Chemistry 180 Determination of pH for Solutions of Salts Handout (Chapter 15)
Determination of pH for Solutions of Salts
Solve the following problems. You will need to use the tables in your text or Appendix C to
retrieve information to solve some of these problems.
1. What is the Kb of the cyanate anion? Note, that the cyanate anion can be found in cyanic
acid, HOCN.
HOCN
H+ + OCN–
Ka = 3.5×10-4
Now, Kw= Ka × Kb. So, the cyanate ion Kb will be… K w  K a  K b


1.0  10 14  3.5  10 4 K b
K b  2.9  10
2. What is Kb for the azide anion (N3– )?
N3 H
H+ + N3–
11
Ka = 1.9×10-5
Now, Kw= Ka × Kb. So, the azide ion Kb will be…
K w  K a  Kb


1.0  10 14  1.9  10 5 K b
K b  5.3  10
10
3. Will a solution of ammonium bromide be acidic, basic or neutral?
First, identify the acid/base these ions come from: ammonia
(weak base) and hydrobromic acid (strong acid). Thus, no
acid will form. Just deal with ammonium ion.
K  K K
w
NH3
NH4
+
NH4 + H2O
+
+ OH
NH3 + H3O+
–
Kb = 1.8×10
-5
a
1.0  10
14
b

 K a 1.8  10 5
K a  5.6  10

10
Thus, the pH will be slightly acidic as some of the ammonium
reforms ammonia releasing protons into the solution.
4. Will a solution of potassium acetate be acidic, basic or neutral?
First, identify the acid/base these ions come from: potassium
hydroxid (strong base) and acetic acid (weak acid). Thus, no
base will form. Just deal with acetate ion.
K w  K a  Kb
CH3CO2H
CH3CO2– + H+
Ka = 1.8×10-5
CH3CO2– + H2O
CH3CO2H + OH–

K b  5.6  10 10
Thus, the pH will be slightly basic as some of the acetate ion
reforms acetic acid releasing hydroxide into the solution.
5. What is the pH of a 0.010 M KBr solution?
potassium hydroxide – strong base; therefore, neutral pH from
hydrobromic acid – strong acid; therefore, neutral pH from
Thus, the pH will be 7.0 assuming water used to make solution
was neutral!
pH of salt soln (107) key.doc
page 17&18 from Steve Pruett’s Che 107 coursepack

1.0  10 14  1.8  10 5 K b
Page 1 of 2
Chemistry 180 Determination of pH for Solutions of Salts Handout (Chapter 15)
6. What is the pH of a 0.01 solution of sodium benzoate?
sodium hydroxide – strong base; therefore, neutral pH from
benzoic acid – weak acid; therefore, basic pH from salt of
i:
C6H5CO2H
C6H5CO2– + H+
0
0.010 M
Ka = 6.5×10-5
0
K w  K a  Kb

i:
∆:
e:
C6H5CO2– + H2O
0.01 M
–x
0.01– x
Kb 
C6H5CO2H + OH–
0M
0M
+x
+x
+x
+x
C 6 H 5 CO 2 HOH- 
C H CO 
6
1.6  10 10 
5
2
( x)( x)
0.01  x 

1.0  10 14  3.5  10 4 K b
Unfortunately, this equation won’t work.
We must rewrite equation and constant as a Kb.
K b  2.9  10 11
Kb = 1.6×10-10
OH   x  1.310
pOH   logOH    log1.3 10

Solve for x using
quadratic or
assumption.
6

6
 5.9
pH  pOH  14
Use x to
determine final
answer!
pH  14  5.9
pH  8.1
x  1.3  10 6
7. What is the pH of a solution that contains 0.010 M pyridinium ion, C5H5NH+?
C5H5N + H2O
C5H5NH+ + OH–
Kb = 1.5×10-9
i:
0
0.010 M
0
Unfortunately, this equation form won’t work. We
must rewrite equation and constant as a Ka.
C5H5NH+
i:
∆:
e:
C5H5N + H–+
0.01 M
–x
0.01– x
Ka 
C 5 H 5 NH 
C H NH 

5
5
0M
+x
+x

1.0  10 14  Ka 1.5  10 9
Ka = 6.7×10-6
K a  6.7  10

6
0M
+x
+x
Solve for x using
quadratic or
assumption.
Use x to determine final answer!
H   x  2.6 10
pH   logH    log 2.6 10

( x)( x)
0.01  x  x  2.6 104
8. What is the pH of a 0.01 M solution of ammonium acetate?
6.7  10 6 
K w  K a  Kb
4

4
 3.6
Both ammonium and acetate are salts of weak base/acid.
Here is what we figured out on the front of the paper:
NH4+
NH3 + H+
Ka = 5.6×10-10
CH3CO2– + H2O
CH3CO2H + OH– Kb = 5.6×10-10
However, note that the constants have equal numerical values.
Thus, the OH– produced by acetate should equally offset the H+
produced by the ammonium!
Thus, the pH should be 7.0.
pH of salt soln (107) key.doc
page 17&18 from Steve Pruett’s Che 107 coursepack
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