# Question 2627e

Oct 29, 2017

$\text{180 g CH"_3"COONa}$

#### Explanation:

You know that you're dealing with a weak acid-conjugate buffer, so you can say that its $\text{pH}$ can be calculated using this version of the Henderson - Hasselbalch equation.

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

Now, you know that the $\text{pH}$ of the buffer is higher than the $\text{p} {K}_{a}$ of the weak acid, so right from the start, you can say that

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

This basically means that the buffer contains more conjugate base than weak acid, i.e. more acetate anions--delivered to the solution by the sodium acetate--than acetic acid.

Use the $\text{pH}$ and the $\text{p} {K}_{a}$ of the weak acid to find the ratio that exists between the concentration of the acetate anions and the acetic acid in this buffer.

$6 = 4.75 + \log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right)$

This gets you

$\log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = 1.25$

You can rewrite this as

${10}^{\log} \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = {10}^{1.25}$

which is equivalent to

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) = 17.78$

You can thus say that in this solution, you have

$\left[\text{CH"_3"COO"^(-)] = 17.78 * ["CH"_3"COOH}\right]$

Now, I assume that you have

["CH"_3"COOH"] = "0.25 M"

This means that the concentration of the acetate anions would be

["CH"_3"COO"^(-)] = 17.78 * "0.25 M"

["CH"_3"COO"^(-)] = "4.445 M"

Next, use the volume of the solution to find the number of moles of acetate anions.

500 color(red)(cancel(color(black)("mL solution"))) * ("4.445 moles CH"_3"COO"^(-))/(10^3color(red)(cancel(color(black)("mL solution")))) = "2.223 moles CH"_3"COO"^(-)#

As you know, when dissolved in water, sodium acetate dissociates in a $1 : 1$ mole ratio to produce sodium cations and acetate anions, so you can say that in order to have $2.223$ moles of acetate anions in the buffer, you need to dissolve $2.223$ moles of sodium acetate to make this solution.

Finally, use the molar mass of sodium acetate to find the number of grams of salt.

$2.223 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CH"_3"COONa"))) * "82 g"/(1color(red)(cancel(color(black)("mole NaCH"_3"COONa")))) = color(darkgreen)(ul(color(black)("180 g}}}}$

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