The chemical equation is

#"CH"_3"CHOHCOOH" + "H"_2"O" ⇌"H"_3"O"^"+" + "CH"_3"CHOHCOO"^"-"#

Let's rewrite this as

#color(white)(mmmmmml)"HA" + "H"_2"O" ⇌ "H"_3"O"^(+) + "A"^"-"; color(white)(mm)K_text(a) = 1.4 × 10^"-4"#

#"E/mol·L"^"-1":color(white)(l) 0.120color(white)(mmmmmmmml)"0.021 32"#

#"Moles of NaCH"_3"CHOHCOO" = "moles of A"^"-" = 2.39 color(red)(cancel(color(black)("g"))) × "1 mol"/(112.06 color(red)(cancel(color(black)("g")))) = "0.021 33 mol"#

#["A"^"-"] = "0.021 33 mol"/"0.500 L" = "0.042 66 mol/L"#

We can now use the **Henderson-Hasselbalch equation** to calculate the #"pH"#.

#"pH" = "p"K_"a" + log(("[A"^"-""]")/"[HA]") = -log(1.4 × 10^"-4") + log("0.042 66"/0.120) = "3.85 - 0.45" = 3.40#