Question

You titrate 10.0 mL of 0.30 M acetic acid with 0.10 M sodium hydroxide.

a. What is the pH of the solution after you have added 10.00 mL of NaOH?

b.   What is the pH of the solution at the equivalence point?

Answer 1

pKa = 4.75

V = 10 ml

0.30 M of acid reats with 0.1 M of base to form 0.1 M of water and 0.1 M of acetate; leaving only 0.2 M of acid so

pH = pKa + log(acetate/acid)

pH = 4.75 + log(0.1/0.2) =4.4489

b)

in the equivalence point

mol of acid = mol of base

mol of acid = MV = 0.3*10 = 3 mmol of acid

so we added 3 mmol of base

V of base needed = 3/(0.1) = 30 ml

then

total volume at end = 10+30 = 40 ml

[acetate] = mmol/VT = 3/(40) = 0.075

in equivalence point

HA + NaOH = H2O + NaA

then

NaA = NA+ A-

A- + H2O <--> HA + OH-

then

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

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

5.55*10^-10 = (x*x)/(0.075-x)

x = [Oh-] = 6.45*10^-6

pOH = -log( 6.45*10^-6 =5.19

pH = 14-5.19 = 8.81