Q19.19P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the pH of a buffer that consists of 0.25 M ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$ and 0.15 M ${\mathbf{\text{NH}}}_{\mathbf{\text{4}}}\mathbf{\text{Cl}}$ ( ${\mathbf{\text{pK}}}_{\mathbf{\text{b}}}$of ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}\mathbf{\text{ = 4.75}}$)?

Step-by-Step Solution

Answer

The pH value is 9.47.

Step 1: Define Ionic Equilibrium

An ionic equilibrium refers to an equilibrium that exists in the solution of weak electrolytes between unionized molecules and ions.

Step 2: The pH of the given solution

Write the 0.25 M ${\text{NH}}_{\text{3}}$ dissociation equation first.

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

Then write the 0.15 M ${\text{NH}}_{\text{4}}\text{Cl}$ dissociation equation.

${\text{NH}}_{\text{4}}\text{Cl}\to \text{N}{{\text{H}}_{\text{4}}}^{\text{ + }}{\text{ + Cl}}^{\text{ - }}.$

We now have the initial ammonia and ammonium concentrations.

$$\begin{gathered} {\text{ NH}}_{\text{3}}\text{ = 0.25 M.} \\ \text{N}{{\text{H}}_{\text{4}}}^{\text{ + }}\text{ = 0.15 M.} \end{gathered}$$

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

$$\begin{gathered} {\text{pK}}_{\text{w}}{\text{ = pK}}_{\text{a}}{\text{ + pK}}_{\text{b}} \\ {\text{pK}}_{\text{a}}{\text{ = pK}}_{\text{w}}{\text{ + pK}}_{\text{b}} \\ \text{ = 14 - 4.75} \\ \text{ = 9.25.} \\ \text{pH = pKa + log}\frac{\left[\text{base}\right]}{\left[\text{acid}\right]} \\ \text{ = pKa + log}\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[\text{N}{{\text{H}}_{\text{4}}}^{\text{ + }}\right]} \\ \text{ = 9.25 + log}\frac{\text{0.25 M}}{\text{0.15 M}} \\ \text{ = 9.47.} \end{gathered}$$

Therefore, the pH value is 9.47.