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  4. 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 [Given: K text SP ( AgBr )=4.9 × 10-13 at 298 K λ Ag +0=6 × 10-3 S m 2 mol -1 λ Br -0=8 × 10-3 S m 2 mol -1 λ NO 3-0=7 × 10-3 S m 2 mol -1 ]

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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 ___
[Given : $K _{\text {SP }}( AgBr )=4.9 \times 10^{-13}$ at $298\, K$
$\lambda_{ Ag ^{+}}^0=6 \times 10^{-3} S\, m ^2\, mol ^{-1} $
$ \lambda_{ Br ^{-}}^0=8 \times 10^{-3} S\, m ^2\, mol ^{-1} $
$ \lambda_{ NO _3^{-}}^0=7 \times 10^{-3} \,S\, m ^2 \,mol ^{-1}$ ]

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Solution:

$ {\left[ Ag ^{+}\right]=10^{-5}}$
${\left[ NO _3^{-}\right]=10^{-5}} $
$[Br^-]= \frac{Ksp}{[Ag^+]}= 4.9 \times 10^{-8}$
$\Lambda_{ m }=\frac{ k }{1000 \times M } $
For $Ag ^{+}$
$ 6 \times 10^{-3}=\frac{ K _{ Ag ^{+}}}{1000 \times 10^{-5}}$
$ K _{ Ag +}=6 \times 10^{-5} $
$ \Rightarrow 6000 \times 10^{-8} $
$ \text { for } Br ^{-} $
$ 8 \times 10^{-3}=\frac{ K _{ Br ^{-}}}{1000 \times 4.9 \times 10^{-8}}$
$ K _{ Br -}=39.2 \times 10^{-8} $
$ \text { for } NO _3^{-}$
$7 \times 10^{-3} =\frac{ K _{ NO _3^{-}}}{1000 \times 10^{-5}} $
$K _{ NO _3} =7 \times 10^{-5} $
$ =7000 \times 10^{-8}$
Conductivity of solution
$ \Rightarrow(6000+7000+39.2) \times 10^{-8} $
$\Rightarrow 13039.2 \times 10^{-8} S m ^{-1}$