If the solubility of silver sulphate is 1.5×10-2 mol/L, the solubility product in mol3L-3 is

Ag2SO4(s)⇌2Ag++SO42-KSP=(2s)2s=4s3 =4×(1.5×10-2)3=13.5×10-6= 1.35×10-5mol3L-3

If the solubility of silver sulphate is 1.5×10-2 mol/L, the solubility product in mol3L-3 is

Ag2SO4(s)⇌2Ag++SO42-KSP=(2s)2s=4s3 =4×(1.5×10-2)3=13.5×10-6= 1.35×10-5mol3L-3

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If the solubility of silver sulphate is $1.5 \times 10^{- 2}$ mol/L, the solubility product in ${mol}^{3} L^{- 3}$ is

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$1.35 \times 10^{- 5}$

$1.15 \times 10^{- 4}$

$3.375 \times 10^{- 6}$

$1.82 \times 10^{- 6}$

answer is A.

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Detailed Solution

${Ag}_{2} {SO}_{4} (s) ⇌ 2 {Ag}^{+} + {SO}_{4}^{2 -}$

$K_{SP} = {(2 s)}^{2} s = 4 s^{3}$ $= 4 \times {(1.5 \times 10^{- 2})}^{3} = 13.5 \times 10^{- 6}$ = $1 .35 \times 10^{- 5} {mol}^{3} L^{- 3}$

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If the solubility of silver sulphate is 1.5×10-2 mol/L, the solubility product in mol3L-3 is

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If the solubility of silver sulphate is $1.5 \times 10^{- 2}$ mol/L, the solubility product in ${mol}^{3} L^{- 3}$ is