Chapter 15, Problem 77QAP

Interpretation Introduction

(a)

Interpretation:

The value of K for equilibrium reaction needs to be determined.

$\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}\left(\text{aq}\right)+{\text{4OH}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\left(\text{aq}\right)+4{\text{NH}}_{\text{3}}\left(\text{aq}\right)$

Concept introduction:

K is known as equilibrium constant of a gaseous mixture and it is equilibrium concentration of products over the equilibrium concentration of reactants.

Answer to Problem 77QAP

The value of K is $8\times {10}^5$

Explanation of Solution

Given the reaction is as follows:

$\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}\left(\text{aq}\right)+{\text{4OH}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\left(\text{aq}\right)+4{\text{NH}}_{\text{3}}\left(\text{aq}\right)$

For this reaction the value of K can be calculated as follows:

$\text{K}=\frac{\left[\text{Zn}{\left(\text{OH}\right)}_{\text{4}}{}^{\text{2-}}\right]{\left[{\text{NH}}_{\text{3}}\right]}^4}{\left[\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}\right]\left[{\text{OH}}^{-}\right]}$   ....... (1)

From the above reaction,

$\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}\left(\text{aq}\right)\rightleftharpoons {\text{Zn}}^{\text{2+}}\left(\text{aq}\right){\text{+4NH}}_{\text{3}}{\text{K}}_{\text{1}}=\frac{1}{{\text{K}}_{\text{fZn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}}}$

${\text{Zn}}^{\text{2+}}\left(\text{aq}\right){\text{+4OH}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\left(\text{aq}\right){\text{K}}_{\text{fZn}{\left(\text{OH}\right)}_4{}^{2-}}$

$\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}\left(\text{aq}\right)+4{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\left(\text{aq}\right){\text{+4NH}}_{\text{3}}\left(\text{aq}\right)\text{K=}{\text{K}}_{\text{1}}{\text{×K}}_{\text{fZn}{\left(\text{OH}\right)}_4{}^{2-}}$

From the overall equation value of K will be

$\begin{array}{c}\text{K}=\frac{1}{{\text{K}}_{\text{fZn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}}}\times {\text{K}}_{\text{fZn}{\left(\text{OH}\right)}_4{}^{2-}}\\ =\frac{1}{3.6\times {10}^8}\times 3\times {10}^{14}\\ =8.3\times {10}^5\\ \approx 8\times {10}^5\end{array}$

Interpretation Introduction

(b)

Interpretation:

The ratio $\left[\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\right]/\left[\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\right]$ in a solution 1.0 M in ${\text{NH}}_{\text{3}}$ needs to be calculated for equilibrium reaction as follows:

$\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{2+}}\left(\text{aq}\right)+{\text{4OH}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\left(\text{aq}\right)+4{\text{NH}}_{\text{3}}\left(\text{aq}\right)$

Concept introduction:${\text{K}}_{\text{b}}$ is known as base equilibrium constant. For the reaction,

${\text{NH}}_{\text{3}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right)+{\text{OH}}^{-}$

Base equilibrium constant is

${\text{K}}_{\text{b}}=\frac{\left[{\text{NH}}_{\text{4}}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{\text{3}}\right]}$

Answer to Problem 77QAP

$2.8\times {10}^{-4}$.

Explanation of Solution

Reaction of ${\text{NH}}_{\text{3}}$ with ${\text{H}}_{\text{2}}\text{O}$ is represented as follows:

${\text{NH}}_{\text{3}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right)+{\text{OH}}^{-}$

Here,

${\text{K}}_{\text{b}}=\frac{\left[{\text{NH}}_{\text{4}}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{\text{3}}\right]}$   ....... (2)

Here, ${\text{K}}_{\text{b}}$ is base equilibrium constant.

$\left[{\text{NH}}_{\text{4}}{}^{\text{+}}\right]=\left[{\text{OH}}^{-}\right]$ for the stoichiometry of the reaction

Put $\left[{\text{OH}}^{-}\right]$ for $\left[{\text{NH}}_{\text{4}}{}^{\text{+}}\right],1.0M$ for $\left[{\text{NH}}_{\text{3}}\right]\text{and}1.8\times {10}^{-5}\text{for}{\text{K}}_{\text{b}}$ in equation (1) and solve for $\left[{\text{OH}}^{-}\right]$

$\begin{array}{c}{\text{K}}_{\text{b}}=\frac{\left[{\text{NH}}_{\text{4}}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{\text{3}}\right]}\\ 1.8\times {10}^{-5}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{OH}}^{-}\right]}{1.0}\\ \left[{\text{OH}}^{-}\right]=\sqrt{\left(1.8\times {10}^{-5}\right)\left(1.0\right)}\\ =4.2\times {10}^{-3}\text{M}\end{array}$

Put $4.2\times {10}^{-3}\text{M}$ for $\left[{\text{OH}}^{-}\right],1.0\text{M}\text{for}\left[{\text{NH}}_{\text{3}}\right]\text{and}8.3\times {10}^5$ for K in equation (1) and solve for the ratio, $\left[\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\right]/\left[\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\right]$

$\begin{array}{c}K=\frac{\left[\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\right]{\left[{\text{NH}}_{\text{3}}\right]}^4}{\left[\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^4}\\ 8.3\times {10}^5=\frac{\left[\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\right]{\left(1.0\right)}^4}{\left[\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\right]\left(4.2\times {10}^{-3}\right)}\\ \frac{\left[Zn{\left(N{H}_3\right)}_4{}^{2+}\right]}{\left[Zn{\left(OH\right)}_4{}^{2-}\right]}=\frac{{\left(1.0\right)}^4}{\left(8.3\times {10}^5\right)\left(4.2\times {10}^{-3}\right)}\\ =2.8\times {10}^{-4}\end{array}$

The ratio, $\left[\text{Zn}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\right]/\left[\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\right]$ in solution 1.0M in ${\text{NH}}_{\text{3}}\text{is }2.8\times {10}^{-4}$