# 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.

Mar 21, 2017

There are two ways to approach this problem:

• Construct an ICE table to see what ${\left[{\text{OAc}}^{-}\right]}_{e q}$ is after using the small $x$ approximation, and include ${\left[{\text{OAc}}^{-}\right]}_{e q}$ 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 ${n}_{{\text{OH}}^{-}}$ will overshadow ${n}_{{\text{OAc}}^{-}}$ 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).

METHOD 1

The acid placed into water reacts as follows:

${\text{HOAc"(aq) + "H"_2"O"(l) rightleftharpoons "OAc"^(-)(aq) + "H"_3"O}}^{+}$

$\text{I"" "" "0.200" "" "-" "" "" "0" "" "" "" "" } 0$
$\text{C"" "" "-x" "" "-" "" "" "+x" "" "" "" } + x$
$\text{E"" "0.200-x" "-" "" "" "x" "" "" "" "" } x$

The ${K}_{a}$ expression would then be:

${K}_{a} = \left(\left[\text{OAc"^(-)]["H"_3"O"^(+)])/(["HOAc}\right]\right)$

$= {x}^{2} / \left(0.200 - x\right)$

Since ${K}_{a}$ $\approx$ ${10}^{- 5}$, we can use the small $x$ approximation to negligible error to get:

${K}_{a} \approx {x}^{2} / \left(0.200\right) \implies x = \sqrt{0.200 {K}_{a}} = \text{0.001897 M}$

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

${x}^{2} + {K}_{a} x - 0.200 {K}_{a} = 0$

which would end up being $x = \text{0.001888 M}$.

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:

$\left({\left[\text{OAc"^(-)]_(buffer))/(["HOAc}\right]}_{b u f f e r}\right) = {10}^{4.50 - 4.7447} \approx {0.5692}_{099788}$

Since the $\text{mol}$s of weak acid neutralized by $\text{NaOH}$ generates the same number of $\text{mol}$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 $\text{NaOH}$, $\text{HOAc}$, and ${\text{OAc}}^{-}$ share the same container. Thus, the total volume is shared and it can be cancelled out.

$\implies \left({\left[\text{OAc"^(-)]_(buffer))/(["HOAc}\right]}_{b u f f e r}\right) = \frac{{n}_{O A {c}^{-}} + {n}_{O {H}^{-}}}{{n}_{H O A c} - {n}_{O {H}^{-}}} = 0.5692$

So, the $\text{mol}$s of acetate and acetic acid in the starting solution are:

${n}_{O A {c}^{-}} = \text{0.001897 M" xx "0.5000 L" = 9.487 xx 10^(-4) "mols}$

${n}_{H O A c} = \left(0.200 - 0.001897\right) \text{M" xx "0.5000 L" = "0.0991 mols}$

We do indeed know both before adding $\text{NaOH}$. Thus, solve for ${n}_{O {H}^{-}}$:

$0.5692 \left({n}_{H O A c} - {n}_{O {H}^{-}}\right) = {n}_{O A {c}^{-}} + {n}_{O {H}^{-}}$

$0.5692 {n}_{H O A c} - 0.5692 {n}_{O {H}^{-}} = {n}_{O A {c}^{-}} + {n}_{O {H}^{-}}$

$0.5692 {n}_{H O A c} - {n}_{O A {c}^{-}} = 1.5692 {n}_{O {H}^{-}}$

$\implies {n}_{O {H}^{-}} = \frac{0.5692 {n}_{H O A c} - {n}_{O A {c}^{-}}}{1.5692}$

$=$ ${0.0353}_{250487}$ $\text{mols NaOH}$

Without the small $x$ approximation we would get ${n}_{O {H}^{-}} = {0.0353}_{279096}$ $\text{mols}$ still, which are pretty much the same. In the end, we get:

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

$=$ $\textcolor{b l u e}{\text{11.78 mL NaOH}}$

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

METHOD 2

Ignoring the ICE table, we assume that:

${\left[\text{HOAc"]_(eq) ~~ ["HOAc}\right]}_{i}$

${n}_{H O A c} \approx \text{0.200 M" xx "0.500 L" = "0.100 mols}$

By ignoring the ICE table, we also assume that ${n}_{O A {c}^{-}}$ $\text{<<}$ ${n}_{{\text{OH}}^{-}}$. As a result, we can say the following:

$\implies \left({\left[\text{OAc"^(-)]_(buffer))/(["HOAc}\right]}_{b u f f e r}\right) \approx \frac{{n}_{O {H}^{-}}}{{n}_{H O A c} - {n}_{O {H}^{-}}} = 0.5692$

Therefore, this method gives:

$0.5692 {n}_{H O A c} - 0.5692 {n}_{O {H}^{-}} = {n}_{O {H}^{-}}$

$1.5692 {n}_{O {H}^{-}} = 0.5692 {n}_{H O A c}$

${n}_{O {H}^{-}} \approx \frac{0.5692}{1.5692} {n}_{H O A c}$

$\approx {0.0362}_{7326}$ $\text{mols}$

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

$\textcolor{b l u e}{{V}_{N a O H}} = \text{0.03627326 mols"/("3.0 M") ~~ "0.01209 L}$

$=$ $\textcolor{b l u e}{\text{12.09 mL NaOH}}$

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.