Q. The solubility product ($K_{sp}$) of the following compounds are given at 25$℃$ [table] Compound | $\mathbf{K}_{\mathbf{sp}}$ ; $AgCl$ | $1.1\ \ 10^{10}$ ; $AgI$ | $1.0\ \ 10^{16}$ ; $PbCrO_{4}$ | $4.0\ \ 10^{14}$ ; $Ag_{2}CO_{3}$ | $8.0\ \ 10^{12}$ [/table] The most soluble and least soluble compounds are
A) AgCl and ${PbCrO}_{4}$
B) $AgI\ and\ {Ag}_{2}{CO}_{3}$
C) $AgCl\ and\ {Ag}_{2}{CO}_{3}$
D) ${Ag}_{2}{CO}_{3}\ and\ AgI$
Explanation
(d)
$AgCl \rightleftharpoons Ag^{+} + Cl^{-}$
$s$ $s$
$K_{sp} = \left\lbrack Ag^{+} \right\rbrack\left\lbrack Cl^{-} \right\rbrack = s^{2}$
$s = \sqrt{K_{sp}} = 1.1 \times 10^{- 10} = 10^{- 5}\$
$Similarly\ s\ for\ AgI\ is\ 10^{- 8},\$
$s\ for\ PbCrO_{4}is\ 2 \times 10^{- 7}and\ s\ for\ Ag_{2}CO_{3}is\ 1.26 \times 10^{- 4}.$
$Most\ soluble - Ag_{2}CO_{3},\ least\ soluble\ AgI$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.