Question

Consider 1.0 L of a solution that is 0.85 M ${\mathbf{\text{HOC}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}$, and 0.80 M${\mathbf{\text{NaOC}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}$. (${\mathbf{\text{K}}}_{\mathbf{a}}{\mathbf{\text{\hspace{0.17em}for\hspace{0.17em}HOC}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{=}\mathbf{1}\mathbf{.6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$.)
a.Calculate the pH of this solution.
b. Calculate the pH after 0.10 mole of HCl has been added to the original solution. Assume no volume change on addition of HCI.
c. Calculate the pH after 0.20 mole of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

Short answer

ThepH of the given solutions is:

1. $\mathrm{pH}=9.77$
   
2. $\mathrm{pH}=9.67$
   
3. $\mathrm{pH}=9.99\simeq 10$
   

Step-by-step solution

Subpart (a) The pH of the solution.

The pKa value can be calculated as,

$\begin{array}{l}{\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}\\ {\mathrm{pK}}_{\mathrm{a}}=-\log (1.6\times {10}^{-10})\\ {\mathrm{pK}}_{\mathrm{a}}=9.8\end{array}$

The pH can be calculated using the Henderson-Hasselbalch equation.

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}\\ \mathrm{pH}=9.8+\log \frac{[0.8\mathrm{M}]}{[0.85\mathrm{M}]}\end{array}$

$\begin{array}{l}\mathrm{pH}=9.8+\log (0.94)\\ \mathrm{pH}=9.8-0.03\\ \mathrm{pH}=9.77\end{array}$

Subpart (b) The pH after 0.10 mole of HCl has been added to the original solution

The relation between the strong acid and base can be given as,

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{aq}\right)$

The initial concentration can be denoted as,

$\begin{array}{l}\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}\right]=0.8\mathrm{M}\\ \left[{\mathrm{H}}^{+}\right]=\frac{0.1\mathrm{mol}}{1\mathrm{L}}=0.1\mathrm{M}\end{array}$

The pH can be calculated using the Henderson-Hasselbalch equation.

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}\\ \mathrm{pH}=9.8+\log \frac{\left[0.8\mathrm{M}\right]}{\left[0.85\mathrm{M}\right]}\end{array}$

$\begin{array}{l}\mathrm{pH}=9.8+\log (0.94)\\ \mathrm{pH}=9.8-0.03\\ \mathrm{pH}=9.67\end{array}$

Subpart (c) The pH after 0.20 mole of NaOH has been added to the original buffer solution. 

The reaction between strong base and acid can be denoted as,

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}$

The initial concentration can be denoted as,

$\begin{array}{l}\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\right]=0.85\mathrm{M}\\ \left[{\mathrm{OH}}^{-}\right]=\frac{0.2\mathrm{mol}}{1\mathrm{L}}=0.2\mathrm{M}\end{array}$

The pH can be calculated using Henderson-Hasselbalch equation.

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}\\ \mathrm{pH}=9.8+\log \frac{\left[1\mathrm{M}\right]}{\left[0.65\mathrm{M}\right]}\end{array}$

$\begin{array}{l}\mathrm{pH}=9.8+\log \left(1.54\right)\\ \mathrm{pH}=9.8+0.19\\ \mathrm{pH}=9.99\simeq 10\end{array}$