We have taken a saturated solution of AgBr. of AgBr is . If moles of is added to 1 litre of this sol — Electrochemistry Chemistry Question
Question
We have taken a saturated solution of AgBr. $K_{sp}$ of AgBr is $12 \times 10^{-14}$. If $10^{-7}$ moles of $AgNO_3$ is added to 1 litre of this solution, find conductivity (specific conductance) of this solution in terms of $10^{-7} \text{ S m}^{-1}$ units. Given $\lambda^\circ(Ag^+) = 6 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}$, $\lambda^\circ(Br^-) = 8 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}$, $\lambda^\circ(NO_3^-) = 7 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}$.
💡 Solution & Explanation
Let the new solubility be $y$, so $[Br^-] = y$ and $[Ag^+] = y + 10^{-7}$. $K_{sp} = y(y + 10^{-7}) = 12 \times 10^{-14}$. Solving the quadratic equation $y^2 + 10^{-7}y - 12 \times 10^{-14} = 0$ yields $y = 3 \times 10^{-7}$ M. So, $[Br^-] = 3 \times 10^{-7}$ M, $[Ag^+] = 4 \times 10^{-7}$ M, $[NO_3^-] = 10^{-7}$ M. Converting to mol/m$^3$ (multiply by 1000): $C_{Br} = 3 \times 10^{-4}$, $C_{Ag} = 4 \times 10^{-4}$, $C_{NO3} = 1 \times 10^{-4}$. Conductivity $\kappa = \Sigma \lambda_i C_i = (4 \times 10^{-4})(6 \times 10^{-3}) + (3 \times 10^{-4})(8 \times 10^{-3}) + (1 \times 10^{-4})(7 \times 10^{-3}) = (24 + 24 + 7) \times 10^{-7} = 55 \times 10^{-7} \text{ S m}^{-1}$. The value is 55. Therefore, correct answer is 0055.