## General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal

# Chapter 15 - Priciples of Chemical Equilibrium - Example 15-1 - Relating Equilibrium Concentrations of Reactants and Products - Page 661: Practice Example B

#### Answer

$$[Hg_2^{2+}]_{eq} = 0.0099 \space M$$

#### Work Step by Step

1. Write the K expression: $$K = \frac{[Fe^{2+}]^2[Hg^{2+}]^2}{[Fe^{3+}]^2[Hg_2^{2+}]}$$ 2. Solve for $[Hg_2^{2+}]$ $$[Hg_2^{2+}] =\frac{[Fe^{2+}]^2[Hg^{2+}]^2}{[Fe^{3+}]^2K}$$ 3. Substitute the values and calculate: $$[Hg_2^{2+}] = \frac{(0.0018)^2(0.0025)^2}{(0.015)^2(9.14 \times 10^{-6})} M = 0.0099 \space M$$

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