Question

a. What is the theoretical pH at the stoichiometric equivalence point for the titration of the weak acid? ACETIC ACID REACTING WITH NaOH

Hint: Your volume at the stoichiometric point should include the amount of base you added to reach the stoichiometric point (4.174 mL) as well as the initial amount of stock HAc and the water added to cover the pH meter bulb. You should use the actual concentration of the HAc (from question 8a) in this calculation. Again, please use the simplifying assumption, not the quadratic equation. Ka(acetic acid) = 1.76 x 10-5    HAc = .088739 WATER ADDED = 55 ML 5 ML of ACETIC ACID WAS ADDED TO THE WATER, MOLARITY OF THE WATER WITH THE ACETIC ACID ADDED .0073949

Answer 1

Acetic acid + NaOH = NaAcetate + H2O

so..

Vbase = 4.174 mL

Ka = 1.76*10^-5

HA = 0.088739 M

V water = 55 mL

Vtotal = 55+5 = 60 mL

[HA] = M1V1/(Vtotal = (0.088739)(5)/60 = 0.0073949 M

so...

In equivalence point:

Vtotal = Vacid + Vbase = 4.174 + 60 = 64.174

[A-] = M1V1/(Vtotal) = 0.0073949*60/64.174 = 0.006913 M

for pH in equivalence point:

A- + H2O <-> HA + OH-

Kb = [HA][OH-]/[A-]

Kb = 10^-14/1.8*10^-5 = 5.55*10^-10

5.55*10^-10 = x*x/(0.006913 -x)

x = 1.958*10^-6

so

[OH-] = x = 1.958*10^-6

pOH= -log(1.958*10^-6 = 5.708

pH = 14-5.708 = 8.292

makes sense, since this is an acidic salt, meaning it will hydrolyse to form a basic solution