The minimum mass of NaBr which should be added in 200 ml of solution just to start the precipitation — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

To start precipitation, $[\text{Ag}^+][\text{Br}^-] = K_{sp}$. With $[\text{Ag}^+] = 4 \times 10^{-4}\text{ M}$, $[\text{Br}^-] = \frac{4 \times 10^{-13}}{4 \times 10^{-4}} = 10^{-9}\text{ M}$. In 200 ml (0.2 L), the moles of NaBr needed $= 0.2 \times 10^{-9} = 2 \times 10^{-10}\text{ moles}$. Molar mass of NaBr $= 23 + 80 = 103\text{ g/mol}$. Minimum mass $= 2 \times 10^{-10} \times 103 = 2.06 \times 10^{-8}\text{ g}$.