$$1 \times 10^{-5}$$ M $$AgNO_3$$ is added to 1 L of saturated solution of AgBr. The conductivity of this solution at 298 K is ______ $$\times 10^{-8} \text{ S m}^{-1}$$.
[Given : $$K_{sp}(AgBr) = 4.9 \times 10^{-13}$$ at 298 K, $$\lambda^0_{Ag^+} = 6 \times 10^{-3} \text{ Sm}^2 \text{ mol}^{-1}$$, $$\lambda^0_{Br^-} = 8 \times 10^{-3} \text{ Sm}^2 \text{ mol}^{-1}$$, $$\lambda^0_{NO_3^-} = 7 \times 10^{-3} \text{ Sm}^2 \text{ mol}^{-1}$$]

The saturated solution of $$AgBr$$ contains equal concentrations of $$Ag^+$$ and $$Br^-$$ ions. If the molar solubility is $$s\; \text{mol L}^{-1}$$, then by definition

$$K_{sp} = [Ag^+][Br^-] = s^2$$

$$\Rightarrow\; s = \sqrt{K_{sp}} = \sqrt{4.9 \times 10^{-13}} \approx 7 \times 10^{-7}\; \text{mol L}^{-1}$$

1 L of this saturated solution is treated with $$1 \times 10^{-5}\; \text{mol}$$ of $$AgNO_3$$, giving an additional $$1 \times 10^{-5}\; \text{mol}$$ of $$Ag^+$$ and the same amount of $$NO_3^-$$. Because $$[Ag^+]$$ now far exceeds the solubility, almost all the original $$Br^-$$ is precipitated. At the new equilibrium let the $$Br^-$$ concentration be $$b$$.

Using the solubility-product relation again,

$$[Ag^+][Br^-] = 4.9 \times 10^{-13}$$

Since $$[Ag^+] \approx 1 \times 10^{-5}\; \text{M}$$ after precipitation,

$$b = \frac{4.9 \times 10^{-13}}{1 \times 10^{-5}} = 4.9 \times 10^{-8}\; \text{mol L}^{-1}$$

Hence the equilibrium ionic concentrations are

$$[Ag^+] \approx 1.0 \times 10^{-5}\; \text{M},\qquad [Br^-] = 4.9 \times 10^{-8}\; \text{M},\qquad [NO_3^-] = 1.0 \times 10^{-5}\; \text{M}$$

To evaluate the conductivity $$(\kappa)$$ we use

$$\kappa = \sum \lambda_i^{0}\,c_i$$

but the concentrations must be in $$\text{mol m}^{-3}$$:

$$1\; \text{mol L}^{-1} = 1000\; \text{mol m}^{-3}$$

Therefore

$$c_{Ag^+}=1.0 \times 10^{-5}\,(\text{M}) \times 1000 = 1.0 \times 10^{-2}\; \text{mol m}^{-3}$$ $$c_{Br^-}=4.9 \times 10^{-8}\,(\text{M}) \times 1000 = 4.9 \times 10^{-5}\; \text{mol m}^{-3}$$ $$c_{NO_3^-}=1.0 \times 10^{-5}\,(\text{M}) \times 1000 = 1.0 \times 10^{-2}\; \text{mol m}^{-3}$$

Now compute each ionic contribution (units: $$\text{S m}^{-1}$$):

$$\kappa_{Ag^+} = 6 \times 10^{-3} \times 1.0 \times 10^{-2} = 6.0 \times 10^{-5}$$ $$\kappa_{Br^-} = 8 \times 10^{-3} \times 4.9 \times 10^{-5} = 3.92 \times 10^{-7}$$ $$\kappa_{NO_3^-}= 7 \times 10^{-3} \times 1.0 \times 10^{-2} = 7.0 \times 10^{-5}$$

Total conductivity:

$$\kappa = 6.0 \times 10^{-5} + 7.0 \times 10^{-5} + 3.92 \times 10^{-7}$$ $$\kappa = 1.3039 \times 10^{-4}\; \text{S m}^{-1}$$

Expressing the result as $$\kappa = \text{(number)} \times 10^{-8}\; \text{S m}^{-1}$$:

$$1.3039 \times 10^{-4}\; \text{S m}^{-1} = 13039 \times 10^{-8}\; \text{S m}^{-1}$$

Hence the required conductivity is 13039 $$\times 10^{-8}\; \text{S m}^{-1}$$.