Question

The pH of a 0.016 M aqueous solution of p-toluidine (CH₃C₆H₄NH₂) is 8.60. Calculate K_b.

Short answer

K_b for the given p-toluidine solution is 9.9×10^-10.

Step-by-step solution

[OH-] Concentration Calculation of p-toluidine

It has to be calculated as:

$\begin{array}{l}\text{pOH=pOH=14-pH}\\ \text{pOH}=14-8.6=5.4\\ {\text{[OH}}^{\text{-}}{\text{]=\hspace{0.17em}10}}^{\text{-pH}}={\text{10}}^{\text{-5.4}}=3.98\times {10}^{-6}\end{array}$

Calculation of Kb for p-toluidine 

Reaction of p-tolidine with water:

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{NH}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}\hspace{1em}CH}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{{\text{NH}}_3}^{\text{+}}{\text{+\hspace{0.17em}OH}}^{\text{-}}\\ 0.016-00\\ 0.016-\text{x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{1em}\hspace{1em}}-\text{\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}}3.98\times {10}^{-6}\end{array}$

K_bcan be calculated as:

$\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.016-x}}\\ \text{=}\frac{{{\text{(3.98×10}}^{\text{-6}}\text{)}}^{\text{2}}}{{\text{0.016-(3.98×10}}^{\text{-6}}\text{)}}\\ {\text{=9.9×10}}^{\text{-10}}\end{array}$