Asked by YBB

What is the concentrations of HOAc and OAc- in a 0.2 M “acetate” buffer, pH 5.0? The Ka for acetic acid 1.70x10-5 (pKa 4.77)
Answered by DrBob222
You solve two equation simultaneously. Eqn 1 is this.
pH = pKa + log (base)/(acid). Base is OAc and acid is HOAc. You know pH and pKa and you solve for the ratio of base/acid. Eqn 2 is this.
base + acid = 0.2
Simplified:
eqn 1..........base/acid = ? some number
eqn 2 ........ base + acid = 0.2
Post your work if you get stuck.
Answered by DAR
could you please say why base+acide = 0.2 ??
Answered by DrBob222
Let's take acetic acid, HAc. It ionizes as HAc ==> H^+ + Ac^-. When you say 0.2 M you are measuring that out as HAc. However, as soon as water is added it ionizes and form some H^+ and some Ac^-. So the total is still 0.2 M which means if you add the acid (in this case HAc) and the base (in this case Ac-) it will add up to 0.2 M. so (acid) + ( base) = total molarity.
Answered by MMS
Describe the preparation of 5 liters of a 0.3 M acetate buffer, pH 4.47, starting from a 2 M solution of acetic acid and a 2.5 M solution of KOH (pKa 4.77).
Answered by SAM
so the OAC= 0.9999 M
and the HOAC= 1x10^-5 that right ?
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