Question

Draw the ph vs volume of NaOH for titrating HOOCCH2CH2OH with NaOH. What is the ph at the endpoint? What will be different when HOOCCH2COOH is titrated with NaOH? Please explain what you are doing. Thank you

Answer 1

Titration plot for monoprotic acid

lets say we are titrating 50 ml of 1 M HOOCCH2CH2OH with 1 M NaOH

initial pH

Acid dissociates in solution,

HOOCCH2CH2OH <==> H+ + -OOCCH2CH2OH

let x amount dissociated

Ka = 3.1 x 10^-5 = x^2/1

x = [H+] = 5.57 x 10^-3 M

pH = -log[H+] = 2.25

After 12.5 ml NaOH added

acid remaining in solution = (1 M x 50 ml - 1 M x 12.5 ml)/62.5 ml = 0.6 M

acid salt formed = 12.5 mmol/62.5 ml = 0.2 M

pH = pKa + log(base/acid)

      = 4.51 + log(0.2/0.6) = 4.03

after 25 ml NaOH added

half-equivalence point

pH = pKa = -log(Ka) = 4.51

after 37.5 ml NaOH added

acid remaining in solution = (1 M x 50 ml - 1 M x 37.5 ml)/87.5 ml = 0.143 M

acid salt formed = 37.5 mmol/87.5 ml = 0.43 M

pH = pKa + log(base/acid)

      = 4.51 + log(0.43/0.143) = 5.0

After 50 ml NaOH added

equivalence point

conjugate acid formed hydrolyzes

[conjugate acid] = 1 M x 50 ml/100 ml = 0.5 M

let A- be the conjugate acid

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

let x amount reacted

Kb = 1 x 10^-14/3.1 x 10^-5 = x^2/0.5

x = [OH-] = 1.27 x 10^-5 M

[H+] =1 x 10^-14/1.27 x 10^-5 = 7.9 x 10^-10 M

pH = -log[H+] = 9.10

after 150 ml NaOH added

excess [OH-] = 1 M x 150 ml/200 ml = 0.75 M

pOH = -log[OH-] = 0.125

pH = 14 - pOH = 13.87

Plot shown below

When the diacid HOOCCH2COOH is titrated instead, 2 moles of NaOH is used to reach equivalence point as opposed to 1 mole as seen in the previous case.