Question

Chloropropionic acid, CICH2CH2COOH is a weak monoprotic acid with K, 7.94 x 105 M. Calculate the pH at the equivalence point in a titration 24.9 mL of 1.79 M chloropropionic acid with 0.3 M KOH. Answer:

Answer 1

find the volume of KOH used to reach equivalence point

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

1.79 M *24.9 mL = 0.3M *V(KOH)

V(KOH) = 148.57 mL

Given:

M(ClCH2CH2COOH) = 1.79 M

V(ClCH2CH2COOH) = 24.9 mL

M(KOH) = 0.3 M

V(KOH) = 148.57 mL

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

mol(ClCH2CH2COOH) = 1.79 M * 24.9 mL = 44.571 mmol

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

mol(KOH) = 0.3 M * 148.57 mL = 44.571 mmol

We have:

mol(ClCH2CH2COOH) = 44.571 mmol

mol(KOH) = 44.571 mmol

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

ClCH2CH2COO- here is strong base

ClCH2CH2COO- formed = 44.571 mmol

Volume of Solution = 24.9 + 148.57 = 173.47 mL

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

concentration ofClCH2CH2COO-,c = 44.571 mmol/173.47 mL = 0.2569M

ClCH2CH2COO- dissociates as

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

0.2569 0 0

0.2569-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)*0.2569) = 5.689*10^-6

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

so, x = 5.689*10^-6 M

[OH-] = x = 5.689*10^-6 M

use:

pOH = -log [OH-]

= -log (5.689*10^-6)

= 5.245

use:

PH = 14 - pOH

= 14 - 5.245

= 8.755

Answer: 8.76