Answer: pH of the given solution is 8.7.
Explanation:
For the buffer mixture, initial pOH will be given as follows.
pOH = [tex]pK_{b} + log \frac{salt}{base}[/tex]
= [tex]-log(1 \times 10^{-6}) + log \frac{0.25}{0.75}[/tex]
= [tex]6 - 0.477[/tex]
= 5.523
When 0.05 mol NaOH is added to this buffer solution then concentration of species present will be as follows.
[tex]NaOH + HB^{+}Cl^{-} \rightarrow BOH + NaCl[/tex]
0.05 mol 0.25 0.75 0
0 0.20 (0.75 + 0.25) 0.05
Hence, the volume of solution will be 1 liter.
[BOH] = [tex]\frac{1.00}{1}[/tex], [Salt] = [tex]\frac{0.20}{1}[/tex]
So, pOH = [tex]-log(1 \times 10^{-6}) + log (\frac{0.20}{1})[/tex][/tex]
= 6 - 0.69
= 5.30
Now, we will calculate the pH as follows.
pH = 14 - 5.30
= 8.7
Thus, we can conclude that pH of the given solution is 8.7.
