Chapter 17, Problem 93RQ

Interpretation Introduction

Interpretation:

The $\text{pH}$ range of the solution that allows selective precipitation of ${\text{Cu}}^{2+}$ as $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ from the given solution, is to be determined.

Concept introduction:

According to the law of chemical equilibrium, the equilibrium constant for an equilibrium reaction is the ratio of the product of molar concentrations of the products to the product of molar concentrations of the reactants, each raised to the power of their stochiometric coefficient in the overall balanced equilibrium reaction.

For a general equilibrium reaction, $\text{aA + bB }\rightleftharpoons \text{cC+dD}$, and the equilibrium constant is represented as:

$K\text{ = }\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

When partially soluble salts (like $\text{AB}$) are dissolved in water, the equilibrium established between the undissolved salt and the dissolved salt is represented as:

$\text{AB}\rightleftharpoons {\text{A}}^{n+}+{\text{B}}^{n-}$

The expression for the solubility product of the salt $\text{AB}$ is represented as:

$\begin{array}{l}K\text{ = }\frac{\left[{\text{A}}^{n+}\right]\left[{\text{B}}^{n-}\right]}{\left[\text{AB}\right]}\\ K\times \left[\text{AB}\right]=\left[{\text{A}}^{n+}\right]\left[{\text{B}}^{n-}\right]\\ K_{\text{sp}}=\left[{\text{A}}^{n+}\right]\left[{\text{B}}^{n-}\right]\end{array}$

The relation between $\text{pH}$ and $\text{pOH}$ is shown as:

$\text{pH}+\text{pOH}=14$.