An amount of 0.1 millimole of is present in 10 ml acid solution of 0.08 M – HCl. Now is passed to pr — Ionic Equilibrium Chemistry Question
Question
An amount of 0.1 millimole of $\text{CdSO}_4$ is present in 10 ml acid solution of 0.08 M – HCl. Now $\text{H}_2\text{S}$ is passed to precipitate all the $\text{Cd}^{2+}$ ions. What would be the pH of solution after filtering off precipitate, boiling off $\text{H}_2\text{S}$ and making the solution 100 ml by adding water?
💡 Solution & Explanation
Initial $\text{H}^+$ from HCl $= 10\text{ ml} \times 0.08\text{ M} = 0.8\text{ mmol}$. The precipitation reaction is $\text{Cd}^{2+} + \text{H}_2\text{S} \rightarrow \text{CdS} \downarrow + 2\text{H}^+$. Precipitating $0.1\text{ mmol}$ of $\text{Cd}^{2+}$ generates $0.2\text{ mmol}$ of $\text{H}^+$. Total $\text{H}^+ = 0.8 + 0.2 = 1.0\text{ mmol}$. When diluted to $100\text{ ml}$, the final concentration $[\text{H}^+] = \frac{1.0\text{ mmol}}{100\text{ ml}} = 0.01\text{ M}$. The $\text{pH} = -\log(10^{-2}) = 2.0$.