A solution is 0.010 M in each of Ba^2+, Sr^2+ and Ca^2+. Solid Na2SO4 is added gradually (no volume change). Take K_sp: BaSO4 = 1.1x10^-10, SrSO4 = 3.2x10^-7, CaSO4 = 4.9x10^-5. Which statements are correct?

When SrSO4 just starts to precipitate, [Ba^2+] approx 3.4x10^-6 M.

CaSO4 starts to precipitate at [SO4^2-] = 4.9x10^-3 M.

When SrSO4 just starts to precipitate, [Ba^2+] approx 3.4x10^-6 M.

CaSO4 starts to precipitate at [SO4^2-] = 4.9x10^-3 M.

When CaSO4 just starts to precipitate, [Sr^2+] is below 1.0x10^-5 M.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3qc149

A solution is 0.010 M in each of $Ba X^{2 +}$ , $Sr X^{2 +}$ and $Ca X^{2 +}$ . Solid $Na X_{2} SO X_{4}$ is added gradually (no volume change). Take $K_{s p}$ : $BaSO X_{4}$ = $1.1 \times 1 0^{- 10}$ , $SrSO X_{4}$ = $3.2 \times 1 0^{- 7}$ , $CaSO X_{4}$ = $4.9 \times 1 0^{- 5}$ . Which statements are correct?

A $BaSO X_{4}$ precipitates first.
BWhen $SrSO X_{4}$ just starts to precipitate, $[Ba X^{2 +}] \approx 3.4 \times 1 0^{- 6} M$ .
C $CaSO X_{4}$ starts to precipitate at $[SO X_{4} X^{2 -}] = 4.9 \times 1 0^{- 3} M$ .
DWhen $CaSO X_{4}$ just starts to precipitate, $[Sr X^{2 +}]$ is below $1.0 \times 1 0^{- 5} M$ .

Required

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

: Ba

1.1 \times 1 0^{- 8} M

, Sr

3.2 \times 1 0^{- 5} M

, Ca

4.9 \times 1 0^{- 3} M

, so Ba precipitates first (A true). When Sr starts,

[Ba X^{2 +}] = 1.1 \times 1 0^{- 10} /3.2 \times 1 0^{- 5} = 3.4 \times 1 0^{- 6}

M (B true). Ca starts at

4.9 \times 1 0^{- 3} M

(C true). Then

[Sr X^{2 +}] = 3.2 \times 1 0^{- 7} /4.9 \times 1 0^{- 3} = 6.5 \times 1 0^{- 5}

M, which is above

1.0 \times 1 0^{- 5}

M, so D is false.