# Question 816f3

May 6, 2017

The pH is 5.76.

#### Explanation:

The amount of sodium acetate ($\text{A"^"-}$) is

0.100 color(red)(cancel(color(black)("L solution"))) × ("0.1 mol A"^"-")/(1 color(red)(cancel(color(black)("L solution")))) = "0.01 mol A"^"-"

The amount of acetic acid ($\text{HA}$) added is

1000 color(red)(cancel(color(black)("µL solution"))) × (1 color(red)(cancel(color(black)("L solution"))))/(10^6 color(red)(cancel(color(black)("µL solution")))) × "1 mol HA"/(1 color(red)(cancel(color(black)("L solution")))) = "0.001 mol HA"

We have a solution that contains $\text{0.001 mol HA}$ and $\text{0.01 mol A"^"-}$.

A solution of a weak acid and its salt is a buffer.

We can therefore apply the Henderson-Hasselbalch equation:

color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "

Since both the acetic acid and the sodium acetate are in the same solution, the ratio of their concentrations is the same as the ratio of their moles.

"pH" = 4.76 + log((0.01 color(red)(cancel(color(black)("mol"))))/(0.001 color(red)(cancel(color(black)("mol"))))) = 4.76 + log(10) = 4.76 + 1 = 5.76#