Question 97ac9

Mar 12, 2017

Here's what I got.

Explanation:

The first thing you need to do here is to look up the $\text{p} {K}_{a}$ of acetic acid, which you'll find listed as

$\text{p} {K}_{a} = 4.75$

http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Now, when you mix acetic acid, a weak acid, and sodium acetate, the salt of the acetate anion, which is the conjugate base of acetic acid, you will get a buffer solution.

Right from the start, the fact that you have

$\text{pH" > "p} {K}_{a}$

should tell you that the buffer contains more conjugate base than weak acid.

The pH of a weak acid/conjugate base buffer can be calculated using the Henderson - Hasselbalch equation

color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"])))))

In your case, you will have

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

Use the pH of the solution to calculate the ratio between the concentration of the conjugate base and the concentration of the weak acid in the buffer

$4.85 = 4.75 + \log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right)$

$\log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = 4.85 - 4.75$

This will be equivalent to

${10}^{\log} \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = {10}^{0.10}$

which will get you

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) = 1.26$

Now, let's say that ${V}_{1}$ represents the volume of the acetic acid solution and ${V}_{2}$ represents the volume of the sodium acetate.

The total volume of the buffer will be equal to ${V}_{1} + {V}_{2}$, which means that you have

$\text{ "["CH"_ 3"COOH"] = n_"acetic acid"/(V_1 + V_2)" }$ and $\frac{\text{ "["CH"_ 3"COO"^(-)] = n_"acetate}}{{V}_{1} + {V}_{2}}$

Consequently, you can say that

(["CH"_ 3"COO"^(-)])/(["CH"_ 3"COOH"]) = (n_"acetate"/color(red)(cancel(color(black)(V_1 + V_2))))/(n_"acetic acid"/color(red)(cancel(color(black)((V_1 + V_2))))) = 1.26

This is equivalent to

n_"acetate" = 1.26 * n_"acetic acid"" " " "color(darkorange)("(*)")

Now, use the concentrations of the two solutions to write

n_"acetate" = "0.15 mol" color(red)(cancel(color(black)("L"^(-1)))) * V_2 color(red)(cancel(color(black)("L")))

${n}_{\text{acetate" = (0.15 * V_2)color(white)(.)"moles CH"_3"COO}}^{-}$

Do the same for the acetic acid

n_"acetic acid" = "0.10 mol" color(red)(cancel(color(black)("L"^(-1)))) * V_2 color(red)(cancel(color(black)("L")))#

${n}_{\text{acetic acid" = (0.10 * V_2)color(white)(.)"moles CH"_3"COOH}}$

You can now rewrite equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ as

$\left(0.15 \cdot {V}_{2}\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) = 1.26 * (0.10 * V_1)color(red)(cancel(color(black)("moles}}}}$

which will get you

$0.15 \cdot {V}_{2} = 0.126 \cdot {V}_{1}$

You know that

${V}_{1} + {V}_{2} = \text{0.020 L}$

which means that you now have a system of two equations with two unknowns.

${V}_{1} = \text{0.020 L} - {V}_{2}$

Plug this into the first equation to get

$0.15 \cdot \left(\text{0.020 L} - {V}_{2}\right) = 0.126 \cdot {V}_{2}$

$\text{0.003 L} - 0.15 \cdot {V}_{2} = 0.126 \cdot {V}_{2}$

You will end up with

${V}_{2} = \text{0.003 L"/(0.126 + 0.15) = "0.0109 L}$

This means that

${V}_{1} = \text{0.020 L" - "0.0109 L" = "0.0091 L}$

Therefore, you can say that in order to make $\text{20 mL}$ of an acetic acid/sodium acetate buffer that has a pH equal to $4.85$, you must mix

$\text{9.0 mL " -> " 0.10 M}$ acetic acid solution

$\text{11 mL " -> " 0.15 M}$ sodium acetate solution

I'll leave the answers rounded to two sig figs, but keep in mind that you only have one significant figure for the volume of the buffer.