Question

The solubility of $BaS{O}_4$ in water, is $2.33\times {10}^{-3}$ $g{L}^{-1}$. Its solubility product will be: (molecular weight of $BaS{O}_4=233$).

• A. $1\times {10}^{-5}$
• B. $1\times {10}^{-10}$
• C. $1\times {10}^{-15}$
• D. $1\times {10}^{-20}$

Answer 1

Answer: (B)

$1\times {10}^{-10}$

Let us consider the problem.

Hence,

$SolubilityofBaS{O}_4=2.33\times {10}^{-3}g{L}^{-1}$

Since,

$s=\frac{2.33\times {10}^{-3}}{233}=1.0\times {10}^{-5}mol{L}^{-1}$

$K_{SP}=s^2=(1.038\times {10}^{-5}{)}^2=1.08\times {10}^{-10}mo{l}^2{L}^{-2}$