A solution is 0.020 M in each of Ba^2+ and Sr^2+. Solid Na2SO4 is added gradually with no change in volume. Take K_sp(BaSO4) = 1.0x10^-10 and K_sp(SrSO4) = 3.0x10^-7. Which sulphate precipitates first, and at what [SO4^2-] does the second one begin to precipitate?

BaSO4 first; second begins at 1.5x10^-5 M

SrSO4 first; second begins at 5.0x10^-9 M

BaSO4 first; second begins at 3.0x10^-7 M

BaSO4 first; second begins at 5.0x10^-9 M

BaSO4 first; second begins at 1.5x10^-5 M

JEE Main levelSingle correct optionDifficulty 2 of 3qc132

A solution is 0.020 M in each of $Ba X^{2 +}$ and $Sr X^{2 +}$ . Solid $Na X_{2} SO X_{4}$ is added gradually with no change in volume. Take $K_{s p} (BaSO X_{4}) = 1.0 \times 1 0^{- 10}$ and $K_{s p} (SrSO X_{4}) = 3.0 \times 1 0^{- 7}$ . Which sulphate precipitates first, and at what $[SO X_{4} X^{2 -}]$ does the second one begin to precipitate?

Required

[SO X_{4} X^{2 -}]

K_{s p} / c

: for

BaSO X_{4}

it is

5.0 \times 1 0^{- 9} M

, for

SrSO X_{4}

it is

1.5 \times 1 0^{- 5} M

. The smaller requirement means

BaSO X_{4}

precipitates first, and

SrSO X_{4}

begins when

[SO X_{4} X^{2 -}]

reaches

1.5 \times 1 0^{- 5} M