Question

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

Short answer

The pH value is 9.47.

Step-by-step solution

Define Ionic Equilibrium

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

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:

role="math" localid="1663396782915" $$\begin{gathered} {\text{pK}}_{\text{w}}{\text{= pK}}_{\text{a}}{\text{+ pK}}_{\text{b}} \\ {\text{pK}}_{\text{a}}{\text{= pK}}_{\text{w}}{\text{+ pK}}_{\text{b}} \\ \text{= 14 - 4.75} \\ \text{= 9.25.} \\ \text{pH = pKa + log}\frac{\left[\text{base}\right]}{\left[\text{acid}\right]} \\ \text{= pKa + log}\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}\right]} \\ \text{= 9.25 + log}\frac{\text{0.25 M}}{\text{0.15 M}} \\ \text{= 9.47.} \end{gathered}$$

Therefore, the pH value is 9.47.