First slide

Solubility product(KSP)

Question

If $S_{1}, S_{2}, S_{3} and S_{4}$ are the solubilities of AgCl in water, in 0.01 M ${CaCl}_{2}$ , in 0.01 M NaCl and in 0.05 M ${AgNO}_{3}$ respectively at a certain temperature, the correct order of solubilities is

Difficult

A

$S_{1} > S_{2} > S_{3} > S_{4}$
B

$S_{1} > S_{3} > S_{2} > S_{4}$
C

$S_{1} > S_{2} = S_{3} > S_{4}$
D

$S_{1} > S_{3} > S_{4} > S_{2}$

Solution

$0.01 M {CaCl}_{2} = 0.02 M {Cl}^{-},$
$0.01 NaCl = 0.01 M {Cl}^{-}$
$0.05 M {AgNO}_{3} = 0.05 M {Ag}^{+}$
$[{Ag}^{+}] [{Cl}^{-}] = K_{sp} (const.) Hence,$
$S_{2} = \frac{K_{sp}}{0.02} = 50 K_{sp}$
$S_{3} = \frac{K_{sp}}{0.01} = 100 K_{sp}$
$S_{4} = \frac{K_{sp}}{0.05} = 20 K_{sp} or (S_{1} = \sqrt{K_{sp}} = maximum)$

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