Question

Suppose you have `"500.0 mL"` of `"0.200 M"` acetic acid (`K_a = 1.8 xx 10^(-5)`). To make a `"pH"` `4.50` buffer, how many `"mL"` of `"3.0 M"` `"NaOH"` must be added?

This is a fun challenge question this week in the chemistry class I'm TAing. I thought I might as well put it up here and answer it.

Answer 1

There are two ways to approach this problem:

• Construct an ICE table to see what `["OAc"^(-)]_(eq)` is after using the small `x` approximation, and include `["OAc"^(-)]_(eq)` in the calculations. This gives the most accurate answer, unless you wish to find the true `x` without the small `x` approximation!
• Ignore the ICE table altogether and therefore assume that `n_("OH"^(-))` will overshadow `n_("OAc"^(-))` in solution. This is the fastest way to the answer, but gives a `2.63%` error in comparison to the first method with the small `x` approximation.

I'll show the results of both methods (including the exact quadratic formula answer).

DISCLAIMER: LONG ANSWER!

METHOD 1

The acid placed into water reacts as follows:

`"HOAc"(aq) + "H"_2"O"(l) rightleftharpoons "OAc"^(-)(aq) + "H"_3"O"^(+)`

`"I"" "" "0.200" "" "-" "" "" "0" "" "" "" "" "0`
`"C"" "" "-x" "" "-" "" "" "+x" "" "" "" "+x`
`"E"" "0.200-x" "-" "" "" "x" "" "" "" "" "x`

The `K_a` expression would then be:

`K_a = (["OAc"^(-)]["H"_3"O"^(+)])/(["HOAc"])`

`= x^2/(0.200 - x)`

Since `K_a` `~~` `10^(-5)`, we can use the small `x` approximation to negligible error to get:

`K_a ~~ x^2/(0.200) => x = sqrt(0.200K_a) = "0.001897 M"`

whereas the true `x` would be from the quadratic formula on:

`x^2 + K_ax - 0.200K_a = 0`

which would end up being `x = "0.001888 M"`.

Nonetheless, we will proceed with the small `x` approximation. From the Henderson-Hasselbalch equation, we can find the ratio of base to acid we need to generate for the buffer:

`"pH" = stackrel(-log(K_a))overbrace"pKa" + log\frac(["OAc"^(-)]_(buffer))(["HOAc"]_(buffer))`

`4.50 = -log(1.8 xx 10^(-5)) + log\frac(["OAc"^(-)]_(buffer))(["HOAc"]_(buffer))`

Thus, the ratio of concentrations of acetate to acetic acid is:

`(["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) = 10^(4.50 - 4.7447) ~~ 0.5692_(099788)`

Since the `"mol"`s of weak acid neutralized by `"NaOH"` generates the same number of `"mol"`s of its conjugate base, we can rewrite this ratio more conveniently as:

`(["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) = (n_(OAc^(-))"/"V_(t ot) + n_(OH^(-))"/"V_(t ot))/(n_(HOAc)"/"V_(t ot) - n_(OH^(-))"/"V_(t ot))`

This is because the `"NaOH"`, `"HOAc"`, and `"OAc"^(-)` share the same container. Thus, the total volume is shared and it can be cancelled out.

`=> (["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) = (n_(OAc^(-)) + n_(OH^(-)))/(n_(HOAc) - n_(OH^(-))) = 0.5692`

So, the `"mol"`s of acetate and acetic acid in the starting solution are:

`n_(OAc^(-)) = "0.001897 M" xx "0.5000 L" = 9.487 xx 10^(-4) "mols"`

`n_(HOAc) = (0.200 - 0.001897) "M" xx "0.5000 L" = "0.0991 mols"`

We do indeed know both before adding `"NaOH"`. Thus, solve for `n_(OH^(-))`:

`0.5692(n_(HOAc) - n_(OH^(-))) = n_(OAc^(-)) + n_(OH^(-))`

`0.5692n_(HOAc) - 0.5692n_(OH^(-)) = n_(OAc^(-)) + n_(OH^(-))`

`0.5692n_(HOAc) - n_(OAc^(-)) = 1.5692n_(OH^(-))`

`=> n_(OH^(-)) = (0.5692n_(HOAc) - n_(OAc^(-)))/(1.5692)`

`=` `0.0353_(250487)` `"mols NaOH"`

Without the small `x` approximation we would get `n_(OH^(-)) = 0.0353_(279096)` `"mols"` still, which are pretty much the same. In the end, we get:

`color(blue)(V_(NaOH)) = n_("OH"^(-))/(["NaOH"]) = "0.01178 L"`

`=` `color(blue)("11.78 mL NaOH")`

This would be the most exact answer using the most reasonable approximations.

METHOD 2

Ignoring the ICE table, we assume that:

`["HOAc"]_(eq) ~~ ["HOAc"]_i`

`n_(HOAc) ~~ "0.200 M" xx "0.500 L" = "0.100 mols"`

By ignoring the ICE table, we also assume that `n_(OAc^(-))` `"<<"` `n_("OH"^(-))`. As a result, we can say the following:

`=> (["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) ~~ (n_(OH^(-)))/(n_(HOAc) - n_(OH^(-))) = 0.5692`

Therefore, this method gives:

`0.5692n_(HOAc) - 0.5692n_(OH^(-)) = n_(OH^(-))`

`1.5692n_(OH^(-)) = 0.5692n_(HOAc)`

`n_(OH^(-)) ~~ (0.5692)/(1.5692)n_(HOAc)`

`~~ 0.0362_(7326)` `"mols"`

So, the volume under this approximation that the weak acid dissociates negligibly is:

`color(blue)(V_(NaOH)) = "0.03627326 mols"/("3.0 M") ~~ "0.01209 L"`

`=` `color(blue)("12.09 mL NaOH")`

This answer, from ignoring the ICE table, is off by a bit, but if you work it out, it's a `2.63%` error, which isn't that bad.