Acetic acid (HC2H3O2) has a pKa = 4.745. Determine the acetate concentration, (C2H3O2-), of a solution prepar

Question

Acetic acid (HC2H3O2) has a pKa = 4.745. Determine the acetate concentration, (C2H3O2-), of a solution prepared by diluting 20.0 mL of a 0.500 M acetic acid solution with 2.39 x 10^2 mL of 1.90 x 10^-1 M HCl(aq). HCl is a strong acid. (assume volumes are additive)

I'm having a lot of trouble with this chemistry problem. Any help would be appreciated. Thank you...

Answer 1

First find the value of the concentrations of the two acids just after mixing.
C1V1=C'1Vtotal
=>C'1=C1*V1/(V1+V2) =0.500*20/(20+239) = 0.0386 M CH3COOH (it is acetic acid written in a different way)
Here I didn't convert mL to L as I should because the conversion factor is simplified in the ratio and thus it would be a bit redundant. However in order to be 100% accurate you would need to multiply each volume value with 0.001

similarly for HCl
C2V2= C'2Vfinal
=> C2'= C2V2/(V1+V2)= 0.190*239/(20+239) =0.175 M

HCl is a stong acid thus it dissociates 100% according to the reaction HCl -> H+ +Cl-
Since the stoichiometry is 1 HCl :1 H+ you have for the H+ that come from HCl C=C'2= 0.175 M

Acetic acid is a weak acid. It has to dissociate in the presence of H+ coming from the strong HCl. Thus the ICE table is set up as follows

.. .. .. .. .. CH3COOH <=> CH3COO- + H+
Initial .. .. .. 0.0386 .. .. .. .. .. .. .. .. .. .. .. 0.175
Dissociate .. ..x
Produce .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. ..x
At Equil. .. 0.0386-x .. .. .. .. .. .. x .. .. 0.175+x

Ka= [CH3COO-][H+] / [CH3COOH]
Ka=10^-pKa ,so
x(0.175+x) / (0.0386-x) =10^-4.745 =1.80*10^-5 =>
x^2+ 0.175x = 0.0386*1.80*10^-5 - 1.8*10^-5 x =>
x^2+ 0.175018x- 6.948*10^-7 =0
x1= 3.97*10^-6 M
x2= -0.17502 <0 rejected

so [CH3COO-]= x= 3.97*10^-6 M

Check for numerical errors