Question

Consider the titration of $\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{ml}$of a solution that contains a mixture of$\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{M}$${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$and $\mathbf{0}\mathbf{.}\mathbf{20}$${\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{6}}$. Calculate the pH

1. before any$\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{M}\mathbf{NaOH}$has been added.
2. after a total of $\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{ml}$of$\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{M}\mathbf{NaOH}$has been added.
3. after a total of$\mathbf{300}\mathbf{.}\mathbf{0}\mathbf{ml}$of $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{M}\mathbf{NaOH}$has been added.
4. after a total of $\mathbf{500}\mathbf{.}\mathbf{0}\mathbf{ml}$of $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{M}\mathbf{NaOH}$ has been added.

Short answer

1. The pH of the solution of a is 1.36.
2. The pH of the solution b is 1.52.
3. The pH of the solution c is 2.01.
4. The pH of the solution d is 11.683.

Step-by-step solution

Subpart (a) The solution of before .

According to the ICE dissociation of${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{SO}}_4^{2-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HSO}}_4^{2-}\right]}$

$1.2\times {10}^{-2}=\frac{\left(\mathrm{x}\right)\left(0.05+\mathrm{X}\right)}{0.05+\mathrm{X}}$

$\mathrm{X}=0.00722$

At equilibrium of$0.05-0.00722=0.04278$

Before adding$0.1\mathrm{M}\mathrm{NaOH}$

$\mathrm{pH}=-\log (0.04278)$

$\mathrm{pH}=1.36$

Subpart (b) The solution of after 100.0ml of 0.10M  NaOH.

Assume that 1L of acids be titrated with 100mL of NaOH.

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{acid}=0.04278\times 1\\ =0.04278\end{array}$

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{NaOH}=0.10\times 100\times {10}^{-3}\\ =0.01\mathrm{mol}\end{array}$

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{H}}_3{\mathrm{O}}^{+}\mathrm{ion}\mathrm{in}\mathrm{solution}=0.04278-1.01\\ =0.03278\end{array}$

$\begin{array}{l}\mathrm{Total}\text{\hspace{0.17em}}\mathrm{volume}\text{\hspace{0.17em}}\mathrm{of}\mathrm{solution}=1000\mathrm{ml}+100\mathrm{ml}\\ =1100\mathrm{ml}\\ =1.1\mathrm{L}\end{array}$

The Concentration of${\mathrm{H}}_3{\mathrm{O}}^{+}\mathrm{ion}=\frac{0.03278}{1.1\mathrm{L}}$

$=0.0298\mathrm{M}$

$\begin{array}{l}\mathrm{pH}=-\log (0.0298)\\ =1.52\end{array}$

Subpart (c) The solution of after 300.0ml of 0.10M  NaOH.

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{NaOH}=0.10\times 200\times {10}^{-3}\\ =0.02\mathrm{mol}\end{array}$

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{H}}_3{\mathrm{O}}^{+}\mathrm{ion}=0.03278\mathrm{mol}-0.02\mathrm{mol}\\ =0.01278\mathrm{mol}\end{array}$

The Concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}\mathrm{ion}=\frac{0.01278}{1.3\mathrm{L}}$

$\begin{array}{l}\mathrm{pH}=-\log (0.0983)\\ =2.01\end{array}$

Subpart (d) The solution of after 500.0ml of 0.10M  NaOH.

Here, the ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ions are completely consumed

${\text{n}}_{{\text{OH}}^{\text{-}}}{\text{-n}}_{\text{H}}\text{=\hspace{0.17em}0.00722mol}$

$\begin{array}{l}\mathrm{Total}\text{\hspace{0.17em}}\mathrm{volume}\text{\hspace{0.17em}}\mathrm{of}\mathrm{the}\mathrm{solution}=1300\mathrm{mL}+200\mathrm{mL}\\ =1500\mathrm{mL}\\ =1.5\mathrm{L}\end{array}$

The concentration of

$\begin{array}{c}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{ion\hspace{0.33em}=\hspace{0.33em}}\frac{\text{0.00722}}{\text{1.5L}}\\ =0.00481\mathrm{M}\end{array}$

The pH is

$\begin{array}{l}\mathrm{pH}=14-(-\log (0.0983\left)\right)\\ =14-2.317\\ =11.683\end{array}$