The pH of the resulting solution on adding 0.15 mole HCl in 500 ml of the buffer solution — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Adding $0.15\text{ moles}$ of HCl to $500\text{ ml}$ (0.5 L) corresponds to an added acid concentration of $0.15 / 0.5 = 0.30\text{ M}$. This strong acid reacts with the weak base $\text{NH}_4\text{OH}$, reducing its concentration and increasing the salt $\text{NH}_4\text{Cl}$. New $[\text{Base}] = 0.8 - 0.30 = 0.50\text{ M}$. New $[\text{Salt}] = 0.2 + 0.30 = 0.50\text{ M}$. The new pH is $\text{pH} = pK_a + \log(\frac{0.50}{0.50}) = 9.3 + \log(1) = 9.3$. Therefore, correct answer is A.