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

Answer 1

Step-by-step 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 * 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)` =
`([H^(+)]^(2))/([Aspirin])`

=
`([13.8 * 10^(-3)]^(2))/(0.50)`

=
`3.8 * 10^(-4)`

Also,
`pK_(a) = -log [K_(a)]`

=
`-log [3.8 * 10^(-4)]`

= 3.41

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

pH =
`pK_(a) + log ([CH_(3)COO^(-)])/([Aspirin])`

= 3.41 +
`log ([0.25])/([0.5])`

= 3.11

Determine the hydronium ion concentration of buffer as follows.

`[H^(+)] = 10^(-pH)`

=
`10^(-3.11)`

=
`7.76 * 10^(-4)` M

Therefore, we calculate the recent dissociation as follows.

% dissociation =
`\frac{\text{Amount dissociated}}{\text{Initial concentration}} * 100%`

=
`(0.000776)/(0.50 M) * 100%`

=
`1.5 * 10^(-3)` × 100

= 0.15%

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