# Question 7f273

Apr 7, 2017

$\text{249 mL KOH}$

#### 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, $\text{CH"_3"COOH}$, is the weak acid and the acetate anion, ${\text{CH"_3"COO}}^{-}$, is its conjugate base.

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

Now, acetic acid has

$\text{p} {K}_{a} = 4.76$

The buffer must have a pH of $6.48$, significantly higher than the $\text{p} {K}_{a}$ 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

$\left[\text{CH"_3"COO"^(-)] > ["CH"_3"COOH}\right]$

which implies

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) > 1$

You start with $\text{875 mL}$ of a $\text{0.621 M}$ 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")))) 

$= \text{0.5434 moles CH"_3"COOH}$

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.

${\text{CH"_ 3"COOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 2"O}}_{\left(l\right)}$

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

$6.48 = 4.76 + \log \left(\frac{y}{x}\right)$

This is equivalent to

$\log \left(\frac{y}{x}\right) = 1.72$

${10}^{\log} \left(\frac{y}{x}\right) = {10}^{1.72}$

which gets you

$\frac{y}{x} = 52.48$

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

$x = 0.5434 - n \to$ the number of moles of acetic acid

$y = 0 + n \to$ the number of moles of acetate anions

You can thus say that

$\frac{y}{x} = \frac{n}{0.5434 - n} = 52.48$

Solve this for $n$ to get

$n = 28.52 - 52.48 \cdot n$

$n \cdot \left(52.48 + 1\right) = 28.52 \implies n = \frac{28.52}{53.48} = 0.5333$

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{volume KOH = 249 mL}}}}$

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 $\text{pH}$ of the buffer and at the $\text{p} {K}_{a}$ of the acid.

$6.48 = 4.76 + \log \left(\left(0.4745 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))))/(0.008986color(red)(cancel(color(black)("M}}}}\right)\right)$

If we round the calculation, we will get

$6.48 = 4.76 + 1.72 \text{ } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$

or

$6.48 \approx 4.76 + 1.7227 \text{ } \to$ close enough!