Question

Question #7f273

Answer 1

`"249 mL KOH"`

Explanation:

The idea here is that you can use the Henderson - Hasselbalch equation to figure out the ratio that must exist between the concentration of conjugate base and the concentration of weak acid in this buffer.

In your case, acetic acid, `"CH"_3"COOH"`, is the weak acid and the acetate anion, `"CH"_3"COO"^(-)`, is its conjugate base.

`color(blue)(ul(color(black)("pH" = "p"K_a + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])))))`

Now, acetic acid has

`"p"K_a = 4.76`

The buffer must have a pH of `6.48`, significantly higher than the `"p"K_a` of the acid. This tells you that the buffer will contain a lot more conjugate base than weak acid.

So right from the start, you should expect to see

`["CH"_3"COO"^(-)] > ["CH"_3"COOH"]`

which implies

`(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) > 1`

You start with `"875 mL"` of a `"0.621 M"` acetic acid solution. The number of moles of acetic acid present in the initial solution will be equal to

`875 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.621 moles CH"_3"COOH")/(1color(red)(cancel(color(black)("L"))))`

`= "0.5434 moles CH"_3"COOH"`

When you add potassium hydroxide, a strong base, the hydroxide anions delivered to the solution by the salt will react with the acid to form acetate anions.

`"CH"_ 3"COOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 2"O"_ ((l))`

Notice that it takes `1` mole of hydroxide anions, i.e. `1` mole of potassium hydroxide, to react with `1` mole of acetic acid in order to produce `1` mole of acetate anions. Keep this in mind.

Now, let's say that `x` represents the number of moles of acetic acid and `y` represents the number of moles of acetate anions present in the target solution.

If we take `V` to be the volume of potassium hydroxide solution, expressed in milliliters, added to the initial acetic acid solution, we can say that, at equilibrium, the buffer will contain

`["CH"_3"COOH"] = "x moles"/( (875 + V) * 10^(-3)"L") = ((10 ^ 3 * x)/(875 + V))color(white)(.)"M"`

`["CH"_3"COO"^(-)] = "y moles"/((875 + V) * 10^(-3)"L") = ( (10^3 * y)/(875 + V))color(white)(.)"M"`

This means that the ratio that must exist between the conjugate base and the weak acid in the buffer will be equal to

`(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = ( ( (color(red)(cancel(color(black)(10^3))) * y)/color(red)(cancel(color(black)(875 + V))))color(red)(cancel(color(black)("M"))))/( ((color(red)(cancel(color(black)(10^3))) * x)/color(red)(cancel(color(black)(875 + V))))color(red)(cancel(color(black)("M")))) = y/x`

According to the Henderson - Hasselbalch equation, you will have

`6.48 = 4.76 + log(y/x)`

This is equivalent to

`log(y/x) = 1.72`

`10^log(y/x) = 10^1.72`

which gets you

`y/x = 52.48`

Let's get back to the balanced chemical equation that describes the neutralization reaction. If we use `n` as the number of moles of hydroxide anions added to the initial solution, you can say that the buffer will contain

`x = 0.5434 - n ->` the number of moles of acetic acid

`y = 0 + n ->` the number of moles of acetate anions

You can thus say that

`y/x = n/(0.5434 - n) = 52.48`

Solve this for `n` to get

`n = 28.52 - 52.48 * n`

`n * (52.48 + 1) = 28.52 implies n= 28.52/53.48 = 0.5333`

Therefore, you must add `0.5333` moles of hydroxide anions to the initial acetic acid solution.

Use its molarity to figure out the volume that would contain this number of moles of potassium hydroxide

`0.5333 color(red)(cancel(color(black)("moles KOH"))) * "1 L solution"/(2.14 color(red)(cancel(color(black)("moles KOH")))) = "0.2492 L solution"`

Expressed in milliliters and rounded to three sig figs, the answer will be

`color(darkgreen)(ul(color(black)("volume KOH = 249 mL")))`

You can double-check the result by calculating the concentrations of the two species in the buffer

`["CH"_3"COOH"] = ((0.5434 - 0.5333)"moles")/((875 + 249) * 10^(-3)"L") = "0.008986 M"`

`["CH"_3"COO"^(-)] = "0.5333 moles"/((875 + 249) * 10^(-3)"L") = "0.4745 M"`

As you cans ee, the buffer does contain significantly more conjugate base than weak acid, just as we predicted by looking at the `"pH"` of the buffer and at the `"p"K_a` of the acid.

`6.48 = 4.76 + log ((0.4745 color(red)(cancel(color(black)("M"))))/(0.008986color(red)(cancel(color(black)("M")))))`

If we round the calculation, we will get

`6.48 = 4.76 + 1.72 " "color(darkgreen)(sqrt())`

or

`6.48 ~~ 4.76 + 1.7227 " " ->` close enough!