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#