Chapter U6, Problem 5RE

Interpretation Introduction

Interpretation:

The concentration of ${\text{Cu}}^{\text{2 +}}$ ions in an aqueous solution of copper (II) carbonate needs to be determined.

Concept introduction:

A constant that defines the solubility equilibrium that exists between a compound in itssolid state and its ions in the aqueous state is said to be solubility product constant. Its general expression is written as:

$\begin{array}{l}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\text{(s)}\rightleftharpoons {\text{xA}}^{\text{+}}{\text{(aq) + yB}}^{\text{-}}\text{(aq)}\\ $ {\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{]}}^{\text{x}}{{\text{[B}}^{\text{-}}\text{]}}^{\text{y}}$\end{array}$

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 5RE

The concentration of ${\text{Cu}}^{\text{2 +}}$ ions is $\text{1}\text{.58}\times {\text{10}}^{\text{-5}}\text{ mol/L}$.

Explanation of Solution

Given:

${\text{K}}_{\text{sp}}=\text{ 2}\text{.5}\times {\text{10}}^{\text{-10}}$

The dissociation reaction of ${\text{CuCO}}_3$ is:

${\text{CuCO}}_3\text{(s) }\rightleftharpoons {\text{ Cu}}^{\text{2 +}}{\text{(aq) + CO}}_3^{2-}\text{(aq) }$

The expression for ${\text{K}}_{\text{sp}}$ of ${\text{CuCO}}_3$ is:

${\text{K}}_{\text{sp}}{\text{= [ Cu}}^{\text{2 +}}{\text{][CO}}_3^{2-}\text{]}$

Let the solubility of ${\text{Cu}}^{\text{2 +}}$ be $\text{s mol/L}$ so, the solubility of each is represented as:

$\begin{array}{l}{\text{CuCO}}_3\text{(s) }\rightleftharpoons {\text{ Cu}}^{\text{2 +}}{\text{(aq) + CO}}_3^{2-}\text{(aq) }\\ \text{ s s}\end{array}$

Substituting the values:

$\begin{array}{l}\text{2}\text{.5}\times {\text{10}}^{\text{-10}}\text{ = }\left(\text{s}\right)\left(\text{s}\right)\\ \text{2}\text{.5}\times {\text{10}}^{\text{-10}}{\text{ = s}}^2\\ \text{s = }\sqrt{\text{2}\text{.5}\times {\text{10}}^{\text{-10}}}\\ \text{s = 1}\text{.58}\times {\text{10}}^{\text{-5}}\text{ mol/L}\end{array}$

Thus, the concentration of ${\text{Cu}}^{\text{2 +}}$ ions $\text{1}\text{.58}\times {\text{10}}^{\text{-5}}\text{ mol/L}$.

Conclusion

$\text{1}\text{.58}\times {\text{10}}^{\text{-5}}\text{ mol/L}$ is the concentration of ${\text{Cu}}^{\text{2 +}}$ ions.