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

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Redox Reactions and Electrochemistry

1 \times 1 0^{- 5}

AgNO_{3}

is added to 1 L of saturated solution of AgBr. The conductivity of this solution at 298 K is ______

\times 1 0^{- 8} S m^{- 1}

.
[Given:

K_{sp} (AgBr) = 4.9 \times 1 0^{- 13}

at 298 K,

λ_{Ag^{+}}^{0} = 6 \times 1 0^{- 3} Sm^{2} mol^{- 1}

λ_{Br^{-}}^{0} = 8 \times 1 0^{- 3} Sm^{2} mol^{- 1}

λ_{NO_{3}^{-}}^{0} = 7 \times 1 0^{- 3} Sm^{2} mol^{- 1}

]

Answer: 13039.2

Adding

1 0^{- 5}

M AgNO

_{3}

supplies a common ion (Ag

^{+}

) that suppresses AgBr dissolution. With

[A g^{+}] \approx 1 0^{- 5}

[B r^{-}] = K_{s p} / [A g^{+}] = 4.9 \times 1 0^{- 13} /1 0^{- 5} = 4.9 \times 1 0^{- 8}

M and

[N O_{3}^{-}] = 1 0^{- 5}

M. Converting to

mol m^{- 3}

(x1000) and using

κ = \sum λ_{i}^{0} c_{i}

κ = 6 \times 1 0^{- 3} (1 0^{- 2}) + 8 \times 1 0^{- 3} (4.9 \times 1 0^{- 5}) + 7 \times 1 0^{- 3} (1 0^{- 2}) = 1.30392 \times 1 0^{- 4} S m^{- 1}

, i.e.

13039.2 \times 1 0^{- 8} S m^{- 1}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Numerical answer