Question
The conductivity of a saturated solution containing AgA (Ksp = 3 × 10−14) and AgB (Ksp = 1 × 10−14) is 3.75 × 10−8 Ω−1cm−1. If the limiting molar conductivity of Ag+ and A− ion is 60 and 80 Ω−1 cm2 mol−1, respectively, the limiting molar conductivity of B− (in Ω−1 cm2 mol−1) is
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Solution
$AgA(S) \rightleftharpoons \underset{(x + y)\text{ M}}{Ag^{+}} + \underset{x\text{ M}}{A^{-}}$
$AgB(S) \rightleftharpoons \underset{(x + y)\text{ M}}{Ag^{+}} + \underset{y\text{ M}}{B^{-}}$
$\text{Now, }(x + y).x = 3 \times 10^{- 14}\text{ and }(x + y).y = 1 \times 10^{- 14} $$${\therefore\left\lbrack Ag^{+} \right\rbrack = x + y = 2 \times 10^{- 7}\text{ M,} }{\left\lbrack A^{-} \right\rbrack = x = 1.5 \times 10^{- 7}\text{ M,} }{\left\lbrack B^{-} \right\rbrack = y = 0.5 \times 10^{- 7}\text{ M} }{\text{Now, }\kappa_{\text{solution}} = 3.75 \times 10^{- 8} = 2 \times 10^{- 7} \times 10^{- 3} \times 60 + 1.5 \times 10^{- 7} \times 10^{- 3} \times 80 + 0.5 \times 10^{- 7} \times 10^{- 3}\lambda_{B}^{o} }{\therefore\lambda_{B}^{o} = 135\text{ oh}\text{m}^{- 1}\text{ c}\text{m}^{2}\text{ mo}\text{l}^{- 1}}$$
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