Question

Find the pH and concentrations of${\left({\mathrm{CH}}_3\right)}_{\mathbf{3}}$Nand${\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}{\mathbf{NH}}^{\mathbf{+}}$ in a 0.060M solution of trimethylamine.

Short answer

Concentration of${\left({\mathrm{CH}}_3\right)}_3$Nis 0.058M, Concentration of${\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}$is 1.91×10^-3 M and the pH is 11.28

Step-by-step solution

Introduction

We need to write the chemical equation that describes the ionization of trimethylamine

$$\begin{gathered} {\left({\mathrm{CH}}_3\right)}_3\mathrm{N}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}+{\mathrm{OH}}^{-} \\ \left[{\left({\mathrm{CH}}_3\right)}_3\mathrm{N}\right]=0.06-\mathrm{x} \\ \left[{\left({\mathrm{CH}}_3\right)}_3{\mathrm{NH}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]=\mathrm{x} \end{gathered}$$

Calculate the value of Kb

We have to calculate value of${\mathrm{K}}_{\mathrm{b}}$from the value of${\mathrm{K}}_{\mathrm{w}}$and${\mathrm{K}}_{\mathrm{a}}$because the product is${\mathrm{OH}}^{-}$

${\mathrm{K}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{w}}/{\mathrm{K}}_{\mathrm{a}}=6.3\times {10}^{-5}$

Find out the value of x

Now calculate the value of x by using value of${\mathrm{K}}_{\mathrm{b}}$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{x}}^2}{0.06-\mathrm{x}} \\ 6.3\times {10}^{-5}=\frac{{\mathrm{x}}^2}{0.06-\mathrm{x}}\Rightarrow \mathrm{x}=1.91\times {10}^{-3} \end{gathered}$$

Calculate the pH of solution

Since our product isso first we will have to calculate$\mathrm{pOH}$and then$\mathrm{pH}$

$$\begin{gathered} \mathrm{pOH}=-\mathrm{logx}=-\log \left(1.91\times {10}^{-3}\right)=2.72 \\ \mathrm{pH}=14-\mathrm{pOH}=14-2.72=11.28 \end{gathered}$$

Calculate the concentration of (CH3)3N

$$\begin{gathered} \left[{\left({\mathrm{CH}}_3\right)}_3\mathrm{N}\right]=0.06-\mathrm{x} \\ \left[{\left({\mathrm{CH}}_3\right)}_3\mathrm{N}\right]=0.06-\left(1.91\times {10}^{-3}\right) \\ \left[{\left({\mathrm{CH}}_3\right)}_3\mathrm{N}\right]=0.058\mathrm{M} \end{gathered}$$