# Question 20193

Jun 19, 2017

The pH of the buffer is 3.78.

You have a weak acid (${\text{HC"_2"H"_3"O}}_{2}$) and its salt (${\text{NaC"_2"H"_3"O}}_{2}$), so you have the components of an "acetate" buffer.

The equation for the ionization of acetic acid is

$\text{HC"_2"H"_3"O"_2 + "H"_2"O" ⇌ "H"_3"O"^"+" + "C"_2"H"_3"O"_2^"-}$

For simplicity, let's rewrite this as

$\text{HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$

The ${K}_{\textrm{a}}$ expression is

K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = 1.8 × 10^"-5"

We can manipulate the ${K}_{\textrm{a}}$ expression to get the famous Henderson-Hasselbalch equation:

"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"]))

"p"K_"a" = -log(1.8 × 10^"-5") = 4.74
["A"^"-"] = "0.16 mol/L"
["HA"] = "1.45 mol/L"
"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#