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Question: The specific conductivity of a saturated solution of AgClis \(\text{3.40} \times \text{1}\text{0}^{\...

The specific conductivity of a saturated solution of AgClis 3.40×10-6 ohm-1 cm-1 at 250C\text{3.40} \times \text{1}\text{0}^{\text{-6}}\text{ oh}\text{m}^{\text{-1}}\text{ c}\text{m}^{\text{-1}}\text{ at 2}\text{5}^{0}C. If

λAg+= 62.3 ohm-1 cm2 mol-1 & λCl = 67.7ohm-1cm2mol−1,\lambda_{\text{Ag+}}\text{= 62.3 oh}\text{m}^{\text{-1}}\text{ c}\text{m}^{2}\text{ mo}\text{l}^{\text{-1}}\text{ \& }\lambda_{\text{Cl}}\text{ = 67.7oh}\text{m}^{\text{-1}}cm^{2}mol^{- 1}, the solubility of AgCl at 25 ŗC is :

A

2.6 × 10-5 M\text{2.6 } \times \text{ 1}\text{0}^{\text{-5}}\text{ M}

B

4.5 × 10-3 M\text{4.5 } \times \text{ 1}\text{0}^{\text{-3}}\text{ M}

C

3.6 ×10-5 M\text{3.6 } \times \text{1}\text{0}^{\text{-5}}\text{ M}

D

3.6 × 10  -3 M\text{3.6 } \times \text{ 10 }\text{ }^{\text{-3}}\text{ M}

Answer

2.6 × 10-5 M\text{2.6 } \times \text{ 1}\text{0}^{\text{-5}}\text{ M}

Explanation

Solution

λAg+\lambda _ { \mathrm { Ag } ^ { + } }= 62.3 Scm2 mol–1 , = 67.7 Scm2 mole–1

KAgcl = 3.4 × 10–6 Scm–1 = (62.3 + 67.5)

= 1000×3.4×10−6S\frac { 1000 \times 3.4 \times 10 ^ { - 6 } } { S }

S = 3.4×10−3(62.3+67.5)\frac { 3.4 \times 10 ^ { - 3 } } { ( 62.3 + 67.5 ) } = 2.6 × 10–5 M