$$1 \times 10^{-5}$$ M AgNO$$_3$$ is added to $$1$$ L of saturated solution of AgBr. The conductivity of this solution at 298 K is _______ $$\times 10^{-8}$$ S m$$^{-1}$$.
[Given: K$$_{sp}$$(AgBr) $$= 4.9 \times 10^{-13}$$ at 298 K, $$\lambda^{0}_{\text{Ag}^+} = 6 \times 10^{-3}$$ Sm$$^2$$ mol$$^{-1}$$, $$\lambda^{0}_{\text{Br}^-} = 8 \times 10^{-3}$$ Sm$$^2$$ mol$$^{-1}$$, $$\lambda^{0}_{\text{NO}_3^-} = 7 \times 10^{-3}$$ Sm$$^2$$ mol$$^{-1}$$]

We need to find the conductivity of the solution when $$1 \times 10^{-5}$$ M AgNO$$_3$$ is added to 1 L of saturated AgBr solution at 298 K.

Identify ions in solution.

AgNO$$_3$$ dissociates completely: AgNO$$_3$$ $$\to$$ Ag$$^+$$ + NO$$_3^-$$.

AgBr dissolves sparingly: AgBr $$\rightleftharpoons$$ Ag$$^+$$ + Br$$^-$$.

The ions present are Ag$$^+$$, Br$$^-$$, and NO$$_3^-$$.

Find ion concentrations using common ion effect.

$$[\text{Ag}^+]$$ from AgNO$$_3$$ = $$1 \times 10^{-5}$$ M. The Ag$$^+$$ from AgBr dissolution is negligible compared to this, so total $$[\text{Ag}^+] \approx 1 \times 10^{-5}$$ M.

From the solubility product:

$$[\text{Br}^-] = \frac{K_{sp}}{[\text{Ag}^+]} = \frac{4.9 \times 10^{-13}}{1 \times 10^{-5}} = 4.9 \times 10^{-8} \text{ mol/L}$$

$$[\text{NO}_3^-] = 1 \times 10^{-5}$$ mol/L (from AgNO$$_3$$).

Convert concentrations to mol/m$$^3$$.

Since 1 mol/L = 1000 mol/m$$^3$$:

$$[\text{Ag}^+] = 1 \times 10^{-5} \times 1000 = 1 \times 10^{-2}$$ mol/m$$^3$$

$$[\text{Br}^-] = 4.9 \times 10^{-8} \times 1000 = 4.9 \times 10^{-5}$$ mol/m$$^3$$

$$[\text{NO}_3^-] = 1 \times 10^{-5} \times 1000 = 1 \times 10^{-2}$$ mol/m$$^3$$

Calculate conductivity using $$\kappa = \sum c_i \lambda_i^0$$.

Where $$c_i$$ is in mol/m$$^3$$ and $$\lambda_i^0$$ is in S m$$^2$$ mol$$^{-1}$$, the product gives $$\kappa$$ in S m$$^{-1}$$.

$$\kappa = [\text{Ag}^+]\lambda^0_{\text{Ag}^+} + [\text{Br}^-]\lambda^0_{\text{Br}^-} + [\text{NO}_3^-]\lambda^0_{\text{NO}_3^-}$$

$$= (1 \times 10^{-2})(6 \times 10^{-3}) + (4.9 \times 10^{-5})(8 \times 10^{-3}) + (1 \times 10^{-2})(7 \times 10^{-3})$$

$$= 6 \times 10^{-5} + 3.92 \times 10^{-7} + 7 \times 10^{-5}$$

Convert to the required units ($$\times 10^{-8}$$ S m$$^{-1}$$).

$$6 \times 10^{-5} = 6000 \times 10^{-8}$$ S m$$^{-1}$$

$$3.92 \times 10^{-7} = 39.2 \times 10^{-8}$$ S m$$^{-1}$$

$$7 \times 10^{-5} = 7000 \times 10^{-8}$$ S m$$^{-1}$$

$$\kappa = (6000 + 39.2 + 7000) \times 10^{-8} = 13039.2 \times 10^{-8} \text{ S m}^{-1}$$

Round to nearest integer.

$$\kappa \approx 13039 \times 10^{-8}$$ S m$$^{-1}$$.

The correct answer is 13039.