Q19.17P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

The pH value of the given buffer solution is 10.03.

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 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.