Question 16.16: What is the pH of a 0.040 M ammonia solution at 25°C?
What is the \text{pH} of a 0.040 M ammonia solution at 25°\text{C}?
Strategy Construct an equilibrium table, and express equilibrium concentrations in terms of the unknown x. Plug these equilibrium concentrations into the equilibrium expression, and solve for x. From the value of x, determine the \text{pH}.
Setup \text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{NH}_4+(aq) + \text{OH}^–(aq)
| Initial concentration (M): | 0.040 | 0 | 0 | |
| Change in concentration (M): | –x | +x | +x | |
| Equilibrium concentration (M): | 0.040 − x | x | x |
The equilibrium concentrations are substituted into the equilibrium expression to give
K_b = \frac{[\text{NH}_4^+][\text{OH}^–]}{[\text{NH}_3]} = \frac{(x)(x)}{0.040 − x} = 1.8 × 10^{–5}
Assuming that 0.040 − x ≈ 0.040 and solving for x gives
\frac{(x)(x)}{0.040 − x} ≈ \frac{(x)(x)}{0.040} = 1.8 × 10^{–5}
x^2 = (1.8 × 10^{–5})(0.040) = 7.2 × 10^{–7}
x = \sqrt{7.2 × 10^{–7}} = 8.5 × 10^{–4} M^*
According to the equilibrium table, x = [\text{OH}^–]. Therefore, \text{pOH} = –\text{log} (x):
–\text{log} (8.5 × 10^{–4}) = 3.07
and \text{pH} = 14.00 − \text{pOH} = 14.00 − 3.07 = 10.93. The \text{pH} of a 0.040-M solution of \text{NH}_3 at 25°\text{C} is 10.93
^*Student Annotation: Applying the 5 percent test indicates that the approximation shortcut is valid in this case: (8.5 × 10^{–4}/0.040) × 100\% ≈ 2\%.