Among AgCl⁡Ks=1.7×10−10M2,SrF2Ks=8.0×10−11M3,SrCO3⁡Ks=7×10−10M2 and Fe(OH)2Ks=1.8×10−15M3 the most soluble salt is

Since KsSrCO3>Ks(AgCl) and KsSrF2>KsFe(OH)2we compute solubility of SrCO3 and SrF2 only . SrCO3⇌Sr2+s+CO32−; Ks=Sr2+CO32−=s2 s=Ks=7×10−10M21/2=2.65×10−5MSrF2⇌Sr2++2F−; Ks=Sr2+F−2=4s3 s=Ks/41/3=8.0×10−11M3/41/2=2.71×10−4MThus, SrF2 has largest solubility.

Among AgCl⁡Ks=1.7×10−10M2,SrF2Ks=8.0×10−11M3,SrCO3⁡Ks=7×10−10M2 and Fe(OH)2Ks=1.8×10−15M3 the most soluble salt is

Since KsSrCO3>Ks(AgCl) and KsSrF2>KsFe(OH)2we compute solubility of SrCO3 and SrF2 only . SrCO3⇌Sr2+s+CO32−; Ks=Sr2+CO32−=s2 s=Ks=7×10−10M21/2=2.65×10−5MSrF2⇌Sr2++2F−; Ks=Sr2+F−2=4s3 s=Ks/41/3=8.0×10−11M3/41/2=2.71×10−4MThus, SrF2 has largest solubility.

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Among $AgCl (K_{s} = 1.7 \times 10^{- 10} M^{2}), {SrF}_{2} (K_{s} = 8.0 \times 10^{- 11} M^{3}), {SrCO}_{3} (K_{s} = 7 \times 10^{- 10} M^{2})$ and $Fe (OH)_{2} (K_{s} = 1.8$ $\times 10^{- 15} M^{3})$ the most soluble salt is

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$Fe (OH)_{2}$

${SrF}_{2}$

$AgCl$

${SrCO}_{3}$

answer is B.

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Detailed Solution

Since $K_{s} ({SrCO}_{3}) > K_{s} (AgCl) and K_{s} ({SrF}_{2}) > K_{s} (Fe (OH)_{2})$ we compute solubility of ${SrCO}_{3}$ and ${SrF}_{2}$ only .

$\begin{array}{l} {SrCO}_{3} ⇌ \underset{s}{{Sr}^{2 +}} + {CO}_{3}^{2 -}; K_{s} = [{Sr}^{2 +}] [{CO}_{3}^{2 -}] = s^{2} s = \sqrt{K_{s}} = {(7 \times 10^{- 10} M^{2})}^{1 / 2} = 2.65 \times 10^{- 5} M \\ {SrF}_{2} ⇌ {Sr}^{2 +} + 2 F^{-}; K_{s} = [{Sr}^{2 +}] {[F^{-}]}^{2} = 4 s^{3} s = {(K_{s} / 4)}^{1 / 3} = {(8.0 \times 10^{- 11} M^{3} / 4)}^{1 / 2} = 2.71 \times 10^{- 4} M \end{array}$ Thus, ${SrF}_{2}$ has largest solubility.