from another
               Exploit differences in the solubility of their compounds
               The Ksp of the less soluble compound is much smaller
                than the Ksp of the more soluble compound
               Add solution of precipitating ion until the Qsp value of
                the more soluble compound is almost equal to its Ksp
                value
               The less soluble compound continues to precipitate
                while the more soluble compound remains dissolved,
                i.e., the Ksp of the less soluble compound is always
                being exceeded, i.e. precipitation is ocurring
               At equilibrium, most of the ions of the less soluble
                compound have been removed as the precipitate

4/30/2010                                                                  63
                     Practice Problem
     A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2
     Calculate the [OH-] that would separate the metals ions as
     their hydroxides
       Ksp of Mg(OH)2 = 6.3 x 10-10 Ksp of Cu(OH)2 = 2.2 x 10-20
     Mg(OH)2 is 1010 more soluble than Cu(OH)2
     Solve for [OH-] that will just give a saturated solution of
     Mg(OH)2
     This concentration of [OH-] is the maximum available to
     the cation of the less soluble compound
     Since the [OH-] is far greater than required for the less
     soluble compound to reach its Ksp, precipitation occurs




4/30/2010                                                          64
                            Practice Problem
     Write the equations and ion-product expressions
            Mg(OH)2 (s)        Mg 2+ (aq) + 2OH - (aq)    K sp = [Mg 2+ ][OH - ]2

       Cu(OH)2 (s)  Cu 2+ (aq) + 2OH - (aq) K sp = [Cu 2+ ][OH - ]2
     Calculate [OH-] of saturated Mg(OH)2 solution

                            -
                                   K sp 6.3 x 10-10
                     [OH ] =          =             = 5.6 x 10-5 M
                             [Mg 2+ ]      0.20
     This is the maximum amount of [OH-] that will not
     precipitate Mg2+ ion
     Calculate [Cu2+] remaining in solution
                                  K sp   2.2 x 10-20
                    [Cu 2+ ] =         =               = 7.0 x 10-12 M
                               [OH - ]2 (5.6 x 10-5 )2
     Since initial [Cu2+] was 0.10M, virtually all Cu2+ is
     precipitated
4/30/2010                                                                            65
      Equilibria - Slightly Soluble Ionic Compounds
           Equilibria Involving Complex Ions
             The product of any Lewis acid-base reaction is called
              an “Adduct”, a single species that contains a new
              covalent bond
                        A + :B           A - B (Adduct)
             A complex ion consists of a central metal ion
              covalently bonded to two or more anions or
              molecules, called ligands
                         Ionic ligands    – OH- Cl- CN-
                        Molecular ligands – H2O CO NH3
                                    Ex Cr(NH3)63+
            Cr+3 is central metal ion; NH3 are molecular ligands
             All complex ions are Lewis adducts
             Metal acts as Lewis Acid; ligands acts Lewis base by
              donating an electron pair

4/30/2010                                                             66
      Equilibria - Slightly Soluble Ionic Compounds
           Complex Ions
             Acidic Hydrated metal ions are complex ions with
              water molecules as ligands
             When the hydrated cation is treated with a solution of
              another ligand, the bound water molecules exchange
              for the other ligand
                M(H 2O)4 2+ (aq) + 4NH 3 (aq)        M(NH 3 )4 2+ (aq) + 4H 2O(l)
               At equilibrium          [M(NH 3 ) 4 2+ ][H 2 O]4
                                   Kc =
                                        [M(H 2 O) 4 2+ ][NH 3 ]4

               Water is constant in aqueous reactions
               Incorporate in Kc to define Kf (formation constant)
                                     Kc          [M(NH 3 ) 4 2+ ]
                              Kf =        4
                                            =
                                   [H 2 O]    [M(H 2 O) 4 2+ ][NH 3 ]4

4/30/2010                                                                            67
      Equilibria - Slightly Soluble Ionic Compounds
           Complex Ions (Con’t)
             The actual process is stepwise, with ammonia
              molecules replacing water molecules one at a time to
              give a series of intermediate species, each with its own
              formation constant (Kf)
                   M(H 2O)4 2+ (aq) + NH 3 (aq)         M(H 2O)3 (NH 3 ) 2+ (aq) + H 2O(l)
                                       [M(H 2O)3 (NH 3 )2+
                              K f1   =
                                       [M(H 2O)4 2+ ][NH 3 ]
               The sum of the 4 equations gives the overall equation
               The product of the individual formation constants gives
                the overall formation constant
               The Kf for each step is much larger than “1” because
                “ammonia” is a stronger Lewis base than H2O
                          K f = K f1 x K f2 x K f3 x K f4
               Adding excess NH3 replaces all H2O and all M2+ exists
                as M(NH3)42+
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                             Practice Problem
     What is the final [Zn(H2O)42+] after mixing 50.0 L of
     0.002 M Zn(H2O)42+ and 25 L of 0.15 M NH3
                      K f of Zn(NH 3 )4 2+ = 7.8 x 108
            Zn(H 2O)4 2+ + 4NH 3 (aq)              Zn(NH 3 )4 2+ + 4H 2O(l)
                                     [Zn(NH 3 )4 2
                             Kf =
                                  [Zn(H 2O)4 2 ][NH 3 ]4
     Determine Limiting Reagent
             Moles Zn(H 2O)4 2+ = 50.0 L × 0.002 mol / L = 0.100 mol
                     Moles NH 3 = 25 L × 0.15 mol / L             = 3.75 mol
     moles NH 3 (3.75) >> 4  moles Zn(H 2O)4 2+ (0.1); thus Zn(H 2O)4 2+ is limiting

     Find the initial reactant concentrations:
                                          50.0 L x 0.0020 M
                 [Zn(H 2O)4 2+ ]init =                      = 1.3 x 10-3 M
                                           50.0 L + 25.0 L
                                         25.0 L x 0.15 M                            Con’t
                        [NH 3 ]init =                    = 5.0 x 10-2 M
4/30/2010                                50.0 L + 25.0 L                                69
                      Practice Problem (Con’t)
     Assume all Zn(H2O)42+ is converted to Zn(NH3)42+
                [NH 3 ]reacted  4(1.3 x 10-3 M)  5.2 x 10-3 M

                       [Zn(NH 3 )4 2+ ]  1.3 x 10-3 M
            Concentration (M)   Zn(H2O)42+(aq) +    4NH3(aq)        ⇄   Zn(NH3)42+(aq)   +   4H2O(l)
                 Initial          1.3 x 10-3        5.0 x 10-2                0                
                Change          (-1.3 x 10-3)     (-5.2 x 10-3)       (+1.3 x 10-3)         
               Equilibrium            X             4.5 x 10-2            1.3 x 10-3           


     Solve for x, the [Zn(H2O]42+ remaining at equilibrium
                         [Zn(NH 3 )4 2+ ]                   1.3  10-3
                 Kf =                       = 7.8  10 
                                                      8
                                2+
                      [Zn(H 2O)4 ][NH 3 ] 4
                                                         (x) (4.5  10-2 )4

                  x = [Zn(H 2O)4 2+ ]  4.1  10-7 M

              Kf is very large;  [Zn(H2O)42+ should be “low”
4/30/2010                                                                                              70
     Complex Ions – Solubility of Precipitates
           Increasing [H3O+] increases solubility of slightly soluble
            ionic compounds if the anion of the compound is that of
            a weak acid
           A Ligand increases the solubility of a slightly soluble ionic
            compound if it forms a complex ion with the cation
             ZnS(s) + H 2O(l)      Zn 2+ (aq) + HS- (aq) + OH- (aq)
                                K sp = 2.0 x 10-22
           When 1.0 M NaCN is added to the above solution, the
            CN- ions act as ligands and react with the small amount
            of Zn2+(aq) to form a complex ion

                   Zn 2+ (aq) + 4CN- (aq)         Zn(CN)4 2- (aq)

                                K f = 4.2 x 1019


4/30/2010                                                                   71
     Complex Ions – Solubility of Precipitates
           Add the two reactions and compute the overall K
               ZnS(s) + H 2O(l)          Zn 2+ (aq) + HS- (aq) + OH- (aq)
               Zn 2+ (aq) + 4CN- (aq)           Zn(CN)4 2- (aq)

       ZnS(s) + 4CN- (aq) + H 2O(l)            Zn(CN)4 2- + HS- (aq) + OH - (aq)
               K overall = K sp x K f = (2.0 x 10-22 ) (4.2 x 1019 ) = 8.4 x 10-3
           The overall equilibrium constant, Koverall , is more than a
            factor of 1019 larger than the original Ksp (2.0 x 10-22)
           This reflects the increased amount of ZnS in solution as
            Zn(CN42-)




4/30/2010                                                                           72
                          Practice Problem
     Calculate the solubility of Silver Bromide (AgBr) in water (H2O)
     and 1.0 M of photographic “Hypo” – Sodium Thiosulfate
     (Na2S2O3), which forms the “complex” ion Ag(S2O3)23-
            K sp (AgBr) = 5  10-13        K f (Ag(S 2O 3 )2 3- ) = 4.7  1013
     a. Solubility AgBr in Water
                             K sp = 5 x 10-13

                  AgBr(s)         Ag + (aq) + Br - (aq)
                    K sp = [Ag + ][Br - ] = 5.0 x 10-13

                  S = [AgBr]dissolved = [Ag + ] = [Br - ]

               K sp = [Ag + ][Br-] = S x S = 5.0 x 10-13

                           S = 7.1 x 10-7 M                                      Con’t
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                        Practice Problem (con’t)
    Solubility AgBr in 1.0 M Hypo - Sodium Thiosulfate (Na2S2O3)

    Write the overall equation:

                          AgBr(s)  Ag + (aq) + Br - (aq)
                    Ag + (aq) + 2S 2O 3 2- (aq)  Ag(S 2O 3 )2 3- (aq)
        AgBr(s) + 2S 2O 3 2- (aq)                Ag(S 2O 3 )2 3- (aq) + Br - (aq)


    Calculate Koverall (product of Ksp & Kf):

                                  3-
                      [Ag(S 2O 3 )2 ][Br - ]
        K overall   =                        = K sp  K f = (5.0 × 10-13 )(4.7 × 1013 ) = 24
                         [(S 2O 3 )2- ]




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                         Practice Problem (Con’t)
           Set Up the Reaction Table
                             S = [AgBr]dissolved = [Ag(S 2O3 )2 3- ]
            Concentration (M)      AgBr(s)   +   2S2O32-(aq)    ⇄   Ag(S2O3)23-(aq)   +    Br-(l)
                   Initial                          1.0                  0                  0
                 Change                            -2S                   +S                +S
               Equilibrium                       1.0 - 2S                S                  S


                           [Ag(S 2O 3 )2 3- ][Br-]        S2
             K overall   =               2-
                                                   =              2
                                                                     24
                              [(S 2O 3 ) ]           (1.0 M - 2S)
               S
                     = 24 = 4.9                    S = 4.9 - 2S(4.9)                  S + 2S(4.9) = 4.9
            1.0 - 2S
                                                               4.9
             S(1 + 2 × 4.9) = 4.9                  S =                = 0.45 M
                                                             1 + 9.8)

             Solubility Ag(S2O3 )2 3- or AgBr = S                              = 0.45 M
4/30/2010                                                                                                 75
            Ionic Equilibria in Aqueous Solutions
           Equilibria Involving Complex Ions
               Amphoteric Oxides & Hydroxides (Recall Chapter 8)
                   Some metals and many metalloids form oxides or
                    hydroxides that are amphoteric; they can act as acids or
                    bases in water
                   These compounds generally have very little solubility in
                    water, but they do dissolve more readily in acids or
                    bases
                   Ex. Aluminum Hydroxide
                           Al(OH)3(s) ⇆ Al3+(aq) + 3OH-(ag)
                          Ksp = 3 x 10-34 (very insoluble in water)
                   In acid solution, the OH- reacts with H3O+ to form water
                            3H3O+(ag) + 3OH-(aq)  6H2O(l)
                     Al(OH)3(s) + H3O+(aq)  Al3+(aq) + 6H2O(l)
4/30/2010                                                                      76
            Ionic Equilibria in Aqueous Solutions
           Equilibria Involving Complex Ions
               Aluminum Hydroxide in basic solution
                     Al(OH)3(s) + OH-(aq)  Al(OH)4-(aq)
               The above reaction is actually a much more complex
                situation, involving multiple species
               When dissolving an aluminum salt, such as Al(NO3), in
                a strong base (NaOH), a precipitate forms initially and
                then dissolves as more base is added
               The formula for hydrated Al3+ is Al(H2O)63+
               Al(H2O)63+ acts as a “weak polyprotic acid and reacts
                with added OH- in a stepwise removal of the H2O
                ligands attached to the hydrated Al



4/30/2010                                                                 77
            Ionic Equilibria in Aqueous Solutions
           Amphoteric Aluminum Hydroxide in basic solution
     Al(H2O)63+(aq) + OH-(aq) ⇆ Al(H2O)5OH2+(aq)                 + H2O(l)
     Al(H2O)52+(aq) + OH-(aq) ⇆ Al(H2O)4(OH)2+(aq) + H2O(l)
     Al(H2O)4+(aq) + OH-(aq) ⇆ Al(H2O)3(OH)3(s)                  + H2O(l)
           Al(H2O)3(OH)3(s) is more simply written Al(OH)3(s)
           As more base is added, a 4th H+ is removed from a H2O ligand
            and the soluble ion Al(H2O)2(OH)4-(aq) forms
              Al(H2O)3(OH)3(s) + OH-(aq) ⇆        Al(H2O)2(OH)4-(aq)
     The ion is normally written as Al(OH)4-(aq)




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                              Equation Summary
                                      [HA - ][H 3O + ]
                                 Ka =
                                         [HA]
             Concentration-        A(aq)    +   B(aq)       A-(aq)   +   H2O(l)
                 Initial Ci          -            -            -            0
                                    -            -           +X           +X
              Equilibrium Cf         -            -            X            X


                               HA + H 2O      H 3O + + A -
                                         [H 3O + ][A - ]
                                   Ka =
                                             [HA]
                                                  [base]
                                 pH = pKa + log
                                                   [acid]
                    [A - ]
            When :         =1
                   [HA]
                            [A - ] 
            pH = pKa + log          = pKa + log1 = pKa + 0 = pKa
                            [HA] 
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                   Equation Summary
                                   [BH + ][OH - ]
                            Kb   =
                                       [B]
                       K w = K a x K b = 1.0 x 10-14
                       pK w = pH + pOH = 14

                      Qsp = [M n+ ]p [X z- ]q = K sp
                                            +
                       2H2O         H3 O       + OH-

                   [H3O+ ][OH- ]
              Kw =               = [H3O+ ][OH- ] = 1×10   -14

                      [H2O]2
                            +
                     H3 O       + OH-          2H2O

                    [H2O]2          1           1
            Kn =              =              =    = 1×1014
                 [H3O+ ][OH- ] [H3O+ ][OH- ]   Kw
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