JEE Advanced 2013 · Paper 2 · Chemistry Q25

ChemistryMultiple correct+3 / -1Question 25 of 60

The $K_{sp}$ of $\ce{Ag2CrO4}$ is $1.1\times10^{-12}$ at 298 K. The solubility (in mol/L) of $\ce{Ag2CrO4}$ in a 0.1 M $\ce{AgNO3}$ solution is

1.1\times10^{-11}

1.1\times10^{-10}

1.1\times10^{-12}

1.1\times10^{-9}

Answer and worked solution

Official answer: B

Common ion. $[\ce{Ag+}]\approx0.1$ M, so
$s=\frac{K_{sp}}{[\ce{Ag+}]^2}=\frac{1.1\times10^{-12}}{0.01}=1.1\times10^{-10}\ \text{mol/L}.$

(B)

Topics:Common ion effect Solubility product

JEE Advanced 2013 · Paper 2, Chemistry question 25