Question

Calculate the pH of the solution made by adding 0.50 mol of HObr in 0.30 mol of KOBr to 1.00 L of water. The value of Ka for HObr is 2.0 * 10^-9

Answer 1

Concepts and reason

The concept used to solve this problem is based on the chemical equilibrium.

The pH of solution is used to determine acidity and basicity of a solution.

Fundamentals

The pH of solution can be determine by Henderson-Hasselbalch equation as follow.

$pH = p{K_{\rm{a}}} + \log \frac{{\left[ {{\rm{conjugate base}}} \right]}}{{\left[ {{\rm{acid}}} \right]}}$

Here ${\rm{p}}{{\rm{K}}_{\rm{a}}}$ is written as follow.

$p{K_{\rm{a}}} = - \log {K_{\rm{a}}}$

Here, ${K_{\rm{a}}}$ is acid dissociation constant.

The $p{K_{\rm{a}}}$ of the solution can be calculated by formula as follow.

$p{K_{\rm{a}}} = - \log {K_{\rm{a}}}$

Substitute $2.0 \times {10^{ - 9}}$ for ${K_{\rm{a}}}$ .

$\begin{array}{c}\\p{K_{\rm{a}}} = - \log \left( {2.0 \times {{10}^{ - 9}}} \right)\\\\ = 8.7\\\end{array}$

Here, weak acid is ${\rm{HOBr}}$ and its conjugate base is ${\rm{OB}}{{\rm{r}}^ - }$ . In the buffer mixture, ${\rm{HOBr}}$ is act as conjugate base.

The Henderson-Hasselbalch equation is as follow.

$pH = p{K_{\rm{a}}} + \log \frac{{\left[ {{\rm{conjugate base}}} \right]}}{{\left[ {{\rm{acid}}} \right]}}$

Thus, Substitute 8.7 for $p{K_{\rm{a}}}$ , $0.30{\rm{ mol }}{{\rm{L}}^{ - 1}}$ for $\left[ {{\rm{conjugate base}}} \right]$ and $0.50{\rm{ mol }}{{\rm{L}}^{ - 1}}$ for $\left[ {{\rm{weak acid}}} \right]$ .

$\begin{array}{c}\\pH = 8.7 + \log \left( {\frac{{{\rm{0}}{\rm{.30}}}}{{{\rm{0}}{\rm{.50}}}}} \right)\\\\ = 8.5\\\end{array}$

Ans:

The pH value of buffer solution is 8.5.