Chemistry - Ionic Equilibrium Question with Solution | TestHub

The given reaction represents the dissolution of a metal sulfide ($MS$) in an acidic medium:

$$MS(s)+2H^{+}(aq)\rightleftharpoons M^{2+}(aq)+H_{2}S(aq)$$

The equilibrium constant for this dissolution is:

$$K_{eq}=\frac{[M^{2+}][H_{2}S]}{[H^{+}{]}^2}$$

Condition for Precipitation:

For a metal sulfide to precipitate, the reaction must shift to the left. This occurs when the reaction quotient ($Q$) exceeds the equilibrium constant ($K_{eq}$):

$$Q>K_{eq}$$

Step-by-Step Solution

1. Calculate the Reaction Quotient ($Q$):

Given the experimental conditions:

$[M^{2+}]=0.01M=10^{-2}M$

$[H_{2}S]=0.1=10^{-1}M$

$[H^{+}]=1M$

Substitute these values into the expression for $Q$:

$$Q=\frac{[M^{2+}][H_{2}S]}{[H^{+}{]}^2}=\frac{(10^{-2})(10^{-1})}{(1{)}^2}$$

$$Q=10^{-3}$$

2. Compare $Q$ with given $K_{eq}$ values:

We compare $Q=1\times 10^{-3}$ with the $K_{eq}$ of each metal sulfide:

Conclusion:

Only PbS satisfy the condition $Q>K_{eq}$ under the given conditions.