A volume of 20 ml of 0.8 M – HCN solution is mixed with 80 ml of 0.4 M – NaCN solution. Calculate th — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Moles of HCN (acid) $= 20\text{ ml} \times 0.8\text{ M} = 16\text{ mmol}$. Moles of NaCN (salt) $= 80\text{ ml} \times 0.4\text{ M} = 32\text{ mmol}$. This forms a buffer. $\text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) = -\log(2.5 \times 10^{-10}) + \log\left(\frac{32}{16}\right) = (10 - \log 2.5) + \log 2 = (10 - 0.4) + 0.3 = 9.6 + 0.3 = 9.9$.