(a) Using activity coefficients, calculate the pH of a solution containing a 2.00:1.00 mole ratio of ${\mathbf{HC}}^{\mathbf{2}}\mathbf{:}{\mathbf{C}}^{\mathbf{3}\mathbf{-}}$( ${\mathbf{H}}_{\mathbf{3}}\mathbf{C}$= citric acid). The ionic strength is 0.010 M.

(b) What will be the pH if the ionic strength is raised to 0.10 M and the mole ratio ${\mathbf{HC}}^{\mathbf{-}\mathbf{2}}\mathbf{:}{\mathbf{C}}^{\mathbf{3}\mathbf{-}}$ is kept constant?

Citric acid ($\mathrm{HOC}{\left({\mathrm{CH}}_2{\mathrm{CO}}_2\mathrm{H}\right)}_2$ ) is one type of weak organic acid. It is colourless in nature.

In this task we will use activity coefficient in order to calculate the pH of solution that contains a 2:1 mole ratio of (=citric acid). When:

(a)The ionic strength is 0.01M

(b)The ionic strength is 0.1M

First look at the table and find the following values considering that $\mu$=0.01M:

$\mathrm{\gamma }\left({\mathrm{C}}^{3-}\right)=0.405\mathrm{\gamma }\left({\mathrm{HC}}^{2-}\right)=0.665$

Henderson-Hasselbach equation and calculate the pH of a solution:

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ \mathrm{pH}={\mathrm{pK}}_3+\log \left(\left[{\mathrm{C}}^{3-}\right]\mathrm{\gamma }\left({\mathrm{C}}^{3-}\right)/\left[{\mathrm{HC}}^{2-}\right]\mathrm{\gamma }\left({\mathrm{HC}}^{2-}\right)\right) \\ \mathrm{pH}=6.396+\log \left(\frac{1\times 0.405}{2\times 0.665}\right) \end{gathered}$$

First look at the table and find the following values considering that $\mu$=0.1M:

$\mathrm{\gamma }\left({\mathrm{C}}^{3-}\right)=0.115\mathrm{\gamma }\left({\mathrm{HC}}^{2-}\right)=0.370$

Henderson-Hasselbach equation and calculate the pH of a solution:

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ \mathrm{pH}={\mathrm{pK}}_3+\log \left(\left[{\mathrm{C}}^{3-}\right]\mathrm{\gamma }\left({\mathrm{C}}^{3-}\right)/\left[{\mathrm{HC}}^{2-}\right]\right) \\ \mathrm{pH}=6.396+\log \left(\frac{1\times 0.115}{2\times 0.370}\right) \\ \mathrm{pH}=5.59 \end{gathered}$$