Chapter 15, Problem 65QAP

Interpretation Introduction

Interpretation:

The correct statement/s needs to be identified.

Concept introduction:

The reaction between an ion and ligand to form a complex ion is called a complex ion formation reaction. The equilibrium constant for formation reaction is called formation constant ${\text{K}}_{\text{f}}$. If the formation constant ${\text{K}}_{\text{f}}$ is large, the stability of the complex will also high.

Answer to Problem 65QAP

Correct Answer: Option (b) and (e).

Explanation of Solution

Reason for correct option/s:

b. The equation of $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{}^{\text{+}}$ with strong acid is-

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

Dissociation of the complex, $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{}^{\text{+}}$

$\begin{array}{c}\text{Ag}{\left(\text{NH}{\text{3}}_{}\right)}_{\text{2}}{}^{\text{+}}\left(\text{aq}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2NH}}_{\text{3}}\left(\text{aq}\right)\\ {\text{K}}_{\text{1}}\text{=}\frac{\text{1}}{{\text{K}}_{\text{fAg}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{}^{\text{+}}}}\\ =\frac{1}{1.6\times {10}^7}\\ =6.25\times {10}^{-8}\end{array}$

Equation for two moles of ammonia dissociation is-

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

Equation for ${\text{K}}_{\text{2}}$ is two times equation for dissociation for dissociation of ammonia in water with

$\begin{array}{c}{\text{K}}_{\text{b}}{}_{{\text{NH}}_{\text{3}}}=1.8\times {10}^{-5}\\ {\text{K}}_{\text{2}}\text{=}{\left({\text{K}}_{\text{b}}{}_{{\text{NH}}_{\text{3}}}\right)}^{\text{2}}\\ ={\left(1.8\times {10}^{-5}\right)}^2\\ =3.24\times {10}^{-10}\end{array}$

To produce two moles of water the equation for reaction of two moles of ${\text{OH}}^{-}$ with proton is-

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

Equation for ${\text{K}}_{\text{3}}$ is reverse of two times the equation for self ionization of water with

$\begin{array}{c}{\text{K}}_{\text{w}}={10}^{-14}\\ =\frac{\text{1}}{{\text{K}}_{\text{w}}{}^{\text{2}}}\\ =\frac{1}{{\left({10}^{-14}\right)}^2}\\ ={10}^{28}\end{array}$

By adding above three equilibrium equation, the net ionic equation for the equilibrium reaction of $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{}^{\text{+}}$ with strong acid can be obtain.

$\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{}^{\text{+}}\left(\text{aq}\right){\text{+2H}}^{\text{+}}\left(\text{aq}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{K}}_{\text{1}}$

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

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

$\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{+2H}}^{\text{+}}\left(\text{aq}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{K=K}}_{\text{1}}{\text{×K}}_{\text{2}}{\text{×K}}_{\text{3}}$

Now,

$\begin{array}{c}\text{K}\text{=}{\text{K}}_{\text{1}}{\text{×K}}_{\text{2}}{\text{×K}}_{\text{3}}\\ \text{K}=\left(6.25\times {10}^{-8}\right)\times \left(3.24\times {10}^{-10}\right)\times \left({10}^{28}\right)\\ =2\times {10}^{11}\end{array}$

For the reaction of $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{}^{\text{+}}$ with strong acid, the equilibrium constant K is so large that the equilibrium will favor products and overall reaction goes essentially to completion. Hence, the statement that adding a strong acid $\left({\text{HNO}}_{\text{3}}\right)$ to a solution that is $0.010\text{M}$ in $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{}^{\text{+}}$ will tend to dissolve the complex ion into ${\text{Ag}}^{\text{+}}\text{and}{\text{NH}}_{\text{4}}{}^{\text{+}}$ is true.

e. The equilibrium reaction for the formation of ${\text{AgBr}}_2{}^{-}$ is-

$\begin{array}{l}{\text{Ag}}^{\text{+}}\left(\text{aq}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{AgBr}}_2{}^{-}\\ {\text{K}}_{\text{f}{\text{AgBr}}_2{}^{-}}=\frac{\left[{\text{AgBr}}_2{}^{-}\right]}{\left[{\text{Ag}}^{\text{+}}\right]{\left[{\text{Br}}^{-}\right]}^2}\end{array}$

For solution A: Put $0.01\text{M}$ for $\left[{\text{Br}}^{-}\right],1.3\times {10}^7\text{for}{\text{K}}_{\text{f}{\text{AgBr}}_{\text{2}}}{}^{-}$ and solve for the ratio

$\left[{\text{AgBr}}_{\text{2}}{}^{-}\right]/\left[{\text{Ag}}^{+}\right]$

$\begin{array}{c}{\text{K}}_{\text{f}{\text{AgBr}}_2{}^{-}}=\frac{\left[{\text{AgBr}}_2{}^{-}\right]}{\left[{\text{Ag}}^{\text{+}}\right]{\left[{\text{Br}}^{-}\right]}^2}\\ 1.3\times {10}^7=\frac{\left[{\text{AgBr}}_2{}^{-}\right]}{\left[{\text{Ag}}^{\text{+}}\right]{\left(0.10\right)}^2}\\ \frac{\left[{\text{AgBr}}_2{}^{-}\right]}{\left[{\text{Ag}}^{\text{+}}\right]{\left(0.10\right)}^2}=\left(1.3\times {10}^7\right)\times {\left(0.10\right)}^2\\ =1.3\times {10}^5\end{array}$

The equation for the formation of $\text{Ag}{\left(\text{CN}\right)}_2{}^{-}$ is-

${\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2CN}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_2{}^{-}$

For this reaction the formation constant is,

${\text{K}}_{\text{f}\text{Ag}{\left(\text{CN}\right)}_2{}^{-}}=\frac{\left[\text{Ag}{\left(\text{CN}\right)}_2{}^{-}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{CN}}^{-}\right]}^2}$

For solution A: Put $0.01\text{M}$ for $\left[{\text{CN}}^{-}\right],5.6\times {10}^{18}\text{for}{\text{K}}_{\text{f}\text{Ag}{\left(\text{CN}\right)}_2{}^{-}}$ and solve for the ratio $\frac{\left[\text{Ag}{\left(\text{CN}\right)}_2{}^{-}\right]}{\left[{\text{Ag}}^{+}\right]}$

$\begin{array}{c}{\text{K}}_{\text{f}\text{Ag}{\left(\text{CN}\right)}_2{}^{-}}=\frac{\left[\text{Ag}{\left(\text{CN}\right)}_2{}^{-}\right]}{\left[\text{Ag+}\right]{\left[{\text{CN}}^{-}\right]}^2}\\ 5.6\times {10}^{18}=\frac{\left[\text{Ag}{\left(\text{CN}\right)}_2{}^{-}\right]}{\left[\text{Ag+}\right]{\left(0.10\right)}^2}\\ \frac{\left[\text{Ag}{\left(\text{CN}\right)}_2{}^{-}\right]}{\left[\text{Ag+}\right]}={\left(0.10\right)}^2\times 5.6\times {10}^{18}\\ =5.6\times {10}^{16}\end{array}$

Thus, for the complex the ratio of concentration of ${\text{Ag}}^{\text{+}}$ ion in solution A is $1.3\times {10}^5$ and solution B is $5.6\times {10}^{16}$.

Hence, the solution B will have more particles of complex ion per particle of ${\text{Ag}}^{\text{+}}$ than solution A is true.

Conclusion

Reason for incorrect options:

a. Given formation constant $\left({\text{K}}_{\text{f}}\right)$ values for $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$ and $\text{Ag}{\left(\text{CN}\right)}_2{}^{-}$ are $1.6\times {10}^7\text{and}\text{5}\text{.6}\times {\text{10}}^{\text{18}}$ respectively. Since, the value of ${\text{K}}_{\text{f}}$ for $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$ smaller than that of $\text{Ag}{\left(\text{CN}\right)}_2{}^{-}$. So, $\text{Ag}{\left(\text{CN}\right)}_2{}^{-}$ is more stable than $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$. The statement $\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$ is more stable than $\text{Ag}{\left(\text{CN}\right)}_2{}^{-}$ is false.

c. The equilibrium reaction of ${\text{AgBr}}_2{}^{-}$ with strong acid gives ${\text{Ag}}^{\text{+}}\text{and}{\text{Br}}^{-}$ respectively.

$\begin{array}{c}{\text{AgBr}}_2{}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2Br}}^{-}\\ \text{K}\text{=}\frac{1}{{\text{K}}_{{\text{fAgBr}}_2{}^{-}}}\\ =\frac{1}{1.3\times {10}^7}\\ =7.7\times {10}^{-8}\end{array}$

For the reaction of ${\text{AgBr}}_2{}^{-}$ with strong acid, the equilibrium constant is so small that the reactants will extent and overall reaction virtually not all take places. Hence the statement that adding a strong acid $\left({\text{HNO}}_{\text{3}}\right)$ to a solution that is $0.010\text{M}{\text{AgBr}}_2{}^{-}$ will tend to dissociate the complex ion into ${\text{Ag}}^{\text{+}}\text{and}{\text{Br}}^{-}$ is false.

d. Reaction with AgI and NaCN form $\text{Ag}{\left(\text{CN}\right)}_2{}^{-}$ as follows:

$\text{AgI}\left(\text{s}\right){\text{+2CN}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_2{}^{-}+{\text{I}}^{-}\left(\text{aq}\right)$

This reaction occurs in following two steps as follows:

$\begin{array}{l}\text{AgI}\left(\text{aq}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+I}}^{-}\left(\text{aq}\right){\text{K}}_{\text{sp}\text{AgI}}=1\times {10}^{-6}\\ {\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{CN}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_2{}^{-}\left(\text{aq}\right){\text{K}}_{\text{f}\text{Ag}{\left(\text{CN}\right)}_2{}^{-}}=5.6\times {10}^{18}\\ \underset{\_}{\overline{\text{AgI}\left(\text{s}\right)+2{\text{CN}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_2\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\text{K}={\text{K}}_{\text{spAgI}}\times {\text{K}}_{\text{fAg}{\left(\text{CN}\right)}_2{}^{-}}}}\end{array}$

Again,

$\begin{array}{c}\text{K}={\text{K}}_{\text{spAgI}}\times {\text{K}}_{\text{fAg}{\left(\text{CN}\right)}_2{}^{-}}\\ =\left(1\times {10}^{-16}\right)\left(5.6\times {10}^{18}\right)\\ =5.6\times {10}^2\end{array}$

The value of equilibrium constant (K) is so large that the equilibrium will favor products and overall reaction goes typically to completion.

Reaction of AgI with weak acid HCN forming $\text{Ag}{\left(\text{CN}\right)}_2{}^{-}$ is represented as follows:

$\text{AgI}\left(\text{s}\right)\text{+2HCN}\left(\text{aq}\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_2{}^{-}\left(\text{aq}\right){\text{+2H}}^{\text{+}}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)$

Above reaction occurs in following steps-

$\begin{array}{l}\text{AgI}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+I}}^{-}\left(\text{aq}\right){\text{K}}_{\text{sp}\text{AgI}}=1\times {10}^{-16}\\ \text{2HCN}\left(\text{aq}\right)\rightleftharpoons {\text{2H}}^{\text{+}}\left(\text{aq}\right){\text{+2CN}}^{-}\left(\text{aq}\right){\text{K}}_{\text{1}}={\left({\text{K}}_{\text{a}\text{HCN}}\right)}^2\\ {\text{Ag}}^{\text{+}}\left(\text{aq}\right)+{\text{2CN}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_2{}^{-}\left(\text{aq}\right){\text{K}}_{\text{f}\text{Ag}{\left(\text{CN}\right)}_{\text{2}}}=5.6\times {10}^{18}\\ \\ \text{AgI}\left(\text{s}\right)+{\text{2CN}}^{-}\left(\text{aq}\right)\rightleftharpoons \text{Ag}{\left(\text{CN}\right)}_2{}^{-}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\text{K}={\text{K}}_{\text{sp}\text{AgI}}\times {\left({\text{K}}_{\text{a}\text{HCN}}\right)}^2\times {\text{K}}_{\text{f}\text{Ag}{\left(\text{CN}\right)}_{\text{2}}}\end{array}$

Again,

$\begin{array}{c}\text{K}={\text{K}}_{\text{sp}\text{AgI}}\times {\left({\text{K}}_{\text{a}\text{HCN}}\right)}^2\times {\text{K}}_{\text{f}\text{Ag}{\left(\text{CN}\right)}_{\text{2}}}\\ =\left(1\times {10}^{-16}\right)\times {\left(5.8\times {10}^{-10}\right)}^2\times \left(5.6\times {10}^{-18}\right)\\ =1.9\times {10}^{-16}\end{array}$

The value of K is so large that the equilibrium will favor reactants to an extent that the overall reaction virtually will not takes place. Hence, the statement that one can added either NaCN or HCN as a source of the cyanide ion, to dissolve AgI and fewer moles of NaCN would be required is false.