Two buffers, X and Y of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both — Ionic Equilibrium Chemistry Question
Question
Two buffers, X and Y of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are $0.50 \text{ M}$ in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers? $K_a$ of $\text{HA} = 1.0 \times 10^{-5}$. ($\log 5.05 = 0.7$)
💡 Solution & Explanation
For HA, $pK_a = 5.0$. For Buffer X (pH 4.0): $4.0 = 5.0 + \log(\text{Salt}_X / 0.5) \implies \text{Salt}_X = 0.5 \times 10^{-1} = 0.05 \text{ M}$. For Buffer Y (pH 6.0): $6.0 = 5.0 + \log(\text{Salt}_Y / 0.5) \implies \text{Salt}_Y = 0.5 \times 10^1 = 5.0 \text{ M}$. Mixing equal volumes halves the concentrations but we can just use total moles over total volume. The average concentration of HA remains $0.5 \text{ M}$. The average concentration of Salt becomes $(0.05 + 5.0)/2 = 2.525 \text{ M}$. New $\text{pH} = 5.0 + \log(2.525 / 0.5) = 5.0 + \log(5.05) = 5.0 + 0.7 = 5.7$. Therefore, correct answer is D.