What will be the pH of the final solution? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Initially, $[\text{H}^+] = 8.0 \times 10^{-3}\text{ M}$ for a $C = 0.2\text{ M}$ solution. $K_a \approx \frac{[\text{H}^+]^2}{C} = \frac{(8.0 \times 10^{-3})^2}{0.2} = \frac{64 \times 10^{-6}}{0.2} = 3.2 \times 10^{-4}$. After adding sodium formate, the effective concentration of formate ions $[\text{A}^-]$ is the dissociated amount: $1.0\text{ M} \times 0.8 = 0.8\text{ M}$. The acid concentration is $[\text{HA}] = 0.2\text{ M}$. The mixture is a buffer. Using the Henderson-Hasselbalch equation: $\text{pH} = pK_a + \log(\frac{[\text{A}^-]}{[\text{HA}]}) = -\log(3.2 \times 10^{-4}) + \log(\frac{0.8}{0.2}) = 4 - \log(3.2) + \log(4) = 4 - 0.505 + 0.602 \approx 4.1$. Therefore, correct answer is A.