Q19.20P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the pH of a buffer that consists of 0.50 M methylamine (${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}$) and 0.60 M ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{NH}}}_{\mathbf{\text{3}}}\mathbf{\text{Cl}}$ ( ${\mathbf{\text{pK}}}_{\mathbf{\text{b}}}$of ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}\mathbf{\text{ = 3.35}}$)?

Step-by-Step Solution

Answer

The pH value is 10.57.

Step 1: Basic buffer solution

The combination of a weak base with a salt of its conjugate acid forms a basic buffer solution. The mixture of methylamine and methyl ammonium chloride forms a basic buffer solution.

Step 2: The pH of the solution

Write the 0.50 M  ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$dissociation equation first.

${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CH}}_{\text{3}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}{\text{ + OH}}^{\text{ - }}.$

Then write the 0.60 M  ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$dissociation equation.

${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}\to {\text{CH}}_{\text{3}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}{\text{ + Cl}}^{\text{ - }}.$

We now have the initial methylamine and methyl ammonium concentrations.

$$\begin{gathered} {\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{ = 0.50 M.} \\ {\text{CH}}_{\text{3}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\text{= 0.60 M.} \end{gathered}$$

To find the pH, we may utilize the Henderson-Hasselbalch equation. First, solve for the value of ${\text{pK}}_{\text{a}}$ as:

$$\begin{gathered} {\text{pK}}_{\text{w}}{\text{ = pK}}_{\text{a}}{\text{ + pK}}_{\text{b}} \\ {\text{pK}}_{\text{a}}{\text{ = pK}}_{\text{w}}{\text{ + pK}}_{\text{b}} \\ \text{ = 14 - 3.35} \\ \text{ = 10.65.} \\ \text{pH = pKa + log}\frac{\left[\text{base}\right]}{\left[\text{acid}\right]} \\ \text{ = pKa + log}\frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\right]}{\left[{\text{CH}}_{\text{3}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\right]} \\ \text{ = 10.65 + log}\frac{\text{0.50 M}}{\text{0.60 M}} \\ \text{ = 10.57.} \end{gathered}$$

Therefore, the pH value is 10.57.