Question

The solubility of $BaS{O}_4$ in water is 2.42 × ${10}^{–3}$ $g{L}^{–1}$ at 298 K. The value of its solubility product ($Ksp$) will be
(Given molar mass of$BaS{O}_4$ = 233 g mol–1)

• A. 1.08 × ${10}^{–14}$ $mo{l}^2{L}^{-2}$
• B. 1.08 × ${10}^{–12}$ $mo{l}^2{L}^{-2}$
• C. 1.08 × ${10}^{–10}$ $mo{l}^2{L}^{-2}$
• D. 1.08 × ${10}^{–8}$ $mo{l}^2{L}^{-2}$

Answer 1

The correct option is C 1.08 × ${10}^{–10}$ $mo{l}^2{L}^{-2}$

Solubility of $BaS{O}_4,s=\frac{2.42\times {10}^{-3}}{233}\left(mol{L}^{-1}\right)$
$=1.04\times {10}^{-5}\left(mol{L}^{-1}\right)$
$BaS{O}_4\left(s\right)\rightleftharpoons \underset{s}{B{a}^{2+}}\left(aq\right)+\underset{s}{S{O}_4^{2+}}\left(aq\right)$
$K_{SP}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]=s^2$
$=(1.04\times {10}^{-5}{)}^2$
$=1.08\times {10}^{-10}mo{l}^2{L}^{-2}$