The pH of the buffer is 3.78.
You have a weak acid ("HC"_2"H"_3"O"_2HC2H3O2) and its salt ("NaC"_2"H"_3"O"_2NaC2H3O2), 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^"-"HC2H3O2+H2O⇌H3O++C2H3O-2
For simplicity, let's rewrite this as
"HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"HA + H2O⇌H3O++A-
The K_text(a)Ka expression is
K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = 1.8 × 10^"-5"Ka=[H3O+][A-][HA]=1.8×10-5
We can manipulate the K_text(a)Ka expression to get the famous Henderson-Hasselbalch equation:
"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"]))pH=pKa+log([A-][HA])
Your data are
"p"K_"a" = -log(1.8 × 10^"-5") = 4.74pKa=−log(1.8×10-5)=4.74
["A"^"-"] = "0.16 mol/L"[A-]=0.16 mol/L
["HA"] = "1.45 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
