Question 6.13: Calculate the pH of a buffer that is 0.020 M in NH3 and 0.03...
Calculate the pH of a buffer that is 0.020 M in NH_3 and 0.030 M in NH_4Cl. What is the pH after adding 1.00 mL of 0.10 M NaOH to 0.10 L of this buffer?
The acid dissociation constant for {NH_4}^+ \text{is} 5.70 × 10^{–10}; thus the initial pH of the buffer is
pH=9.24+\log \frac{C_{NH_3}}{C_{{NH_4}^+}}=9.24+\log \frac{0.020}{0.030}=9.06
Adding NaOH converts a portion of the {NH_4}^+ \text{to} NH_3 due to the following reaction
{NH_4}^+(aq) + OH^–(aq) \xrightleftharpoons[]{} NH_3(aq) + H_2O(\ell)
Since the equilibrium constant for this reaction is large, we may treat the reaction as if it went to completion. The new concentrations of {NH_4}^+ \text{and} NH_3 are therefore
C_{{NH_4}^+}=\frac{\text{moles} {NH_4}^+- \text{moles} OH^-}{V_{tot}}
=\frac{(0.030 M)(0.10 L)-(0.10 M)(1.00\times 10^{-3} L)}{0.101 L} =0.029 M
C_{NH_3}=\frac{\text{moles} {NH_3}+ \text{moles} OH^-}{V_{tot}}
=\frac{(0.020 M)(0.10 L)+(0.10 M)(1.00\times 10^{-3} L)}{0.101 L}=0.021 M
Substituting the new concentrations into the Henderson–Hasselbalch equation gives a pH of
pH = 9.24 + \log\frac{0.021}{0.029}=9.10