Chapter 4, Problem 4.81P

Interpretation Introduction

Interpretation: Concentration of reaction of monoprotic acid $\text{HA}$ and $\text{HB}$ with $\text{NaOH}$ is to be calculated.

Concept introduction: Concentration of strong base or strong acid is the amount of solute that is dissolved in the solvent. Solute is the component that is present in less quantity in the solution whereas solvent is present in large amount.

Concentration of solution is expressed in terms of molarity. It is the ratio of moles of solute to the volume of solution in litres. The expression used to determine the concentration of solution is as follows:

  $\text{Concentration of solution}=\frac{\text{moles of solute}}{\text{volume of solution}\left(\text{L}\right)}$

Answer to Problem 4.81P

Molarity of monoprotic acid $\text{HA}$ is $0.182M$ and molarity of monoprotic acid $\text{HB}$ is $0.154M$ .

Explanation of Solution

Given Information:

The volume of $\text{HA}$ solution in flask 1 is $43.5\text{ mL}$ , that of in flask 2 is $37.2\text{ mL}$ and final volume in flask 2 is $50.0\text{ mL}$ . Volume of $\text{NaOH}$ solution that is used to titrate $\text{HA}$ solution is $87.3\text{ mL}$ and molarity is $0.0906M$ . Volume of mixture in flask 2 is

  $96.4\text{ mL}$ .

  $\text{HA}$ and $\text{HB}$ on reaction with $\text{NaOH}$ forms $\text{NaA}$ and $\text{NaB}$ respectively. The chemical equation of the reaction is as follows:

  $\begin{array}{l}\text{HA}\left(aq\right)+\text{NaOH}\left(aq\right)\to \text{NaA}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \text{HB}\left(aq\right)+\text{NaOH}\left(aq\right)\to \text{NaB}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

The formula used to calculate the moles of $\text{NaOH}$ in flask 1 is as follows:

  $\text{Moles}\text{of}\text{NaOH}=\left(\text{molarity of NaOH}\left(\text{mol/L}\right)\right)\left(\text{Volume of}\text{NaOH}\left(\text{L}\right)\right)$

The molarity of $\text{NaOH}$ is $0.0906\text{mol/L}$ .

The volume of $\text{NaOH}$ is $87.3\text{mL}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{NaOH}=\left(\text{molarity of NaOH}\left(\text{mol/L}\right)\right)\left(\text{Volume of}\text{NaOH}\left(\text{L}\right)\right)\\ =\left(0.0906\text{mol/L}\right)\left(87.3\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.007909\text{mol}\end{array}$

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

The formula used to calculate the moles of $\text{HA}$ in flask 1 is as follows:

  $\text{Moles}\text{of}\text{HA}=\left(\text{moles}\text{of NaOH}\right)\left(\frac{\text{1 mol}\text{HA}}{\text{1 mol}\text{NaOH}}\right)$

The moles of $\text{NaOH}$ is $0.007909\text{mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{HA}=\left(\text{moles}\text{of NaOH}\right)\left(\frac{\text{1 mol}\text{HA}}{\text{1 mol}\text{NaOH}}\right)\\ =\left(0.007909\text{mol}\right)\left(\frac{\text{1 mol}\text{HA}}{\text{1 mol}\text{NaOH}}\right)\\ =0.007909\text{mol}\end{array}$

The formula used to calculate the molarity of monoprotic acid $\text{HA}$ is as follows:

  $\text{Molarity}\text{of}\text{HA}=\left(\frac{\text{moles}\text{of HA}}{\text{volume of HA solution}\left(\text{L}\right)}\right)$

The moles of $\text{HA}$ is $0.007909\text{mol}$ .

The volume of $\text{HA}$ solution is $43.5\text{mL}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Molarity}\text{of}\text{HA}=\left(\frac{\text{moles}\text{of HA}}{\text{volume of HA solution}\left(\text{L}\right)}\right)\\ =\left(\frac{0.007909\text{mol}}{43.5\text{mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =0.1818248M\\ \approx 0.182M\end{array}$

The formula used to calculate the moles of $\text{HA}$ in the flask 2 of mixture is as follows:

  $\text{Moles}\text{of}\text{HA}=\left(\text{molarity of HA}\right)\left(\text{Volume of}\text{HA}\left(\text{L}\right)\right)$

The molarity of $\text{HA}$ is $0.1818248\text{mol/L}$ .

The volume of $\text{HA}$ acid in flask 2 is $37.2\text{mL}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{HA}=\left(\text{molarity of HA}\right)\left(\text{Volume of}\text{HA}\left(\text{L}\right)\right)\\ =\left(0.1818248\text{mol/L}\right)\left(37.2\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.0067639\text{mol}\end{array}$

To calculate the volume of $\text{NaOH}$ that is used for the titration of $\text{HA}$ in flask 2, the formula used is as follows:

  $\text{Volume}\text{of}\text{NaOH used to titrate HA}=\left(\frac{\text{moles of HA}}{\text{molarity of NaOH}}\right)\left(\frac{1\text{mol}\text{NaOH}}{1\text{mol}\text{HA}}\right)$

The moles of $\text{HA}$ is $0.0067639\text{mol}$ .

The molarity of $\text{NaOH}$ is $0.0906\text{mol/L}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Volume}\text{of}\text{NaOH used to titrate HA}=\left(\frac{\text{moles of HA}}{\text{molarity of NaOH}}\right)\left(\frac{1\text{mol}\text{NaOH}}{1\text{mol}\text{HA}}\right)\\ =\left(\frac{0.0067639\text{mol}}{0.0906\text{mol/L}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\left(\frac{1\text{mol}\text{NaOH}}{1\text{mol}\text{HA}}\right)\\ =0.07465673\text{L}\left(\frac{1000\text{mL}}{1\text{L}}\right)\left(\frac{1\text{mol}\text{NaOH}}{1\text{mol}\text{HA}}\right)\\ =74.6565\text{mL}\end{array}$

To calculate the volume of $\text{NaOH}$ that is used for the titration of $\text{HB}$ in flask 2, the formula used is as follows:

  $\text{Volume}\text{of}\text{NaOH used to titrate HB}=\left(\begin{array}{l}\text{volume of mixture}-\\ \text{volume}\text{of}\text{NaOH used to titrate HA}\end{array}\right)$

The volume of mixture is $96.4\text{mL}$ .

The volume of $\text{NaOH}$ used to titrate $\text{HA}$ is $74.6565\text{mL}$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Volume}\text{of}\text{NaOH used to titrate HB}=\left(\begin{array}{l}\text{volume of mixture}-\\ \text{volume}\text{of}\text{NaOH used to titrate HA}\end{array}\right)\\ =\left(96.4\text{mL}-74.6565\text{mL}\right)\\ =21.7435\text{mL}\end{array}$

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

The formula used to calculate the moles of $\text{HB}$ is as follows:

  $\text{Moles}\text{of}\text{HB}=\left[\begin{array}{l}\left(\text{molarity of NaOH }\right)\\ \left(\text{volume of}\text{NaOH used to titrate HB}\right)\\ \left(\frac{\text{1 mol}\text{HB}}{\text{1 mol}\text{NaOH}}\right)\end{array}\right]$

The molarity of $\text{NaOH}$ is $0.0906\text{mol/L}$ .

The volume of $\text{NaOH}$ used to titrate $\text{HB}$ is $21.7435\text{mL}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{HB}=\left[\begin{array}{l}\left(\text{molarity of NaOH }\right)\\ \left(\text{volume of}\text{NaOH used to titrate HB}\right)\\ \left(\frac{\text{1 mol}\text{HB}}{\text{1 mol}\text{NaOH}}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.0906\text{mol/L}\right)\left(21.7435\text{mL}\right)\\ \left(\frac{\text{1 mol}\text{HB}}{\text{1 mol}\text{NaOH}}\right)\end{array}\right]\\ =0.00196996\text{mol}\end{array}$

To calculate the volume of $\text{HB}$ in the mixture, the formula used is as follows:

  $\text{Volume}\text{of}\text{HB}=\left(\text{volume of mixture}-\text{volume}\text{of}\text{HA}\right)$

The volume of mixture is $50.0\text{mL}$ .

The volume of HA is $37.2\text{mL}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Volume}\text{of}\text{HB}=\left(\text{volume of mixture}-\text{volume}\text{of}\text{HA}\right)\\ =\left(50.0\text{mL}-37.2\text{mL}\right)\\ =12.8\text{mL}\end{array}$

The formula used to calculate the molarity of $\text{HB}$ is as follows:

  $\text{Molarity}\text{of}\text{HB}=\left(\frac{\text{moles}\text{of HB}}{\text{volume of HB}\left(\text{L}\right)}\right)$

The moles of $\text{HB}$ is $0.00196996\text{mol}$ .

The volume of $\text{HB}$ is $12.8\text{mL}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Molarity}\text{of}\text{HB}=\left(\frac{\text{moles}\text{of HB}}{\text{volume of HB}\left(\text{L}\right)}\right)\\ =\left(\frac{0.00196996\text{mol}}{12.8\text{mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =0.153903M\\ \approx 0.154M\end{array}$

Conclusion

Molarity of monoprotic acid $\text{HA}$ is $0.182M$ and molarity of monoprotic acid $\text{HB}$ is $0.154M$ .