SATHEE: Equilibrium Question 849

Equilibrium Question 849

Question: The solubility (in $m o l L^{- 1}$ ) of $A g C l$

$(K_{s p} = 1.0 \times 10^{- 10})$ in a 0.1 M $K C l$ solution will be

Options:

A) $1.0 \times 10^{- 9}$

B) $1.0 \times 10^{- 10}$

C) $1.0 \times 10^{- 5}$

D) $1.0 \times 10^{- 11}$

Show Answer

Answer:

Correct Answer: A

Solution:

Let solubility of $A g C l = x, m o l e / L$

$A g C l \to A g^{+} + C l^{-}$ i.e. $K_{s p (A g c l)} = x \times x$

$K C l \overset{}{\to} K^{+} + \underset{0.1}{C l^{-}},$

$[C l^{-}]$ from $K C l = 0.1 m$ Total $[C l^{-}]$ in solution $= x + 0.1$

$K_{s p} (A g C l) = [A g^{+}] [C l^{-}] = x (x + 0.1)$

$1.0 \times 10^{- 10} = x (x + 0.1)$

$1.0 \times 10^{- 10} = x^{2} + 0.1 x$

$1.0 \times 10^{- 10} = 0.1 x (a s, x^{2} « 1)$

$x = 1.0 \times 10^{- 9} m o l / L$

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Equilibrium Question 849

Question: The solubility (in molL−1 ) of AgCl

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Question: The solubility (in $m o l L^{- 1}$ ) of $A g C l$