JEE Advanced 2019 · Paper 1 · Chemistry Q15

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ChemistryNumerical+3Question 33 of 54

For the following reaction, the equilibrium constant $K_c$ at 298 K is $1.6\times10^{17}$ .

$\ce{Fe^{2+}(aq) + S^{2-}(aq) <=> FeS(s)}$

When equal volumes of 0.06 M $\ce{Fe^{2+}}(aq)$ and 0.2 M $\ce{S^{2-}}(aq)$ solutions are mixed, the equilibrium concentration of $\ce{Fe^{2+}}(aq)$ is found to be $Y\times10^{-17}$ M. The value of $Y$ is ______

Your answer

Answer and worked solution

Official answer: 8.7

After mixing equal volumes. $[\ce{Fe^{2+}}]=0.03$ M and $[\ce{S^{2-}}]=0.1$ M.

Precipitation. $K$ is huge, so FeS precipitates almost completely, leaving $[\ce{S^{2-}}]\approx0.07$ M.

Remaining iron.
$[\ce{Fe^{2+}}]=\frac1{K_c[\ce{S^{2-}}]}=\frac1{1.6\times10^{17}\times0.07}=8.93\times10^{-17}\text{ M}.$
Answer: 8.93 (key: 8.70 to 9.10)

Topics:Law of mass action and Kc, Kp Solubility product

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JEE Advanced 2019 · Paper 1, Chemistry question 15