Question

The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required 27.36 mL of 0.5106 M NaOH for complete reaction. Addition of 15.44 mL of 0.4524 M HCl to the flask containing the aspirin and the sodium hydroxide produced a mixture with $\mathbf{\text{pH\hspace{0.17em}=\hspace{0.17em}3.48}}$. Find the molar mass of acetylsalicylic acid and its${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$value. Acetylsalicylic acid is a monoprotic acid.

Short answer

The molar mass of acetylsalicylic acid is$\mathbf{179}\mathbf{.7}\mathbf{\text{\hspace{0.17em}g/mol}}$ The value of${\mathrm{K}}_{\mathrm{a}}$ is$\mathbf{3.3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{.}$

Step-by-step solution

The molar mass of H+ added.

The reaction of acetylsalicylic acid can be denoted as,

$\mathrm{HA}+{\mathrm{OH}}^{-}\to {\mathrm{A}}^{-}+{\mathrm{H}}_2\mathrm{O}$

The mole number of HA is,

$\begin{array}{c}\mathrm{mmol}\mathrm{HA}\mathrm{present}=27.36\mathrm{mL}\mathrm{OH}\times \frac{05106\mathrm{mmol}\mathrm{OH}}{\mathrm{mL}\mathrm{OH}}\times \frac{1\mathrm{mmol}\mathrm{HA}}{\mathrm{mmol}\mathrm{OH}}\\ =13.97\mathrm{mmol}\end{array}$

The molar mass of HA is calculated as,

$\begin{array}{c}\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{HA}=\frac{\mathrm{mass}\mathrm{of}\mathrm{HA}}{\mathrm{mol}\mathrm{HA}}\\ =\frac{2.51\mathrm{g}\mathrm{HA}}{13.97\times {10}^2\mathrm{mol}\mathrm{HA}}\\ =179.7\mathrm{g}/\mathrm{mol}\simeq 180\mathrm{g}/\mathrm{mol}\end{array}$

Therefore,

$\begin{array}{c}\mathrm{mmol}{\mathrm{H}}^{+}\mathrm{added}=15.44\mathrm{mL}\times \frac{0.4524\mathrm{mmol}{\mathrm{H}}^{+}}{\mathrm{mL}}\\ =6.985\mathrm{mmol}{\mathrm{H}}^{+}\end{array}$

The value of Ka.

The value can be found using Henderson-Hasselbalch equation,

$\begin{array}{c}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}\\ 3.48={\mathrm{pK}}_{\mathrm{a}}+\log \frac{6.985}{6.985}\\ 3.48={\mathrm{pK}}_{\mathrm{a}}+\log 1\end{array}$

Hence,