Chapter 4, Problem 4.46P

An unknown amount of acid can often be determined by adding an excess of base and then “back-titrating” the excess. A 0.3471-g sample of a mixture of oxalic acid, which has two ionizabic protons, and bcnioic acid, which has one, is treated with $\text{100}\text{.0 mL of 0}\text{.1000 M NaOH}\text{.}$ The excess NaOH is titrated with $\text{20}\text{.00 mL of 0}\text{.2000 M HCI}\text{.}$ Find the mass % of benzoic acid.

Interpretation Introduction

Interpretation: Mass percent of benzoic acid $\left({\text{HC}}_7{\text{H}}_5{\text{O}}_2\right)$ in mixture is to be calculated.

Concept introduction: Mass percent is one of the most commonly used concentration terms to determine concentration of any species. The expression for mass percent of any species present in sample is as follows:

  $\text{Mass percent}=\left(\frac{\text{Mass of species}}{\text{Mass of sample}}\right)\left(100\text{ \%}\right)$

Answer to Problem 4.46P

Mass percent of benzoic acid in mixture is $35.13\text{ \%}$ .

Explanation of Solution

Below mentioned chemical reactions take place in given circumstances.

  $\begin{array}{l}\text{NaOH}\left(aq\right)+\text{HCl}\left(aq\right)\to \text{NaCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \text{2NaOH}\left(aq\right)+{\text{H}}_2{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(aq\right)\to {\text{NaC}}_2{\text{O}}_4\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \text{NaOH}\left(aq\right)+{\text{HC}}_{\text{7}}{\text{H}}_5{\text{O}}_{\text{2}}\left(aq\right)\to {\text{NaC}}_2{\text{O}}_4\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

The formula to calculate moles of $\text{NaOH}$ is as follows:

  $\text{Moles}\text{of}\text{NaOH}=\left(\text{Volume}\text{of}\text{NaOH}\right)\left(\text{Molarity of NaOH}\right)$

The volume of $\text{NaOH}$ is $100.0\text{ mL}$ .

The molarity of $\text{NaOH}$ is $0.1000M$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{NaOH}=\left(\text{Volume}\text{of}\text{NaOH}\right)\left(\text{Molarity of NaOH}\right)\\ =\left[\left(100.0\text{ mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(0.1000\text{mol/L}\right)\right]\\ =0.01000\text{mol}\end{array}$

The formula to calculate moles of $\text{HCl}$ is as follows:

  $\text{Moles}\text{of}\text{HCl}=\left(\text{Volume}\text{of}\text{HCl}\right)\left(\text{Molarity of HCl}\right)$

The volume of $\text{HCl}$ is $20.00\text{ mL}$ .

The molarity of $\text{HCl}$ is $0.2000M$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{HCl}=\left(\text{Volume}\text{of}\text{HCl}\right)\left(\text{Molarity of HCl}\right)\\ =\left[\left(20.00\text{ mL}\right)\left(\frac{1\text{L}}{{10}^3\text{mL}}\right)\left(0.2000M\right)\right]\\ =0.004000\text{mol}\end{array}$

One mole of $\text{NaOH}$ reacts with one mole of $\text{HCl}$ .

The excess moles of $\text{NaOH}$ can be calculated as follows:

  $\begin{array}{c}\text{Moles}\text{of}\text{excess NaOH}=\left(0.004000\text{mol}\right)\left(\frac{1\text{mol NaOH}}{1\text{mol HCl}}\right)\\ =0.004000\text{mol}\end{array}$

The formula to calculate moles of $\text{NaOH}$ requiredfor titration of sample is as follows:

  $\text{Moles of NaOH for titartion}=\left[\begin{array}{l}\left(\text{Moles of NaOH added}\right)-\\ \left(\text{Moles of excess NaOH }\right)\end{array}\right]$

The moles of $\text{NaOH}$ added are $0.01000\text{mol}$ .

The moles of $\text{NaOH}$ added are $0.004000\text{mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles of NaOH for titration}=\left[\begin{array}{l}\left(\text{Moles of NaOH added}\right)-\\ \left(\text{Moles of excess NaOH }\right)\end{array}\right]\\ =\left[\left(0.01000\text{mol}\right)-\left(0.004000\text{mol }\right)\right]\\ =0.006000\text{mol}\end{array}$

Consider $x$ to be mass of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ so mass of ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4$ becomes $\left(0.3471-x\right)$ .

One mole of $\text{NaOH}$ reacts with one mole of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ .

The molecular mass of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $122.12\text{g/mol}$

The moles of $\text{NaOH}$ used for titration of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ are calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH }=\left[\left(\frac{x{\text{ g HC}}_7{\text{H}}_5{\text{O}}_2}{122.12\text{g/mol}}\right)\left(\frac{\text{1}\text{mol}\text{NaOH }}{\text{1}\text{mol}{\text{HC}}_7{\text{H}}_5{\text{O}}_2}\right)\right]\\ =0.0081887x\end{array}$

Two moles of $\text{NaOH}$ reacts with one mole of ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4$ .

The molecular mass of ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4$ is $90.04\text{g/mol}$ .

The moles of $\text{NaOH}$ used for titration of ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4$ are calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH }=\left[\left(\frac{0.3471-x}{90.04\text{g/mol}}\right)\left(\frac{\text{2}\text{mol}\text{NaOH }}{\text{1}\text{mol}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4}\right)\right]\\ =0.0077099-0.022212x\end{array}$

The moles of $\text{NaOH}$ used for titration can be calculated as follows:

  $\text{Moles of NaOH for titration}=\left[\begin{array}{l}\left(\text{moles of NaOH used for}{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right)+\\ \left({\text{moles of NaOH used for H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\right)\end{array}\right]$

The moles of $\text{NaOH}$ used for ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ are $0.0081887x$ .

The moles of $\text{NaOH}$ used for ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ are $0.0081887x$ .

The moles of $\text{NaOH}$ are $0.006000\text{mol}$ .

Substitute the values in above equation.

  $\begin{array}{l}\text{Moles of NaOH for titration}=\left[\begin{array}{l}\left(\text{moles of NaOH used for}{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right)+\\ \left({\text{moles of NaOH used for H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\right)\end{array}\right]\\ 0.006000=\left[\left(0.0081887x\right)+\left(0.0077099-0.022212x\right)\right]\\ 0.006000=0.0077099-0.014023x\\ –0.0017099=-0.014023x\end{array}$

Solve for $x$ ,

  $x=0.12195$

The formula to calculate mass percent of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is as follows:

  $\text{Mass percent of}{\text{HC}}_7{\text{H}}_5{\text{O}}_2=\left(\frac{\text{Mass}{\text{of HC}}_7{\text{H}}_5{\text{O}}_2}{\text{Mass of sample}}\right)\left(100\text{ \%}\right)$

The mass of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $0.12195\text{ g}$ .

The molar mass of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $0.3471\text{g}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Mass percent of}{\text{HC}}_7{\text{H}}_5{\text{O}}_2=\left(\frac{\text{Mass}{\text{of HC}}_7{\text{H}}_5{\text{O}}_2}{\text{Mass of sample}}\right)\left(100\text{ \%}\right)\\ =\left(\frac{0.12195}{0.3471\text{g}}\right)\left(100\text{ \%}\right)\\ =35.1348\%\\ \approx 35.13\%\end{array}$

Conclusion

Mass percent of benzoic acid in mixture is $35.13\text{ \%}$ .