Solid BaSO4 and solid SrSO4 are both present in contact with pure water. K_sp(BaSO4) = 1.0 x 10^-10, K_sp(SrSO4) = 3.2 x 10^-7. Find [Ba^2+] in the saturated solution (in M).

JEE Main levelSingle correct optionDifficulty 2 of 3ia061

Solid $BaSO X_{4}$ and solid $SrSO X_{4}$ are both present in contact with pure water. $K_{s p} (BaSO X_{4}) = 1.0 \times 1 0^{- 10}$ , $K_{s p} (SrSO X_{4}) = 3.2 \times 1 0^{- 7}$ . Find $[Ba X^{2 +}]$ in the saturated solution (in $M$ ).

Charge balance gives

[SO X_{4} X^{2 -}] = [Ba X^{2 +}] + [Sr X^{2 +}]

, and

[Ba X^{2 +}] = K_{1} / [SO X_{4} X^{2 -}]

[Sr X^{2 +}] = K_{2} / [SO X_{4} X^{2 -}]

. Hence

[SO X_{4} X^{2 -}]^{2} = K_{1} + K_{2}

, so

[SO X_{4} X^{2 -}] = 5.66 \times 1 0^{- 4} M

and

[Ba X^{2 +}] = 1.0 \times 1 0^{- 10} / (5.66 \times 1 0^{- 4}) = 1.8 \times 1 0^{- 7} M