pH of the solution after adding HCl is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Initial moles of $\text{H}_3\text{PO}_4 = 100\text{ ml} \times 0.1\text{ M} = 10\text{ mmol}$. At the second equivalence point, there are $10\text{ mmol}$ of $\text{HPO}_4^{2-}$. When $10\text{ ml}$ of $0.5\text{ M HCl}$ ($5\text{ mmol}$) is added, it completely reacts with half of the $\text{HPO}_4^{2-}$ to convert it back to $\text{H}_2\text{PO}_4^-$. This forms an equimolar buffer mixture containing $5\text{ mmol}$ of $\text{HPO}_4^{2-}$ and $5\text{ mmol}$ of $\text{H}_2\text{PO}_4^-$. For this buffer, $\text{pH} = pK_2 + \log(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}) = 8.0 + \log(1) = 8.0$. Therefore, correct answer is A.