Question

The solubility product $ ({{K}_{sp}}) $ of the following compounds are given at $ 25{}^\circ C $
The most soluble and least soluble compounds are

Compounds	$K_{sp}$
$AgCl$	$1.1 \times 10^{-10}$
$Agcl$	$1.0 \times 10^{-16}$
$pbCrO_{4}$	$4.0 \times 10^{-14}$
$Ag_{2}CO_{3}$	$8.0 \times 10^{-12}$

• A. $ AgCl $ and $ PbCr{{O}_{4}} $
• B. $ AgI $ and $ A{{g}_{2}}C{{O}_{3}} $
• C. $ AgCl $ and $ A{{g}_{2}}C{{O}_{3}} $
• D. $ A{{g}_{2}}C{{O}_{3}} $ and $ AgI $
• E. $ A{{g}_{2}}C{{O}_{3}} $ and $ PbCr{{O}_{4}} $

Answer 1

Answer: (C)

The correct option is $C Ag A _{2} CO _{3}$ and $Agl$
The salts $AgCl , Agl , PbCrO _{4}$ are of same type $AB$.
So,
Solubility $=\sqrt{ K _{ sp }}$
so,
Solubility $_{ AgCl }=1 \times 10^{-5} M$
Solubility $_{ Agl }=1 \times 10^{-8} M$
Solubility $PbCrO _{4}=2 \times 10^{-7} M$
For $Ag _{2} CO _{3}$,
Solubility $_{ Ag _{2} CO _{3}}=\left(\frac{ K _{ S } P }{4}\right)^{\frac{1}{3}}=10^{-4}$
Thus, most soluble is $Ag _{2} CO _{3}$ and least soluble is $Agl$