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
#"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 #"HA"# and #"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(("moles of A"^"-")/("moles of HA")) = 4.76 + log(x/0.0100)#
#log(x/0.0100) = "4.50 - 4.76" = "-0.26"#
#x/0.0100 = 10^"-0.26" = 0.55#
#x = "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.
