pH of 0.01 M – and 0.02 M – buffer () is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The concentration of ammonium ions ($[\text{NH}_4^+]$) from $0.01\text{ M (NH}_4\text{)}_2\text{SO}_4$ is $2 \times 0.01 = 0.02\text{ M}$. The concentration of the base $[\text{NH}_3] = 0.02\text{ M}$. Using the Henderson-Hasselbalch equation customized for the acid dissociation constant of the conjugate acid: $\text{pH} = pK_a + \log\left(\frac{[\text{Base}]}{[\text{Salt}]}\right) = 9.26 + \log\left(\frac{0.02}{0.02}\right) = 9.26 + \log(1) = 9.26$.