1 × 10 - 5 M AgNO 3 is added to 1 L of saturated solution of AgBr. The conductivity of this solution at 298 K is _________ × 10 - 8 S m - 1 . [Given: K sp ( AgBr ) = 4 . 9 × 10 - 13 at 298 K [2023]

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Solution

Q.
$1 \times 10^{- 5}$ 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

$λ_{{Ag}^{+}}^{\circ} = 6 \times 10^{- 3} {S m}^{2} {mol}^{- 1}$

$λ_{{Br}^{-}}^{\circ} = 8 \times 10^{- 3} {S m}^{2} {mol}^{- 1}$

$λ_{{NO}_{3}^{-}}^{\circ} = 7 \times 10^{- 3} {S m}^{2} {mol}^{- 1}$ ] [2023]

Ans.
(13040)

$10^{- 5} {M AgNO}_{3}$ is added to saturate AgBr.

   $K_{s p} = (s + 10^{- 5}) (s)$

$= s (10^{- 5}) = 4.9 \times 10^{- 13}$

   $s = 4.9 \times 10^{- 8}$

Solution has

$[{Ag}^{+}] = 10^{- 5}, [{NO}_{3}^{-}] = 10^{- 5}$

   $[{Br}^{-}] = 5 \times 10^{- 8}$

   $λ = \frac{K \times 1000}{C}$

   $K_{Total} = \frac{λ_{1} C_{1}}{1000} + \frac{λ_{2} C_{2}}{1000} + \frac{λ_{3} C_{3}}{1000}$

$\Rightarrow \frac{6 \times 10^{- 3} \times 10^{4} \times 10^{- 5}}{1000} + \frac{7 \times 10^{- 3} \times 10^{4} \times 10^{- 5}}{1000} + \frac{8 \times 10^{- 3} \times 10^{4} \times 5 \times 10^{- 8}}{1000}$

$\Rightarrow 6 \times 10^{- 7} + 7 \times 10^{- 7} + 40 \times 10^{- 10}$

$= 13.04 \times 10^{- 7} {S cm}^{- 1} = 13.04 \times 10^{- 5} {S m}^{- 1}$

$= 13040 \times 10^{- 8} {S m}^{- 1}$

$= 13040$

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