Question

I know this question has multiple parts but I would greatly appreciate some help.

Determine the pH during the titration of 27.9 mL of 0.425 M hypochlorous acid (K_a = 3.5×10^-8) by 0.401 M KOH at the following points.

(a) Before the addition of any KOH

(b) After the addition of 7.40 mL of KOH

(c) At the half-equivalence point (the titration midpoint)

(d) At the equivalence point

(e) After the addition of 44.4 mL of KOH

Answer 1

a)when 0.0 mL of KOH is added

HClO dissociates as:

HClO -----> H+ + ClO-

0.425 0 0

0.425-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((3.5*10^-8)*0.425) = 1.22*10^-4

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

so, x = 1.22*10^-4 M

use:

pH = -log [H+]

= -log (1.22*10^-4)

= 3.9138

Answer: 3.91

b)when 7.4 mL of KOH is added

Given:

M(HClO) = 0.425 M

V(HClO) = 27.9 mL

M(KOH) = 0.401 M

V(KOH) = 7.4 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.425 M * 27.9 mL = 11.8575 mmol

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

mol(KOH) = 0.401 M * 7.4 mL = 2.9674 mmol

We have:

mol(HClO) = 11.8575 mmol

mol(KOH) = 2.9674 mmol

2.9674 mmol of both will react

excess HClO remaining = 8.8901 mmol

Volume of Solution = 27.9 + 7.4 = 35.3 mL

[HClO] = 8.8901 mmol/35.3 mL = 0.2518M

[ClO-] = 2.9674/35.3 = 0.0841M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 3.5*10^-8

pKa = - log (Ka)

= - log(3.5*10^-8)

= 7.456

use:

pH = pKa + log {[conjugate base]/[acid]}

= 7.456+ log {8.406*10^-2/0.2518}

= 6.979

Answer: 6.98

c)

Ka = 3.5*10^-8

pKa = - log (Ka)

= - log(3.5*10^-8)

= 7.456

At half equivalence point, pH is same as pKa

Answer: 7.46

d)

find the volume of KOH used to reach equivalence point

M(HClO)*V(HClO) =M(KOH)*V(KOH)

0.425 M *27.9 mL = 0.401M *V(KOH)

V(KOH) = 29.5698 mL

Given:

M(HClO) = 0.425 M

V(HClO) = 27.9 mL

M(KOH) = 0.401 M

V(KOH) = 29.5698 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.425 M * 27.9 mL = 11.8575 mmol

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

mol(KOH) = 0.401 M * 29.5698 mL = 11.8575 mmol

We have:

mol(HClO) = 11.8575 mmol

mol(KOH) = 11.8575 mmol

11.8575 mmol of both will react to form ClO- and H2O

ClO- here is strong base

ClO- formed = 11.8575 mmol

Volume of Solution = 27.9 + 29.5698 = 57.4698 mL

Kb of ClO- = Kw/Ka = 1*10^-14/3.5*10^-8 = 2.857*10^-7

concentration ofClO-,c = 11.8575 mmol/57.4698 mL = 0.2063M

ClO- dissociates as

ClO- + H2O -----> HClO + OH-

0.2063 0 0

0.2063-x x x

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

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 ((2.857*10^-7)*0.2063) = 2.428*10^-4

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

so, x = 2.428*10^-4 M

[OH-] = x = 2.428*10^-4 M

use:

pOH = -log [OH-]

= -log (2.428*10^-4)

= 3.6148

use:

PH = 14 - pOH

= 14 - 3.6148

= 10.3852

Answer: 10.39

e)when 44.4 mL of KOH is added

Given:

M(HClO) = 0.425 M

V(HClO) = 27.9 mL

M(KOH) = 0.401 M

V(KOH) = 44.4 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.425 M * 27.9 mL = 11.8575 mmol

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

mol(KOH) = 0.401 M * 44.4 mL = 17.8044 mmol

We have:

mol(HClO) = 11.8575 mmol

mol(KOH) = 17.8044 mmol

11.8575 mmol of both will react

excess KOH remaining = 5.9469 mmol

Volume of Solution = 27.9 + 44.4 = 72.3 mL

[OH-] = 5.9469 mmol/72.3 mL = 0.0823 M

use:

pOH = -log [OH-]

= -log (8.225*10^-2)

= 1.0848

use:

PH = 14 - pOH

= 14 - 1.0848

= 12.9152

Answer: 12.92