## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693c: 70b

#### Answer

The molar solubility of $ZnCO_3$ in this solution is equal to $2.8 \times 10^{-10}M$

#### Work Step by Step

$0.050M [Zn(NO_3)_2] = 0.050M[Zn^{2+}]$ 1. Write the $K_{sp}$ expression: $ZnCO_3(s) \lt -- \gt 1Zn^{2+}(aq) + 1C{O_3}^{2-}(aq)$ $1.4 \times 10^{-11} = [Zn^{2+}]^ 1[C{O_3}^{2-}]^ 1$ $1.4 \times 10^{-11} = (0.05 + S)^ 1( 1S)^ 1$ 2. Find the molar solubility. Since 'S' has a very small value, we can approximate: $[Zn^{2+}] = 0.05$ $1.4 \times 10^{-11}= 0.05 \times ( 1S)^ 1$ $\frac{1.4 \times 10^{-11}}{0.05} = ( 1S)^ 1$ $2.8 \times 10^{-10} = S$

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