Question

What is the percent ionization in each of the following solutions?

a. 0.10 M NH₃

b. 0.010 M NH₃

c. 0.10 M CH₃NH₂

Short answer

a. Percentage dissociation of 0.10 M NH₃=1.34%

b. Percentage dissociation of 0.010 M NH₃ = 4.24%

c. Percentage dissociation of 0.10 M CH₃NH₂ = 6.26%

Step-by-step solution

Percentage Dissociation of 0.10 M Ammonia  

Reaction of ammonia with water:

$\begin{array}{l}{{\text{NH}}_{\text{3}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ 0.1-00\\ 0.1-\text{x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

[OH^-] ion concentration can be calculated as

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{{\text{x}}^2}{0.1-\mathrm{x}}=1.8\times {10}^{-5}\\ {\text{x}}^2=1.8\times {10}^{-5}\times 0.1\Rightarrow \text{x}=1.34\times {10}^{-3}\end{array}$

$\begin{array}{c}\text{Percent\hspace{0.17em}Dissociation\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Amount\hspace{0.17em}Dissociated}}{\text{Total\hspace{0.17em}Amount}}\times 100\\ =\frac{1.34\times {10}^{-3}}{0.1}\times 100\\ =1.34\%\\ \end{array}$

Percentage Dissociation of 0.01 M Ammonia

Reaction of ammonia with water:

$\begin{array}{l}{{\text{NH}}_{\text{3}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ 0.01-00\\ 0.01-\text{x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{1em}\hspace{1em}\hspace{0.17em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

[OH^-] ion concentration can be calculated as

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{{\text{x}}^2}{0.01-x}=1.8\times {10}^{-5}\\ {\text{x}}^2=1.8\times {10}^{-5}\times 0.01\\ \text{x}=4.24\times {10}^{-4}\end{array}$

$\begin{array}{c}\text{Percent\hspace{0.17em}Dissociation\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Amount\hspace{0.17em}Dissociated}}{\text{Total\hspace{0.17em}Amount}}\times 100\\ =\frac{4.24\times {10}^{-4}}{0.01}\times 100\\ =4.24\%\\ \end{array}$

Percentage Dissociation of 0.1 M CH3NH2

Reaction of Methyl amine with water:

$\begin{array}{l}{\text{CH}}_{\text{3}}{{\text{NH}}_2}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}CH}}_{\text{3}}{\text{NH}}_3{{}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ 0.1-00\\ 0.1-\text{x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{1em}\hspace{0.17em}\hspace{1em}}-\text{\hspace{1em}\hspace{1em}\hspace{0.17em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

[OH^-] ion concentration can be calculated as

role="math" localid="1657282569999" $\begin{array}{l}{\text{K}}_{\text{b}}=\frac{{\mathrm{x}}^2}{0.1-\mathrm{x}}=4.38\times {10}^{-4}(\mathrm{Taken},0.1-\mathrm{x}~0.1)\\ {\text{x}}^2=4.38\times {10}^{-4}\times 0.1\Rightarrow \text{x}=6.62\times {10}^{-3}\end{array}$

$\text{Percent\hspace{0.17em}Dissociation\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Amount\hspace{0.17em}Dissociated}}{\text{Total\hspace{0.17em}Amount}}\text{×100=}\frac{{\text{6.62×10}}^{\text{-3}}}{\text{0.1}}\text{×100=6.62\%}$