Q19.18P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the pH of a buffer that consists of 0.12 M boric acid (${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{BO}}}_{\mathbf{\text{3}}}$) and 0.82 M sodium borate(${\mathbf{\text{NaH}}}_{\mathbf{\text{2}}}{\mathbf{\text{BO}}}_{\mathbf{\text{3}}}$ ) ( ${\mathbf{\text{pK}}}_{\mathbf{\text{a}}}$of boric acid=9.24)?

Step-by-Step Solution

Answer

The pH value of the given solution is 10.07.

Step 1: Acidic buffer

The combination of a weak acid and a salt solution with its conjugate base forms an acidic buffer.

Step 2: The pH of the given solution

Write the 0.12 M ${\text{H}}_{\text{3}}{\text{BO}}_{\text{3}}$ dissociation equation first.

${\text{H}}_{\text{3}}{\text{BO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}\text{B}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }}.$

Then, write the 0.82 M ${\text{NaH}}_{\text{2}}{\text{BO}}_{\text{3}}$ dissociation equation.

${\text{NaH}}_{\text{2}}{\text{BO}}_{\text{3}}\to {\text{H}}_{\text{2}}\text{B}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{ + Na}}^{\text{ + }}.$

We now have the initial boric acid and borate concentrations.

$$\begin{gathered} {\text{ H}}_{\text{3}}{\text{BO}}_{\text{3}}\text{ = 0.12 M.} \\ {\text{H}}_{\text{2}}\text{B}{{\text{O}}_{\text{3}}}^{\text{ - }}\text{ = 0.82 M.} \end{gathered}$$

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

$$\begin{gathered} \text{pH = pKa + log}\frac{\left[{\text{H}}_{\text{2}}\text{B}{{\text{O}}_{\text{3}}}^{\text{ - }}\right]}{\left[{\text{H}}_{\text{3}}{\text{BO}}_{\text{3}}\right]} \\ \text{ = 9.24 + log}\frac{\text{0.82 M}}{\text{0.12 M}} \\ \text{ = 10.07.} \end{gathered}$$

Therefore, the pH value of the given solution is 10.07.