The solubility product of ${\text{BaSO}}_{\text{4}}$ in water is $1.5\times {10}^{-9}$ . The molar solubility of ${\text{BaSO}}_{\text{4}}$ in 0.1 M solution of ${\text{Ba(NO}}_3{)}_2$ is:

Let the solubility of  ${\text{BaSO}}_{\text{4}}$= s mole

${\text{BaSO}}_{\text{4}}\rightleftharpoons \underset{\text{s}}{{\text{Ba}}^{\text{+2}}}\text{+}\underset{\text{s}}{{\text{SO}}_{\text{4}}^{\text{-2}}}$

${\text{BaSO}}_{\text{4}}\rightleftharpoons \underset{0.1}{{\text{Ba}}^{\text{+2}}}\text{+}\underset{2\left(0.1\right)}{{\text{SO}}_{\text{4}}^{\text{-2}}}$

${\text{[Ba}}^{\text{+2}}{\text{]=s+0.1;[SO}}_{\text{4}}^{\text{-2}}\text{]=s}$

${\text{K}}_{\text{sp}}{\text{=BaSO}}_{\text{4}}$

${\text{1.5×10}}^{\text{-9}}{\text{=[s+0.1][S]=s}}^{\text{2}}\text{+0.1s}$

S is very small, neglect  ${\text{s}}^2$

${\text{1.5×10}}^{\text{-9}}\text{=0.1s}$

${\text{S=1.5×10}}^{\text{=8}}\text{M}$