# In your lab you are studying aspirin and its acid/base properties. You find that a 1.00 L of a...

## Question:

In your lab, you are studying aspirin and its acid/base properties. You find that a 1.00 L of a 0.500 M solution of aspirin has a pH of 1.86. You are interested in learning about the % dissociation in a buffered solution of aspirin, so you make a new 1.00 L solution containing 0.500 moles of aspirin and 0.45 moles of the sodium salt of aspirin. What will the % dissociation be in this new buffered solution? Give your answer to three decimal places.

## Henderson-Hasselbalch Equation:

The Henderson-Hasselbalch equation is a useful equation to estimate the pH of a buffer solution by using the {eq}pK_a{/eq} of the weak acid and the concentrations of the weak acid and conjugate base present in the solution.

{eq}pH = pK_a + log \frac {[A^-]}{[HA]}{/eq}

First, we will use the first experiment to calculate the {eq}pK_a{/eq} of aspirin. Using 'HA' to stand for aspirin, we can write its dissociation as:

{eq}HA \rightleftharpoons H^+ + A^-{/eq}

A 0.500 M solution has a pH of 1.86, which means the proton concentration is:

{eq}[H^+] = 10^{-1.86} =0.138 \ M{/eq}

This means the {eq}K_a{/eq} can be expressed as:

{eq}K_a = \dfrac { [H^+][A^- ]}{ [HA] } = \dfrac {(0.138)^2}{(0.500-0.138)} = 0.0526{/eq}

Therefore, {eq}pK_a{/eq} is calculated as follows:

{eq}pK_a = -\log K_a = -\log (0.0526) = 1.279{/eq}

Now, we can apply the {eq}pK_a{/eq} value to the second experiment to calculate the pH of the buffer made:

{eq}pH = pK_a + log \frac {[A^-]}{[HA]}\\ pH = 1.279 + log \frac {0.45}{0.500}=1.233{/eq}

Again, from the pH we can calculate {eq}[H^+]{/eq}:

{eq}[H^+] = 10^{-1.233} =0.05848 \ M{/eq}

Now, we can calculate the percent dissociation of aspirin in the buffer solution:

{eq}\% \ dissociation = \dfrac { [dissociated \ acid] }{ [original \ acid] } \times 100 = \dfrac { 0.05848 \ M }{ 0.500 \ M } \times 100 = \boxed {11.696 \%}{/eq}