Question

Methylamine is a weak base for which K_b is$\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$. Calculate the pH of a solution made by dissolving 0.070 mol of methylamine in water and diluting it to 800.0 mL.

Short answer

The pH of the solution is 11.78.

Step-by-step solution

Methylamine.

When an amine group replaces one hydrogen atom from methane, it forms methylamine. The molecular formula of the methylamine is CH₃NH₂. Methylamine is an organic compound. This methylamine is a colorless gas. This is a primary amine.

 Step 2: Conversion of methylamine mol to M. 

In the given question, moles of methylamine are 0.070. mol. By converting them as below,

[methylamine] $=\frac{0.070mol}{0.800L}$

= 0.085 M

Reaction between methylamine and water.

The following reaction occurswhen methylamine and water are reacted;

$C{H}_{3}N{H}_2\left(aq\right)+H_{2}O\left(l\right)\leftrightharpoons C{H}_{3}N{H}_3^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$

The association constant;

$\begin{array}{c}{K}_b=\frac{\left(H^{+}\right)\left(A^{-}\right)}{HA}\\ K_b=\frac{\left(C{H}_{3}N{H}_3^{-}\right)\left(O{H}^{-}\right)}{C{H}_{3}N{H}_2}\\ K_b=\frac{X^2}{0.0875}-x\\ =4.4\times {10}^{-4}\end{array}$

As;

X= (OH^-)

Therefore;

$\begin{array}{l}{X}^2+4.4\times {10}^{-4}X-3.85\times {10}^{-5}=0\\ X=\left(-4.4\times {10}^{-4}\pm \surd {(4.4\times {10}^{-4})}^2-4\left(1\right)(-3.85\times {10}^{-5})\right)\text{/}2\\ X=6.0\times {10}^{-3}\end{array}$

And;

$\left[O{H}^{-}\right]=6.0\times {10}^{-3}$

Hence;

$\begin{array}{l}pOH=2.22\\ pH=14-2.22\\ =11.78\end{array}$