The pH of the resulting solution on adding 0.5 mole HCl in 500 ml of the buffer solution — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Adding $0.5\text{ moles}$ of HCl to $500\text{ ml}$ corresponds to adding $1.0\text{ M}$ strong acid. The initial base concentration is $0.8\text{ M}$, so the HCl completely neutralizes all the $\text{NH}_4\text{OH}$ and leaves an excess of $1.0 - 0.8 = 0.2\text{ M HCl}$. Because HCl is a strong acid, it completely dictates the pH of the solution. The concentration of $\text{H}^+$ is $0.2\text{ M} = 2 \times 10^{-1}\text{ M}$. $\text{pH} = -\log(2 \times 10^{-1}) = 1 - \log 2 = 1 - 0.30 = 0.70$. Therefore, correct answer is C.