# Question f3c37

##### 1 Answer
Apr 4, 2017

You must add 0.0055 mol of sodium acetate.

#### Explanation:

The moles of acetic acid are

0.100 color(red)(cancel(color(black)("L"))) × "0.100 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.0100 mol"

Let's write the equation for the ionization of acetic acid as

$\text{HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$

We can then write the Henderson-Hasselbalch equation as

"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"]"))#

Because $\text{HA}$ and $\text{A"^"-}$ are in the same solution, the ratio of the concentrations is the same as the ratio of the moles.

$4.50 = 4.76 + \log \left(\left(\text{moles of A"^"-")/("moles of HA}\right)\right) = 4.76 + \log \left(\frac{x}{0.0100}\right)$

$\log \left(\frac{x}{0.0100}\right) = \text{4.50 - 4.76" = "-0.26}$

$\frac{x}{0.0100} = {10}^{\text{-0.26}} = 0.55$

$x = \text{moles of A"^"-" = 0.0100 × 0.55 = "0.0055 mol}$

∴ You must add 0.0055 mol of sodium acetate to the acetic acid solution to get a pH of 4.50.