A solution contains 0.010 M each of Ba^2+ and Sr^2+. Solid K2CrO4 is added slowly (no volume change). Take K_sp: BaCrO4 1.2x10^-10, SrCrO4 2.2x10^-5. The [Ba^2+] remaining at the moment SrCrO4 just begins to precipitate is

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc213

A solution contains $0.010 M$ each of $Ba X^{2 +}$ and $Sr X^{2 +}$ . Solid $K X_{2} CrO X_{4}$ is added slowly (no volume change). Take $K_{s p}$ : $BaCrO X_{4}$ $1.2 \times 1 0^{- 10}$ , $SrCrO X_{4}$ $2.2 \times 1 0^{- 5}$ . The $[Ba X^{2 +}]$ remaining at the moment $SrCrO X_{4}$ just begins to precipitate is

SrCrO4 starts at

[CrO X_{4} X^{2 -}] = 2.2 \times 1 0^{- 5} /0.010 = 2.2 \times 1 0^{- 3} M

. At this

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

[Ba X^{2 +}] = 1.2 \times 1 0^{- 10} /2.2 \times 1 0^{- 3} = 5.5 \times 1 0^{- 8} M

, i.e. barium is removed more than 99.9 percent before strontium begins.