The pH of the buffer is 3.78.
You have a weak acid (#"HC"_2"H"_3"O"_2#) and its salt (#"NaC"_2"H"_3"O"_2#), so you have the components of an "acetate" buffer.
The equation for the ionization of acetic acid is
#"HC"_2"H"_3"O"_2 + "H"_2"O" ⇌ "H"_3"O"^"+" + "C"_2"H"_3"O"_2^"-"#
For simplicity, let's rewrite this as
#"HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#
The #K_text(a)# expression is
#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = 1.8 × 10^"-5"#
We can manipulate the #K_text(a)# expression to get the famous Henderson-Hasselbalch equation:
#"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"]))#
Your data are
#"p"K_"a" = -log(1.8 × 10^"-5") = 4.74#
#["A"^"-"] = "0.16 mol/L"#
#["HA"] = "1.45 mol/L"#
Insert these values into the Henderson-Hasselbalch equation:
#"pH" = 4.74 + log((0.16 cancel("mol/L"))/(1.45 cancel("mol/L"))) = 4.74 + log(0.110) = 4.74 - 0.96 = 3.78#
