Question

1.)A.)What is the pH at the equivalence point in the titration of a 21.1 mL sample of a 0.363 M aqueous hypochlorous acid solution with a 0.397 M aqueous sodium hydroxide solution? pH =

B.)A 23.7 mL sample of a 0.478 M aqueous hypochlorous acid solution is titrated with a 0.353 M aqueous potassium hydroxide solution. What is the pH at the start of the titration, before any potassium hydroxide has been added? pH =

C.)When a 16.1 mL sample of a 0.335 M aqueous hydrofluoric acid solution is titrated with a 0.484 M aqueous potassium hydroxide solution, what is the pH after 16.7 mL of potassium hydroxide have been added?
pH =

Answer 1

Q1.

pH in equivalence point

mmol of acid = MV = 0.397*21.1 = 8.3767 mmol of HClO

V base = mmol / M = 8.3767 / 0.363 = 23.076 mL of base will be required

so

V total = 21.1+23.076 =44.176 mL

HClO + NaOH = Na+ + ClO- + H2O

ClO- + H2O <-> HClO + OH-

Kb = [HClO ][OH-]/[ClO-]

Kb = (10^-14)/(3.5*10^-8) =2.857*10^-7

now

[ClO-] = mmol/ Vtotal = 8.3767 /23.076 = 0.363 m

now

Kb = [HClO ][OH-]/[ClO-]

2.857*10^-7 = x*x/(0.363 -x)

x = 3.218*10^-4

pOH = -log(3.218*10^-4) = 3.4924

pH = 14-3.4924 = 10.5076

B)

find pH of HClO before any addition

HClO <-> H+ + ClO-

Ka = [H+][ClO-]/[HClO]

3.5*10^-8 = x*x/(0.478 - x)

x = H = 1.29*10^-4

pH = -log( 1.29*10^-4) = 3.889

C)

this has excess base

mmol of acid = MV = 16.1*0.335 = 5.3935

mmol of base = MV = 16.7*0.484 = 8.0828

mmol of base left = 8.0828 - 5.3935 = 2.6893

[OH-] = mmol/Vtotal = 2.6893/(16.1+16.7) = 0.0819

pOH = -log(0.0819 = 1.08

pH = 14-1.08 = 12.92