Question

Chloropropionic acid, ClCH2CH2COOH is a weak monoprotic acid with Ka= 7.94 x 10-5 M. Calculate the pH at the equivalence point in a titration 24.1 mL of 1.33 M chloropropionic acid with 0.3 M KOH.

Answer 1

find the volume of KOH used to reach equivalence point

M(ClCH2CH2COOH)*V(ClCH2CH2COOH) =M(KOH)*V(KOH)

1.33 M *24.1 mL = 0.3M *V(KOH)

V(KOH) = 106.8433 mL

Given:

M(ClCH2CH2COOH) = 1.33 M

V(ClCH2CH2COOH) = 24.1 mL

M(KOH) = 0.3 M

V(KOH) = 106.8433 mL

mol(ClCH2CH2COOH) = M(ClCH2CH2COOH) * V(ClCH2CH2COOH)

mol(ClCH2CH2COOH) = 1.33 M * 24.1 mL = 32.053 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.3 M * 106.8433 mL = 32.053 mmol

We have:

mol(ClCH2CH2COOH) = 32.053 mmol

mol(KOH) = 32.053 mmol

32.053 mmol of both will react to form ClCH2CH2COO- and H2O

ClCH2CH2COO- here is strong base

ClCH2CH2COO- formed = 32.053 mmol

Volume of Solution = 24.1 + 106.8433 = 130.9433 mL

Kb of ClCH2CH2COO- = Kw/Ka = 1*10^-14/7.94*10^-5 = 1.259*10^-10

concentration ofClCH2CH2COO-,c = 32.053 mmol/130.9433 mL = 0.2448M

ClCH2CH2COO- dissociates as

ClCH2CH2COO- + H2O -----> ClCH2CH2COOH + OH-

0.2448 0 0

0.2448-x x x

Kb = [ClCH2CH2COOH][OH-]/[ClCH2CH2COO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.259*10^-10)*0.2448) = 5.552*10^-6

since c is much greater than x, our assumption is correct

so, x = 5.552*10^-6 M

[OH-] = x = 5.552*10^-6 M

use:

pOH = -log [OH-]

= -log (5.552*10^-6)

= 5.26

use:

PH = 14 - pOH

= 14 - 5.26

= 8.74

Answer: 8.74