The pH of the solution is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For the basic buffer, $[\text{Base}] = 0.8\text{ M}$ and $[\text{Salt}] = 0.2\text{ M}$. Given $K_a(\text{NH}_4^+) = 5.0 \times 10^{-10}$, the $pK_a = -\log(5.0 \times 10^{-10}) = 10 - \log 5 = 10 - 0.7 = 9.3$. Using the Henderson-Hasselbalch equation customized for the conjugate acid $pK_a$: $\text{pH} = pK_a + \log(\frac{[\text{Base}]}{[\text{Salt}]}) = 9.3 + \log(\frac{0.8}{0.2}) = 9.3 + \log(4) = 9.3 + 0.6 = 9.9$. Therefore, correct answer is A.