Question 7.13: Salts As Weak Acids I Calculate the pH of a 0.10 M NH4Cl sol...
Salts As Weak Acids I
Calculate the pH of a 0.10 M N H_{4} Cl solution. The K_{b} value for N H_{3} is 1.8 × 10^{-5} .
The major species in solution are
N H_{4} ^{+}, Cl^{-}, and H_{2}O
Note that both NH_{4} ^{+} and H_{2} O can produce H ^{+}. The dissociation reaction for the N H_{4} ^{+} ion is
NH_{4} ^{+} (aq) \xrightleftharpoons[]{} NH_{3} (aq) + H ^{+} (aq)
for which
K_{a} = \frac{[NH_{3}] [H ^{+}] }{[NH_{4} ^{+}]}
Note that although the K_{b} value for NH_{3} is given, the reaction corresponding to K_{b} is not appropriate here, since NH_{3} is not a major species in the solution. Instead, the given value of K_{b} is used to calculate K_{a} for NH_{4} ^{+} from the relationship
K_{a} × K_{b} = K_{w}
Thus
K_{a} (for NH_{4} ^{+}) = \frac{ K_{w}}{ K_{b} (for NH_{3} )} = \frac{1.0 × 10 ^{-14} }{1.8 × 10 ^{-5}} = 5.6 × 10 ^{-10}
Although NH_{4} ^{+} is a very weak acid, as indicated by its K_{a} value, it is stronger than H_{2}O and thus will dominate in the production of H ^{+}. Therefore, we will focus on the dissociation reaction of NH_{4} ^{+} to calculate the pH of this solution.
We solve the weak acid problem in the usual way:
| Initial Concentration (mol/L) |
Equilibrium Concentration (mol/L) |
|
| [NH_{4} ^{+}]_{0} = 0.10 |
\xrightarrow[to reach equilibrium]{x mol / L NH_{4} ^{+} dissociates } |
[NH_{4} ^{+}] = 0.10 -x |
| [NH_{3}]_{0} = 0 | [NH_{3} ] = x | |
| [H^{+}]_{0} ≈ 0 | [H^{+}] = x |
Thus
5.6 × 10 ^{-10}= K_{a} = \frac{[H ^{+}] [NH_{3}] }{[NH_{4} ^{+}]} = \frac{(x) (x)}{0.10 -x } =\frac{x²}{0.10}
x≈ 7.5 × 10 ^{-6}
The approximation is valid by the 5% rule, so
[H^{+}] = x= 7.5 × 10 ^{-6} M and pH=5.13