More pH Calculations
1.
CHM 152
The pH of a benzoic acid (C6H5COOH) solution is 4.25. Calculate the concentration of the benzoic acid
−5
solution. [Ka (benzoic acid) = 6.5 × 10 ]
−
+
C6H5COOH(aq) + H2O(l) U H3O (aq) + C6H5COO (aq)
Initial (M)
I
0
0
Change (M)
−x
+x
+x
After Rxn (M)
I−x
x
x
Ka =
[H3O + ][C6 H5 COO− ]
[C6 H5 COOH]
6.5 × 10−5 =
x2
I−x
Since the problem asks you to calculate the concentration of benzoic acid (I), it must indicate the value of x.
x can be calculated from the pH.
+
pH = −log[H3O ]
+
−pH = log[H3O ]
−pH
+
= [H3O ]
10
+
−4.25
[H3O ] = 10
+
−5
[H3O ] = x = 5.6 × 10 M
6.5 × 10−5 =
6.5 × 10−5 =
x2
I−x
(5.6 × 10−5 )2
I − (5.6 × 10−5 )
−5
−9
6.5 × 10 (I) − (3.64 × 10 ) = 3.136 × 10
−5
−9
6.5 × 10 (I) = 6.776 × 10
−4
I = 1.0 × 10 M = [C6H5COOH]
2.
−9
−5
Calculate the pH of a 0.25 M solution of NH3. [Kb (ammonia) = 1.8 × 10 ]
+
Initial (M)
Change (M)
After Rxn (M)
Kb =
−
NH3(aq) + H2O(l) U NH4 (aq) + OH (aq)
0.25
0
0
−x
+x
+x
0.25 − x
x
x
[NH +4 ][OH − ]
[NH3 ]
1.8 × 10−5 =
x2
0.25 − x
Because K is so small, very little reactant goes to product. We make the assumption that 0.25 − x ≈ 0.25.
x2
0.25
1.8 × 10−5 =
−6
2
x = 4.5 × 10
−
−3
x = [OH ] = 2.12 × 10 M
−
pOH = −log[OH ]
−3
pOH = −log(2.12 × 10 )
pOH = 2.67
pH + pOH = 14
pH = 11.33
3.
A 0.10 M weak base has a pH of 9.12. Calculate the equilibrium constant, Kb, for this weak base.
+
Initial (M)
Change (M)
After Rxn (M)
−
B(aq) + H2O(l) U BH (aq) + OH (aq)
0.10
0
0
−x
+x
+x
0.10 − x
x
x
Kb =
[BH + ][OH − ]
[B]
Kb =
x2
0.10 − x
Since the problem asks you to calculate Kb, it must indicate the value of x. x can be calculated from the pH.
pH + pOH = 14
pOH = 14 − 9.12 = 4.88
−
−pOH
−4.88
[OH ] = 10
= 10
−
−5
[OH ] = 1.32 × 10 M = x
Substitute the value of x into the equilibrium constant expression to solve for Kb.
Kb =
Kb =
x2
0.10 − x
(1.32 × 10−5 ) 2
0.10 − (1.32 × 10−5 )
−9
Kb = 1.7 × 10