Question

You need to produce a buffer solution that has a pH of 5.12. You already have a solution that contains 10 mmol (millimoles) 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

`"24 mmol CH"_3"COO"^(-)`

Explanation:

Your buffer solution contains acetic acid, `"CH"_3"COOH"`, a weak acid, and the acetate anion, `"CH"_3"COO"^(-)`, its conjugate base.

You can find the buffer's pH by using the Henderson - Hasselbalch equation, which looks like this

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

Notice that the pH of the solution depends on the ratio that exists between the concentration of the weak acid and the concentration of the conjugate base.

Now, a solution's molarity essentially tells you how many moles of solute you get in one liter of solution.

`color(blue)(c = n_"solute"/V_"solution"`

Since the volume of the solution is the same for both chemical species, the ratio that exist between their concentrations will be equivalent to the ratio that exists between their number of moles.

If you take `V` to be the volume of the buffer solution, you can say that

`["CH"_3"COOH"] = n_text(acetic acid)/V" "` and `" "["CH"_3"COO"^(-)] = n_text(acetate)/V`

This means that you will have

`(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = n_text(acetate)/color(red)(cancel(color(black)(V))) * color(red)(cancel(color(black)(V)))/n_"acetic acid" = n_"acetate"/n_"acetic acid"`

The H- H equation becomes

`"pH" = pK_a + log( n_"acetate"/n_"acetic acid")`

Now, you need the pH of the solution to be equal to `5.12`. Right from the start, you can look at the `pK_a` of the acid and say that since you need

`"pH" > pK_a`

the buffer will contain more moles of conjugate base than of weak acid.

More specifically, you will have

`5.12 = 4.74 + log( n_"acetate"/n_"acetic acid")`

`log(n_"acetate"/n_"acetic acid") = 0.38`

This will be equivalent to

`10^log( n_"acetate"/n_"acetic acid") = 10^0.38`

which will give you

`n_"acetate"/n_"acetic acid" = 2.40`

This means that the solution must contain

`n_"acetate"= 2.40 * "10 mmol" = color(green)(|bar(ul(color(white)(a/a)"24 mmol"color(white)(a/a)|)))`

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