Chapter 4.4, Problem 4.15BFP

Interpretation Introduction

Interpretation:

The mass of acetylsalicylic acid that is needed to react with $14.10\text{mL}$ of $0.128M\text{ NaOH}$ is to be calculated.

Concept introduction:

Strong acids and strong bases are the substance that dissociates completely into its ions when dissolved in the solution. They dissociate completely in water to release ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions.

Weak acids and weak bases are the substance that does not dissociate completely into its ions when dissolved in the solution. They dissociate partially in water to release ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions.

Acetylsalicylic acid $\left({\text{HC}}_9{\text{H}}_7{\text{O}}_4\right)$ is a weak acid and sodium hydroxide $\left(\text{NaOH}\right)$ is a strong base. Sodium hydroxide $\left(\text{NaOH}\right)$ dissociates completely into its respective ions. Acetylsalicylic acid $\left({\text{HC}}_9{\text{H}}_7{\text{O}}_4\right)$ dissociates to some extent and gives only one ${\text{H}}^{+}$ for the reaction. They both react to form ${\text{NaC}}_9{\text{H}}_7{\text{O}}_4$ and a water molecule.

The molecular equation for the acid-base reaction of acetylsalicylic acid and sodium hydroxide is:

${\text{HC}}_9{\text{H}}_7{\text{O}}_4\left(aq\right)+\text{NaOH}\left(aq\right)\to {\text{NaC}}_9{\text{H}}_7{\text{O}}_4\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

Answer to Problem 4.15BFP

The mass of acetylsalicylic acid that is needed to react with $14.10\text{mL}$ of $0.128M\text{ NaOH}$ is $0.325\text{g}$.

Explanation of Solution

The expression to calculate moles of $\text{NaOH}$ is:

$\text{moles}\text{of}\text{NaOH}=\left[\text{volume of NaOH}\left(\text{L}\right)\left(\text{molarity}\text{of}\text{NaOH}\left(\text{mol/L}\right)\right)\right]$ (1)

Substitute $14.10\text{mL}$ for the volume of $\text{NaOH}$ and $0.128\text{ mol/L}$ for molarity of $\text{NaOH}$ in the equation (1).

$\begin{array}{c}\text{moles}\text{of}\text{NaOH}=\left[14.10\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(0.128\text{ mol/L}\right)\right]\\ =0.0018048\text{mol}\end{array}$

One mole of $\text{NaOH}$ reacts with one mole of ${\text{HC}}_9{\text{H}}_7{\text{O}}_4$ to form one mole of ${\text{NaC}}_9{\text{H}}_7{\text{O}}_4$.

The expression to calculate moles of ${\text{HC}}_9{\text{H}}_7{\text{O}}_4$ is:

$\text{moles}\text{of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4=\left[\text{moles of NaOH}\left(\text{mol}\right)\left(\frac{1{\text{mol HC}}_9{\text{H}}_7{\text{O}}_4}{1\text{mol}\text{NaOH}}\right)\right]$ (2)

Substitute $0.0018048\text{mol}$ for moles of $\text{NaOH}$ in the equation (2).

$\begin{array}{c}\text{moles}\text{of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4=\left[0.0018048\text{mol NaOH}\left(\frac{1{\text{mol HC}}_9{\text{H}}_7{\text{O}}_4}{1\text{mol}\text{NaOH}}\right)\right]\\ =0.0018048\text{mol}\end{array}$

The molecular mass of ${\text{HC}}_9{\text{H}}_7{\text{O}}_4$ is $180.15\text{g/mol}$.

The expression to calculate the mass of ${\text{HC}}_9{\text{H}}_7{\text{O}}_4$ is:

$\text{mass}\text{of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4=\left[\begin{array}{c}\text{moles of }{\text{HC}}_9{\text{H}}_7{\text{O}}_4\left(\text{mol}\right)\\ \left(\frac{\text{molecular}\text{mass}\text{of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4\left(\text{g}\right)}{1\text{mol}\text{of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4}\right)\end{array}\right]$ (3)

Substitute $0.0018048\text{mol}$ for moles of ${\text{HC}}_9{\text{H}}_7{\text{O}}_4$ and $180.15\text{g}$ for the molecular mass of ${\text{HC}}_9{\text{H}}_7{\text{O}}_4$ in the equation (3).

$\begin{array}{c}\text{mass}\text{of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4=\left(0.0018048\text{mol of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4\right)\left(\frac{180.15\text{g of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4}{1\text{mol}\text{of}{\text{HC}}_9{\text{H}}_7{\text{O}}_4}\right)\\ =0.3251\text{g}\\ \approx 0.325\text{g}\end{array}$

Conclusion

The mass of acetylsalicylic acid that is needed to react with $14.10\text{mL}$ of $0.128M\text{ NaOH}$ is $0.325\text{g}$.