Chemistry, 12.08.2019 21:20 edfrank6278

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 0.25 moles of the sodium salt of aspirin. what will the % dissociation be in this new buffered solution?

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Answer from: jessnolonger

Explanation:

The given data is as follows.

Initial volume of aspirin = 1.00 L

Initial concentration of aspirin = 0.500 M

Initial pH = 1.86

Hence, hydronium ion concentration will be calculated as follows.

pH = $-log[H^{+}]$

or, $[H^{+}] = 10^{-pH}$

= $10^{-1.86}$

= $13.8 \times 10^{-3} M$

As, buffer containing aspirin and its conjugate base is formed in 1 L solution.

Hence, moles of aspirin are 0.500 moles and moles of the conjugate base are 0.25 moles.

So, upon dissociation the concentration of acetylsalicylate ion and hydronium ion are the same.

Hence, $pK_{a}$ value will be calculated as follows.

$K_{a}$ = $\frac{[H^{+}]^{2}}{[Aspirin]}$

= $\frac{[13.8 \times 10^{-3}]^{2}}{0.50}$

= $3.8 \times 10^{-4}$

Also, $pK_{a} = -log [K_{a}]$

= $-log [3.8 \times 10^{-4}]$

= 3.41

Now, using Henderson-Hasselbalch equation we determine the pH as follows.

pH = $pK_{a} + log \frac{[CH_{3}COO^{-}]}{[Aspirin]}$

= 3.41 + $log \frac{[0.25]}{[0.5]}$

= 3.11

Determine the hydronium ion concentration of buffer as follows.

$[H^{+}] = 10^{-pH}$

= $10^{-3.11}$

= $7.76 \times 10^{-4}$ M

Therefore, we calculate the recent dissociation as follows.

% dissociation = $\frac{\text{Amount dissociated}}{\text{Initial concentration}} \times 100%$

= $\frac{0.000776}{0.50 M} \times 100%$

= $1.5 \times 10^{-3}$ × 100

= 0.15%

Thus, we can conclude that the value of % dissociation of new buffered solution is 0.15%.

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In your lab you are studying aspirin, and its acid/base properties. you find that a 1.00 l of a 0.50...

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