The amount of sodium acetate (#"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 (#"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 #"0.001 mol HA"# and #"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#