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Solubility Product

The solubilit...

Question

The solubility product of ${B a S O}_{4}$ is $1.5 \times 10^{- 9}$ . The precipitation in a $0.01 M {B a}^{2 +}$ ions solution will start on adding $H_{2} {S O}_{4}$ of concentration?

10^{- 9} M

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10^{- 8} M

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1.5 X 10^{- 7} M

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10^{- 6} M

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Solution

The correct option is C $1.5 X 10^{- 7} M$
Solubility in $0.01$ M $B a C l_{2}$
$[B a^{+ 2}] = 0.01 + S$ $[S O_{4} - 2]$
$K_{s p} = [B a^{+ 2}] [S O_{4} - 2]$
$= (0.01 + S) (S)$
$= 0.01 S + S^{2}$
$S^{2}$ being the product of two small quantities is neglected.
$K_{s p} = 0.15$
$0.01 S = 1.5 \times 10^{- 9}$
or $S = 1.5 \times 10^{- 7} m o l L^{- 1}$

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The solubility product of BaSO4 is 1.5× 10−9. The precipitation in a 0.01 M Ba2+ ions solution will start on adding H2SO4 of concentration?

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The solubility product of ${B a S O}_{4}$ is $1.5 \times 10^{- 9}$ . The precipitation in a $0.01 M {B a}^{2 +}$ ions solution will start on adding $H_{2} {S O}_{4}$ of concentration?