# You have a "250.0-mL" sample of "1.00-M" acetic acid. Assuming no volume change, how much "NaOH" must be added to make the best buffer? Calculate the "pH" of the best buffer.

## ${K}_{a} = 1.8 \times {10}^{- 5}$ Thank you!

Apr 8, 2018

$\text{5.00 g NaOH}$

$\text{pH} = 4.74$

#### Explanation:

In this context, the "best buffer" is a buffer that contains equal concentrations of acetic acid, the weak acid, and of acetate anions, the conjugate base.

As you know, the $\text{pH}$ of a weak acid - conjugate base buffer can be calculated using the Henderson - Hasselbalch equation.

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

Here

$\text{p} {K}_{a} = - \log \left({K}_{a}\right)$

Notice that when the buffer contains equal concentrations of the weak acid and of its conjugate base, the $\text{pH}$ of the buffer is equal to the $\text{p} K a$ of the weak acid, since

$\log \left(\left(\left[\text{conjugate base"])/(["weak acid}\right]\right)\right) = \log \left(1\right) = 0$

Your goal here will be to figure out how much sodium hydroxide must be added to your initial solution in order to get

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

Acetic acid and sodium hydroxide, which I'll represent as hydroxide anions, ${\text{OH}}^{-}$, due to the fact that sodium hydroxide is a strong base, react in a $1 : 1$ mole ratio to produce acetate anions and water.

color(white)(overbrace(color(black)("CH"_ 3"COOH"_ ((aq))))^(color(blue)("1 mole consumed"))) + color(white)(overbrace(color(black)("OH"_ ((aq))^(-)))^(color(blue)("1 mole consumed"))) -> color(white)(underbrace(color(black)("CH"_ 3"COO"_ ((aq))^(-)))_ (color(blue)("1 mole produced"))) + "H"_ 2"O"_ ((l))

So in order for the reaction to produce $1$ mole of acetate anions, it must consume $1$ mole of acetic acid.

250.0 color(red)(cancel(color(black)("mL solution"))) * ("1.00 moles CH"_ 3"COOH")/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.250 moles CH"_ 3"COOH"

Now, after the reaction is complete, you want the resulting solution to contain equal numbers of moles of acetic acid and of acetate anions.

So if you start with $0.250$ moles of acetic acid and consume $0.125$ moles, you will be left with $0.125$ moles of acetic acid. At the same time, the reaction will produce $0.125$ moles of acetate anions.

In order to consume $0.125$ moles of acetic acid, you need to add $0.125$ moles of sodium hydroxide to the initial solution.

To convert this to grams, use the molar mass of sodium hydroxide.

$0.125 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("5.00 g}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the concentration of the initial solution.

After the reaction is complete, the buffer will contain--keep in mind that we assume that the volume of the buffer will not change upon the addition of the sodium hydroxide

["CH"_3"COOH"] = "0.125 moles"/(250.0 * 10^(-3) quad "L") = "0.500 mol L"^(-1)

and

["CH"_3"COO"^(-)] = "0.125 moles"/(250.0 * 10^(-3) quad "L") = "0.500 mol L"^(-1)

The $\text{pH}$ of the solution will be

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

which is equal to

$\text{pH} = - \log \left(1.8 \cdot {10}^{- 5}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{4.74}}}$

The answer is rounded to two decimal places because you have two sig figs for the acid dissociation constant.