Barium is separated from strontium and calcium by adding K2CrO4 to a solution buffered with CH3COOH/CH3COONa. The buffer is used because

it keeps [CrO4^2-] low enough for BaCrO4 (small K_sp) to precipitate but not SrCrO4

it converts Ba^2+ to Ba(CH3COO)2 which is insoluble

JEE Main levelSingle correct optionDifficulty 2 of 3qc206

Barium is separated from strontium and calcium by adding $K X_{2} CrO X_{4}$ to a solution buffered with $CH X_{3} COOH$ / $CH X_{3} COONa$ . The buffer is used because

Ait oxidises $Sr X^{2 +}$ to $Sr X^{3 +}$
Bit keeps $[CrO X_{4} X^{2 -}]$ low enough for $BaCrO X_{4}$ (small $K_{s p}$ ) to precipitate but not $SrCrO X_{4}$
Cit reduces $CrO X_{4} X^{2 -}$ to $Cr X^{3 +}$
Dit converts $Ba X^{2 +}$ to $Ba (CH X_{3} COO) X_{2}$ which is insoluble

In acetic acid,

CrO X_{4} X^{2 -}

is partly protonated to

HCrO X_{4} X^{-}

Cr X_{2} O X_{7} X^{2 -}

, giving a low

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

. Only

BaCrO X_{4}

(

K_{s p} \approx 1 0^{- 10}

) exceeds its

K_{s p}

;

SrCrO X_{4}

(about

1 0^{- 5}

) does not.