Ka, Kb by xiw67167

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									Ka, Kb
     Comparing the pH of two acids
1.   Predict the pH of HCl and HF (below)
2.   Calibrate a pH meter
3.   Measure the pH of HCl(aq) and HF(aq)
4.   Complete the chart below
                        HCl (aq)         HF (aq)
[ ] in mol/L (on label)      0.05           0.05
  Net ionic equation      HCl  H+ + HF  H+ + F–
     Predicted [H+]           Cl–
                             0.05           0.05
      Predicted pH      -log(0.05)=1.3 -log(0.05)=1.3
      pH measured             1.3           2.3 ?
       Actual [H+]       10–pH = 0.05 10–pH = 0.005
 Conductivity (demo) Higher / stronger Lower / weaker
Read15.3. (pg. 607+)       Questions
1. Based on your results, which acid ionizes
   (forms ions) to a greater degree?
2. Which two measurements taken in the lab
   support your answer to 1?
3. What is another name for Ka?
4. Solve PE 5, 6
5. Write the Ka equation for HCl (aq) and HF
   (aq) from today’s lab
6. Solve for PE 8, 9 (use this equilibrium for
   butyric acid: HBu  H+ + Bu–)
7. For HF(aq) set up a RICE chart, then solve
   for Ka. How does your value for Ka
   compare to the accepted value (pg. 608)?
8. Try PE 10 (follow example 15.7 on pg. 610)
                 Answers
1. HCl ionizes more than HF
2. HCl has a lower pH (indicating more H+), & a
   higher conductivity (indicating more ions)
3. Ka: acid ionization constant
4. HNO2  H+ + NO2–, Ka=[H+][NO2–]/[HNO2]
   HPO42–  H+ + PO43–,Ka=[H+][PO43–]/[HPO42–
   ]
5. HCl  H+ + Cl–, Ka=[H+][Cl–]/[HCl]
   HF  H+ + F–, Ka=[H+][F–]/[HF]
 PE 8 - pg. 610           HBu  H+ + Bu–
           HBu            H+             Bu–
  R         1               1            1
  I      0.0100             0            0
  C     -0.0004         +0.0004       +0.0004
  E     0.0096           0.0004       0.0004
       [H+] = 10– pH = 10– 3.40 = 3.98 x 10– 4

       [H+][Bu–]         [0.0004]2
Ka =                =                 = 1.67x10 – 5
         [HBu]            [.0096]
 PE 9 - pg. 610           HBu  H+ + Bu–
           HBu            H+             Bu–
  R         1              1              1
  I      0.0100            0              0
  C      -0.001         +0.001         +0.001
  E      0.009           0.001          0.001
       [H+] = 10– pH = 10– 2.98 = 1.05 x 10– 3

       [H+][Bu–]         [0.001]2
Ka =                =                 = 1.1x10 –4
         [HBu]            [.009]
        Question 7: HF  H+ + F–
            HF            H+               F–
  R         1              1                1
  I       0.05             0                0
  C      -0.005         +0.005           +0.005
  E      0.045           0.005            0.005
       [H+] = 10– pH = 10– 2.3 = 0.005
        [H+][F–]         [0.005]2
Ka =                =               = 5.6x10 –4
          [HF]            [.045]
       Accepted value of Ka for HF is 6.4 x 10 – 4
   10: HC2H4NO2  H+ + C2H4NO2–
        HC2H4NO2           H+          C2H4NO2–
   R       1                1              1
   I    0.010               0              0
   C       -x              +x              +x
   E 0.010 - x              x              x
    [H+][C2H4NO2–]              [x]2
Ka =              =                      = 1.4x 10 –5
          [HF]             [0.010 - x]
           Since x is small 0.010 – x = 0.010
 [x]2
        =1.4 x 10 –     x= 3.74 x 10–5 M, pH = 3.43
[0.010]5
               Ka summary
• Ka follows the pattern of other “K” equations
• I.e. for HA(aq) + H2O(l)  H3O+(aq) + A–(aq)
• Ka = [H3O+][A–] / [HA]
• Notice that H2O is ignored because it is liquid
• HA cannot be ignored because it is aqueous
• This is different than with Ksp. In Ksp, solids
  could only be in solution as ions
• Acids can be in solution whether ionized or not
• The solubility of acids makes sense if you
  think back to the partial charges in HCl for ex.
               Ka summary
• Generally Ka tells you about acid strength
• Strong acids have high Ka values
• A “strong” acid is an acid that completely
  ionizes. E.g. HCl + H2O  H3O+ + Cl–
• A “weak” acid is an acid that doesn’t ionize
  completely. E.g. HF + H2O  H3O+ + F–
• Note: don’t get confused between strength
  and concentration. 1 M HCN has a smaller
  [H+], thus a higher pH, than 0.001 M HCl
• In general: Ka < 10 – 3       Weak acid
         10 – 3 < Ka < 1        Moderate acid
                  Ka > 1        Strong acid
      Dissociation vs. Ionization
• Ionization and dissociation indicate ions form
• Dissociation: ions form when a chemical
  comes apart. E.g. NaCl melts to form Na+, Cl–
• Ionization: ions form when two chemicals
  react. E.g. HCl(aq) + H2O  H3O+(aq) + Cl–(aq)
• Even though we write HCl  H+ + Cl– , this is
  just an abbreviation. In reality HCl reacts with
  H2O, thus it is an ionization not a dissociation
• Note that NaCl can also dissociate in water.
  This is not an ionization, since water is only
  required to stabilize ions (it is not needed as a
  reactant involved in forming ions)
     Kb – the last K (I promise)
• Kb is similar to Ka except b stands for base
• The general reaction involving a base can be
  written as B(aq) + H2O  BH+(aq) + OH–(aq)
• Thus Kb = [BH+] [OH–] / [B]
• Recall: shorthand for Ka is HA  H+ + A–
• Kb has no shorthand form
• Read pg. 614 - 617
• Try PE 12 (a-c), 13, 14 (for 13, you do not
  need to know the chemical formula of
  morphine. Symbolize it with M)
                  PE 12
a) CN–(aq) + H2O  HCN(aq) + OH–(aq)
   Kb = [HCN][OH–] / [CN–]
b) C2H3O2–(aq) + H2O  HC2H3O2(aq) + OH–(aq)
   Kb = [HC2H3O2][OH–] / [C2H3O2–]
c) C6H5NH2(aq) + H2O  C6H5NH3+(aq) + OH–(aq)
   Kb = [C6H5NH3+][OH–] / [C6H5NH2]
  PE 13 - pg. 617 M + H2O  MH+ + OH–
           M        MH+       OH–
   R       1          1        1
   I  0.010              0               0
   C -0.00013         +0.00013     +0.00013
   E 0.00987          0.00013      0.00013

pOH = 14 - pH = 14 - 10.10 = 3.90
[OH-] = 10-pOH = 10-3.90 = 1.26 x 10-4
   [MH+] [OH–] [0.00013] [0.00013]   =1.7 x 10-6
Kb =              =
        [M]            [0.00987]
 PE 14 - pg. 617 M + H2O  MH+ + OH–
         NH3       NH4+      OH–
  R       1          1        1
   I    0.020             0           0
   C       -x            +x          +x
   E   0.020 - x          x           x

pOH = -log[OH-] = 3.22
pH = 14 - pOH = 10.78
        [x] [x]            x2    = 1.8 x 10-5
Kb =               =
       [0.020]           [0.020] x= 6.0 x 10-4
          Strength of conjugates
Consider HCl(l) + H2O  Cl–(aq) + H3O+(aq)
The Ka for HCl is [Cl–(aq)] [H3O+(aq)] / [HCl(aq)]
Also, Cl–(aq) + H2O(aq)  HCl(l) + OH–
The Kb for Cl– is [HCl(aq)] / [Cl–(aq)] [H3O+(aq)]
            Relative values of Ka
Recall for HX  H+ + X–, Ka = [H+][X–] / [HX]
Q - what does a large Ka indicate?
A - equilibrium is far to the right (all dissociates)
Thus a large Ka = strong acid
Look at Table 15.4 on page 608
The text uses this definition:
      Ka < 10–3 is a weak acid
      10–3 < Ka < 1 is a moderate acid
       1 < Ka is a strong acid
These definitions are somewhat arbitrary, we
  will not focus on this. Just remember a high
  Ka means the acid is strong.                 For more lessons, visit
                                               www.chalkbored.com

								
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