91Ó°ÊÓ

If 6 mL of HBr have been added, the reaction will be six-tenths completed as V_e (equivalence point)=10.00 mL. The unreacted OH^- fraction will be four-tenths.

Initial volume of OH^- =50 mL

Total volume of Solution= 50+6=56mL

Initial concentration of OH^- =0.02M

The concentration of the remaining OH^-can be calculated using the following equation

=Fraction remaining × Initial concentration × Dilution factor

Concentration of OH^-

$\begin{array}{c}\left[O{H}^{−}\right]=\left(\frac{10−6}{10}\right)\left(0.02\text{ }M\right)\left(\frac{50}{50+6}\right)\\ =0.0071\text{ }M\end{array}$

The concentration of H⁺ will be

$\begin{array}{c}\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{−}\right]}\\ =\frac{1×{10}^{−14}}{0.0071}M\\ =1.41×{10}^{−12}M\end{array}$

The pH of the solution will be

$\begin{array}{l}pH=−\log \left[H^{+}\right]\\ pH=−\log \left[1.41×{10}^{−12}\right]\\ pH=11.85\end{array}$

The pH value obtained from the above calculation and the pH value given in Table11-1 for addition of 6mL HBr in the system exactly matched. Both the pH values are 11.85.