Question

This is a problem about equilibrium between a strong acid mixed with a base, acid = Hc2H3O2 = 1.00 M, 50 ml

NaOH = 1.00 M , also 50 ml , ka for the acid is 1.75x10^-5, Find the pH for this titration?

The answer is 9.228, but i need to know how to do the work and incorporate the I. C. E table

Answer 1

Acid-base titration

With equal volumes and molar concentration of both acid-base

moles of acid = 1 M x 50 ml = 50 mmol

moles of base = 1 M x 50 ml = 50 mmol

Equivalence point

[C2H3O2-] formed = 50 mmol/100 ml = 0.5 M

Salt hydrolyzes

C2H3O2- + H2O <==> HC2H3O2 + OH-

with x amount of reaction

ICE chart

                      C2H3O2- + H2O <==> HC2H3O2 + OH-

I                          0.5                                   0             0

C                          -x                                   +x           +x

E                       0.5 - x                                 x            x

With x being a small number,

we get,

Kb = Kw/Ka = [HC2H3O2][OH-]/[C2H3O2-]

1 x 10^-14/1.8 x 10^-5 = x^2/0.5

x = [OH-] = 1.67 x 10^-5 M

pOH = -log[OH-] = 4.78

pH = 14 - pOH = 9.22