Question

You need to produce a buffer solution that has pH 5.06. You already have a solution that contains 10 mmols of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?

The pKa of acetic acid is 4.74.

Answer 1

`"21 mmoles"`

Explanation:

Your strategy here will be to use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentrations of the weak acid and of the conjugate in order to have a solution that has a pH of `5.06`.

The Henderson - Hasselbalch equation looks like this

`color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))`

Your buffer solution contains acetic acid, `"CH"_3"COOH"`, a weak acid, and the acetate anion, `"CH"_3"COO"^(-)`, its conjugate base. Moreover, you know that the `pK_a` of acetic acid is equal to `4.74`.

This means that you have

`5.06 = 4.74 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))`

Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need relative to the number of mmoles of acetic acid.

Notice that the pH of the target solution is higher than the `pK_a` of acetic acid

`5.06 > pK_a`

This tells you that the target solution must contain more conjugate base than weak acid. Implicitly, this solution must contain more moles of acetate anions than of acetic acid.

Rearrange the H - H equation to isolate the log term on one side

`0.32 = log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))`

This will be equivalent to

`10^0.32 = 10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))`

which will give you

`(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 10^0.32 = 2.09`

As predicted, the concentration of the conjugate base must be higher than that of the weak acid

`color(green)(|bar(ul(color(white)(a/a)color(black)(["CH"_3"COO"^(-)] = 2.09 xx ["CH"_3"COOH"])color(white)(a/a)|)))`

Now, the volume of the solution is the same for both chemical species, let's say `V`, which means that you'll have

`n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))`

Since the solution is said to contain `10` mmoles of acetic acid, it follows that you must add

`n_(CH_3COO^(-)) = 2.09 xx "10 mmoles" = color(green)(|bar(ul(color(white)(a/a)"21 mmoles CH"_3"COO"^(-)color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs.