Ka, Kb
Comparing the pH of two acids
1.
2.
3.
4.
Predict the pH of HCl and HF (below)
Calibrate a pH meter
Measure the pH of HCl(aq) and HF(aq)
Complete the chart below
HCl (aq)
[ ] in mol/L (on label)
Net ionic equation
Predicted [H+]
Predicted pH
pH measured
Actual [H+]
Conductivity (demo)
HF (aq)
0.05
0.05
HCl  H+ + HF  H+ + F–
Cl–
0.05
0.05
-log(0.05)=1.3 -log(0.05)=1.3
1.3
2.3 ?
10–pH = 0.05 10–pH = 0.005
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
R
I
C
E
HBu  H+ + Bu–
HBu
H+
Bu–
1
0.0100
-0.0004
0.0096
1
0
+0.0004
0.0004
1
0
+0.0004
0.0004
[H+] = 10– pH = 10– 3.40 = 3.98 x 10– 4
Ka =
[H+][Bu–]
[HBu]
=
[0.0004]2
[.0096]
= 1.67x10 –5
PE 9 - pg. 610
R
I
C
E
HBu  H+ + Bu–
HBu
H+
Bu–
1
0.0100
-0.001
0.009
1
0
+0.001
0.001
1
0
+0.001
0.001
[H+] = 10– pH = 10– 2.98 = 1.05 x 10– 3
Ka =
[H+][Bu–]
[HBu]
=
[0.001]2
[.009]
= 1.1x10 –4
Question 7: HF  H+ + F–
R
I
C
E
HF
H+
F–
1
0.05
-0.005
0.045
1
0
+0.005
0.005
1
0
+0.005
0.005
[H+] = 10– pH = 10– 2.3 = 0.005
Ka =
[H+][F–]
[HF]
=
[0.005]2
[.045]
= 5.6x10 –4
Accepted value of Ka for HF is 6.4 x 10 – 4
10: HC2H4NO2  H+ + C2H4NO2–
HC2H4NO2
R
1
0.010
I
-x
C
E 0.010 - x
[H+][C2H4NO2–]
Ka =
=
[HF]
H+
C2H4NO2–
1
0
+x
x
1
0
+x
x
[x]2
[0.010 - x]
= 1.4x10 –5
Since x is small 0.010 – x = 0.010
[x]2
=1.4 x 10 –
[0.010]5
x= 3.74 x 10–5 M, pH = 3.43
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
0.010
0
0
I
C -0.00013
E 0.00987
+0.00013
+0.00013
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]
Kb =
[M]
=
=1.7 x 10-6
[0.00987]
PE 14 - pg. 617 M + H2O  MH+ + OH–
NH3
NH4+
OH–
R
1
1
1
0.020
0
0
I
C
E
-x
+x
+x
0.020 - x
x
x
pOH = -log[OH-] = 3.22
pH = 14 - pOH = 10.78
Kb =
[x] [x]
[0.020]
=
x2
= 1.8 x 10-5
[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.
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