Question

Question #4d24c

Answer 1

Here's what I got.

Explanation:

Your tool of choice here will be the Henderson - Hasselbalch equation, which for a weak acid-conjugate base buffer looks like this

`"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))`

Before doing any calculation, take a look at the `"pH"` of the solution, Notice that you have

`"pH " > " p"K_a`

Right from the start, this should tell you that the solution contains more conjugate base than weak acid. In other words, the buffer will contain a higher concentration of acetate anions, `"CH"_3"COO"^(-)`, the conjugate base of acetic acid, then acetic acid.

This implies that

`"moles of CH"_3"COO"^(-)" " > " " "0.01 moles"`

Now, let's assume that the starting solution has a volume of `V` `"L"`. This implies that the molarity of the acetic acid is equal to

`["CH"_3"COOH"] = "0.01 moles"/(Vcolor(white)(.)"L") = (0.01/V)color(white)(.)"mol L"^(-1)`

At this point, you can assume that the volume of the buffer does not change upon the addition of the salt, which dissociates to produce acetate anions in a `1:1` mole ratio.

If you take `x` moles to be the number of moles of sodium acetate that you must add to the buffer, you can say that the concentration of the acetate anions will be

`["CH"_3"COO"^(-)] = (x color(white)(.)"moles")/(Vcolor(white)(.)"L") = (x/V)color(white)(.)"mol L"^(-1)`

Plug this into the Henderson - Hasselbalch equation to get

`5.27 = 4.74 + log [ ((x/color(red)(cancel(color(black)(V))))color(red)(cancel(color(black)("mol L"^(-1)))))/( (0.01/color(red)(cancel(color(black)(V))))color(red)(cancel(color(black)("mol L"^(-1)))))]`

This wil get you

`log(x/0.01) = 5.27 - 4.74`

which is equivalent to

`10^log(x/0.01) = 10^0.53`

`x/0.01 = 3.39 implies x= 0.0339`

You can thus say that you need to dissolve

`color(darkgreen)(ul(color(black)("moles of CH"_3"COONa" = "0.034 moles")))`

in your initial acetic acid solution in order to create a buffer that has `"pH" = 5`.

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the number of moles of acetic acid persent in the solution.

SIDE NOTE Even if you don't assume that the volume of the solution does not change upon the addition of the sodium acetate, you can use the fact that the volume of the solution affects both species.

So if you add the salt and go

`V -> V^'`

you can say that

`["CH"_3"COOH"] = (0.01/V^')color(white)(.)"mol L"^(-1)`

`["CH"_3"COO"^(-)] = (x/V^')color(white)(.)"mol L"^(-1)`

which will get you the same result.