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 34.8 mL of 1.09 M chloropropionic acid with 0.2 M KOH.

Answer 1

8.76

Explanation

ClCH2CH2COOH + KOH --------> ClCH2CH2COOK + ClCH2CH2COOH

stoichiometrically,1mole of ClCH2CH2COOH react with 1mole of KOH

No of mol of ClCH2CH2COOH = (1.09mol/1000ml)×34.8ml = 0.037932

0.037932 mol of ClCH2CH2COOH react with 0.037932mol of KOH

Volume of KOH solution containing 0.037932mol of KOH = (1000ml/0.2mol)×0.037932mol = 189.66ml

Therefore,

Equivalence point = 189.66ml

at equivalence point

No of mole of ClCH2CH2COO- formed = 0.037932

Total volume = 34.8ml + 189.66ml = 224.5ml

[ ClCH2CH2COO- ] = (0.037932mol/224.5ml )×1000ml = 0.1690M

ClCH2CH2COO- partly hydrolysed by water

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

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

Kb = Kw/Ka

= 1.00 ×10^-14/7.94×10^-5

= 1.26 ×10^-10

at equilibrium

[ OH- ] = x

[ ClCH2CH2COOH] = x

[ ClCH2CH2COO-] = 0.1690 - x

Therefore,

x^2/ (0.1690 - x) = 1.26×10^-10

we can assume 0.1690 - x = 0.1690

x^2 = 2.13×10^-11

x = 4.62×10^-6

[OH-]=4.62×10^-6

pOH = 5.34

pH + pOH = 14

pH = 14-5.24

= 8.76