# Question fcc4b

Oct 23, 2015

Yes.

#### Explanation:

You know that a buffer solution must contain either a weak acid and its conjugate base, or a weak base and its conjugate acid, in comparable amounts

Acetic acid, $\text{CH"_3"COOH}$, is a weak acid, which means that if you want to make a buffer solution using acetic acid, you need to add the acetate anion, ${\text{CH"_3"COO}}^{-}$, which is its conjugate base.

Here is where the sodium hydroxide comes in. Sodium hydroxide, $\text{NaOH}$, is a strong base that will react with the acetic acid to form sodium acetate, $\text{CH"_3"COONa}$, and water.

Sodium acetate will exist as sodium cations, ${\text{Na}}^{+}$, and acetate anions, ${\text{CH"_3"COO}}^{-}$, in solution, which means that you now have way of adding acetic acid's conjugate base to the solution.

The balanced chemical eqution for this neutralization rection looks like this - I'll use the net ionic equation

${\text{CH"_3"COOH"_text((aq]) + "OH"_text((aq])^(-) -> "CH"_3"COO"_text((aq])^(-) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you have a $1 : 1$ mole ratio between acetic acid, sodium hydroxide, and the acetate anion.

Now, you don't want to completely neutralize the acid, since that would only leave with its conjugate base in solution.

Instead, you want to convert about half of the moles of acetic acid to moles of acetate anions by adding sodium hydroxide.

This means that every mole of the former will react with one mole of sodium hydroxide and produce one mole of acetate anions.

You're working with a $\text{0.100-M}$ acetic acid solution and with a $\text{0.200-M}$ sodium hydroxide solution.

This means that if you take a volume $V$ of acetic acid, you will have

${n}_{\text{acetic}} = V \cdot 0.100$

This means that you will need a total of

${n}_{\text{NaOH" = 1/2 * n_"acetic}} = \frac{1}{2} \cdot V \cdot 0.100 = 0.050 \cdot V$

moles of sodium hydroxide.

The volume of the sodium hydroxide solution must be

V = (0.050 * V" moles")/"0.200 M" = 0.25 * V = 1/4V#

So, if you mix a volume $V$ of $\text{0.100 M}$ acetic acid solution and a volume ${V}_{4}$ of $\text{0.200 M}$ sodium hydroxide solution, you will produce equal amounts of moles of acetic acid and acetate anions, $0.050 V$ to be precise.

Now the solution will be buffer, since it will have equal concentrations of acetic acid and of acetate anions.