The solubility of AgCl ( s ) with solubility product 1 . 6 × 10 - 10 in 0.1 M NaCl solution would be [2016]

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Q.
The solubility of ${AgCl}_{(s)}$ with solubility product $1.6 \times 10^{- 10}$ in 0.1 M NaCl solution would be [2016]

1 $1.26 \times 10^{- 5}$ $M$

2 $1.6 \times 10^{- 9}$ $M$

3 $1.6 \times 10^{- 11}$ $M$

4 zero

Ans.
(2)

Let $s$ be the solubility of AgCl in moles per litre.

$\underset{s}{{AgCl}_{(aq)}} ⇌ \underset{s}{{Ag}_{(aq)}^{+}} + \underset{(s + 0.1)}{{Cl}_{(aq)}^{-}}$

( $∵$ 0.1 M NaCl solution also provides 0.1 M ${Cl}^{-}$ ion)

$K_{sp} = [{Ag}^{+}] [{Cl}^{-}]; 1.6 \times 10^{- 10} = s (s + 0.1)$

$1.6 \times 10^{- 10} = s (0.1)$ $(∵ s < < < < 0.1)$

$s = \frac{1.6 \times 10^{- 10}}{0.1} = 1.6 \times 10^{- 9}$ $M$

Solution

Q.
The solubility of ${AgCl}_{(s)}$ with solubility product $1.6 \times 10^{- 10}$ in 0.1 M NaCl solution would be [2016]

1 $1.26 \times 10^{- 5}$ $M$

2 $1.6 \times 10^{- 9}$ $M$

3 $1.6 \times 10^{- 11}$ $M$

4 zero

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Solution

Q. The solubility of AgCl(s) with solubility product 1.6×10-10 in 0.1 M NaCl solution would be [2016]

1 1.26×10-5 M

2 1.6×10-9 M

3 1.6×10-11 M

4 zero

Ans. (2) Let s be the solubility of AgCl in moles per litre. AgCl(aq)s⇌Ag(aq)+s+Cl(aq)-(s+0.1) (∵ 0.1 M NaCl solution also provides 0.1 M Cl- ion) Ksp=[Ag+][Cl-] ; 1.6×10-10=s(s+0.1) 1.6×10-10=s(0.1) (∵s<<<<0.1) s=1.6×10-100.1=1.6×10-9 M

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Q.
The solubility of ${AgCl}_{(s)}$ with solubility product $1.6 \times 10^{- 10}$ in 0.1 M NaCl solution would be [2016]

1 $1.26 \times 10^{- 5}$ $M$

2 $1.6 \times 10^{- 9}$ $M$

3 $1.6 \times 10^{- 11}$ $M$