Question

Find the pH of a buffer that consists of 1.3M sodium phenolate (${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{ONa}}$) and 1.2M phenol (${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{OH}}$) (${\mathbf{\text{pK}}}_{\mathbf{\text{a}}}$of phenol=10.00)?

Short answer

The pH value of the given buffer solution is 10.03.

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 1.2 M${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$dissociation equation first.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH +H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}.$

Then, write the 1.3 M${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}$dissociation equation.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{\text{+ Na}}^{\text{+}}.$

We now have the initial phenol and phenolate concentrations.

$$\begin{gathered} {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH = 1.2 M.} \\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\text{= 1.3 M.} \end{gathered}$$

To find the pH, we may utilize the Henderson-Hasselbalch equation.

$$\begin{gathered} \text{pH = pKa + log}\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]} \\ \text{= 10.00 + log}\frac{\text{1.3 M}}{\text{1.2 M}} \\ \text{= 10.03.} \end{gathered}$$

Therefore, the pH value of the given buffer solution is 10.03.