Chapter 17, Problem 17.128QP

The text describes zinc hydroxide as an amphoteric hydroxide, so Zn(OH)₂ is soluble in basic solution. What is the molar solubility of Zn(OH)₂ in 1.00 M NaOH? What is the pH of the equilibrium solution?

Interpretation Introduction

Interpretation:

The molar solubility and $\text{pH}$ of ${\text{Zn(OH)}}_{\text{2}}\text{in}\text{1}\text{.00M}\text{NaOH}$ has to be determined.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution
• Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.

Answer to Problem 17.128QP

The molar solubility and $\text{pH}$ of ${\text{Zn(OH)}}_{\text{2}}\text{in}\text{1}\text{.00M}\text{NaOH}$ is $\text{2}\text{.049}\text{×}{\text{10}}^{\text{-1}}\text{M}$

The $\text{pH}$ of ${\text{Zn(OH)}}_{\text{2}}\text{in}\text{1}\text{.00M}\text{NaOH}$ is $13.77$

Explanation of Solution

To calculate: The molar solubility and $\text{pH}$ of ${\text{Zn(OH)}}_{\text{2}}\text{in}\text{1}\text{.00M}\text{NaOH}$.

When a chemical equation can be derived by taking sum of other equations, then the equilibrium constant of derived equation equals the product of the equilibrium constants of added equations.

$\begin{array}{l}{\text{Zn(OH)}}_{\text{2}}{}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{Zn}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{sp}}\text{ = 2}\text{.1}\text{×}{\text{10}}^{\text{-16}}\\ \\ {\text{Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + 4OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Zn(OH}}^{\text{-}}{\text{)}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{K}}_{\text{f}}\text{ = 2}\text{.8}\text{×}{\text{10}}^{\text{15}}\\ \\ {\text{Zn(OH)}}_{\text{2}}{}_{\text{(s)}}\text{+}{\text{2OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Zn(OH}}^{\text{-}}{\text{)}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{K}}_{\text{c}}\text{ = Ksp}\text{×}{\text{K}}_{\text{f}}\text{=}\text{0}\text{.588}\end{array}$

$\begin{array}{l}{\text{Zn(OH)}}_{\text{2}}{}_{\text{(s)}}\text{+}{\text{2OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Zn(OH}}^{\text{-}}{\text{)}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}\\ \text{Initial }\left(\text{M}\right)\text{: }\text{1}\text{.00 }\text{0}\text{0}\\ \text{Change }\left(\text{M}\right)\text{: }\text{-2x }\text{+x}\text{+x}\\ \text{Equilibrium }\left(\text{M}\right)\text{: }\text{1}\text{.00 - 2x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{c}}\text{ =}\frac{{\text{[Zn(OH}}^{\text{-}}{\text{)}}_{\text{4}}{}^{\text{2-}}\text{]}}{{{\text{[OH}}^{\text{-}}\text{]}}^{\text{2}}}\text{=}\frac{{\text{x}}^{}}{{\text{(1}\text{.00 - 2x)}}^{\text{2}}}\text{=}\text{0}\text{.588}\\ \\ \text{Solve}\text{the}\text{x}\text{by}\text{rearrangining}\text{into}\text{quadratic}\text{equation,}\\ \\ {\text{4x}}^{\text{2}}\text{ - 5}\text{.70x + 1}\text{.00 = 0}\\ \\ \text{x}\text{=}\frac{\text{ 5}\text{.70}\text{±}\sqrt{{\text{(-5}\text{.70)}}^{\text{2}}\text{+}\text{4(4)(1)}}}{\text{2(4)}}\text{=}\text{2}\text{.049}\text{×}{\text{10}}^{\text{-1}}\text{M}\\ \\ \text{The}\text{molar}\text{solubility}\text{of}{\text{Zn(OH)}}_{\text{2}}\text{in}\text{1}\text{.00M}\text{NaOH}\text{is}\text{2}\text{.049}\text{×}{\text{10}}^{\text{-1}}\text{M}\end{array}$

The $\text{pH}$ is calculated using hydroxide ion concentration.

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{] = 1}\text{.00 - 2x = 1}\text{.00 - 2(2}\text{.049 ×}{\text{10}}^{\text{-1}}\text{)}\text{=}\text{0}\text{.590M}\\ \\ {\text{pOH = -log ([OH}}^{\text{-}}\text{]) = -log (0}\text{.590) = 0}\text{.229}\\ \\ \text{pH = 14}\text{.00 - pOH }\\ \text{= 14}\text{.00 - 0}\text{.229 }\\ \text{= 13}\text{.77}\end{array}$

Conclusion

By using the solubility product expression, the molar solubility and $\text{pH}$ of ${\text{Zn(OH)}}_{\text{2}}\text{in}\text{1}\text{.00M}\text{NaOH}$ were determined.