# Ionic Equilibria in Aqueous Solutions

Document Sample

```					            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
4/30/2010                                            1
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

4/30/2010                                                    2
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.

4/30/2010                                                              3
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

4/30/2010                                                                4
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
4/30/2010                                                              5
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
4/30/2010                                                                   6
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
4/30/2010pH is also small                                                        7
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
4/30/2010                                                            8
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
4/30/2010                                                              9
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
4/30/2010                                                                              10
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

4/30/2010                                           11
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
4/30/2010                                                                              12
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
4/30/2010                                                                      13
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
4/30/2010
[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
4/30/2010                                                              15
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

4/30/2010                                                                 16
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
   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
4/30/2010                                                                  17
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

4/30/2010                                                              18
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

4/30/2010                                                               20
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

4/30/2010                                                                     21
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
4/30/2010                                                              22
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
4/30/2010                                                                               23
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
4/30/2010                                                                            24
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.

4/30/2010                                                                  25
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 
4/30/2010                                                                             26
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.
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

4/30/2010                                                                  28
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
4/30/2010                                                             29
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

4/30/2010                                                               30
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

4/30/2010                                                                   31
Titration Curves
Curve for Titration of a Strong Acid by a Strong Base

4/30/2010                                                           32
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
4/30/2010                                                                            33
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
4/30/2010                                                               34
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
4/30/2010                                                                                35
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
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
 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
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+
4/30/2010                                                                                     68
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
4/30/2010                                                                            73
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- ]

4/30/2010                                                                                      74
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)

4/30/2010                                                                   78
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] 
4/30/2010                                                                          79
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
4/30/2010                                                       80

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 472 posted: 4/30/2010 language: English pages: 80