Question

25 mL of 0.12M solution of HClO2 (pKa=1.92) is titrated with a 0.087M solution of KOH. Find the pH

- Initially before adding any KOH

- After 8.5 mL of KOH is added

- At the equivalence point of the titration

- After 39 mL (total) of KOH is added

Answer 1

use:

pKa = -log Ka

1.92 = -log Ka

Ka = 1.202*10^-2

1)when 0.0 mL of KOH is added

HClO2 dissociates as:

HClO2 -----> H+ + ClO2-

0.12 0 0

0.12-x x x

Ka = [H+][ClO2-]/[HClO2]

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 ((1.202*10^-2)*0.12) = 3.798*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.202*10^-2 = x^2/(0.12-x)

1.442*10^-3 - 1.202*10^-2 *x = x^2

x^2 + 1.202*10^-2 *x-1.442*10^-3 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 1.202*10^-2

c = -1.442*10^-3

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 5.914*10^-3

roots are :

x = 3.244*10^-2 and x = -4.446*10^-2

since x can't be negative, the possible value of x is

x = 3.244*10^-2

use:

pH = -log [H+]

= -log (3.244*10^-2)

= 1.4889

Answer: 1.49

2)when 8.5 mL of KOH is added

Given:

M(HClO2) = 0.12 M

V(HClO2) = 25 mL

M(KOH) = 0.087 M

V(KOH) = 8.5 mL

mol(HClO2) = M(HClO2) * V(HClO2)

mol(HClO2) = 0.12 M * 25 mL = 3 mmol

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

mol(KOH) = 0.087 M * 8.5 mL = 0.7395 mmol

We have:

mol(HClO2) = 3 mmol

mol(KOH) = 0.7395 mmol

0.7395 mmol of both will react

excess HClO2 remaining = 2.2605 mmol

Volume of Solution = 25 + 8.5 = 33.5 mL

[HClO2] = 2.2605 mmol/33.5 mL = 0.0675M

[ClO2-] = 0.7395/33.5 = 0.0221M

They form acidic buffer

acid is HClO2

conjugate base is ClO2-

Ka = 1.202*10^-2

pKa = - log (Ka)

= - log(1.202*10^-2)

= 1.92

use:

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

= 1.92+ log {2.207*10^-2/6.748*10^-2}

= 1.435

Answer: 1.44

3)

find the volume of KOH used to reach equivalence point

M(HClO2)*V(HClO2) =M(KOH)*V(KOH)

0.12 M *25.0 mL = 0.087M *V(KOH)

V(KOH) = 34.4828 mL

Given:

M(HClO2) = 0.12 M

V(HClO2) = 25 mL

M(KOH) = 0.087 M

V(KOH) = 34.4828 mL

mol(HClO2) = M(HClO2) * V(HClO2)

mol(HClO2) = 0.12 M * 25 mL = 3 mmol

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

mol(KOH) = 0.087 M * 34.4828 mL = 3 mmol

We have:

mol(HClO2) = 3 mmol

mol(KOH) = 3 mmol

3 mmol of both will react to form ClO2- and H2O

ClO2- here is strong base

ClO2- formed = 3 mmol

Volume of Solution = 25 + 34.4828 = 59.4828 mL

Kb of ClO2- = Kw/Ka = 1*10^-14/1.202*10^-2 = 8.319*10^-13

concentration ofClO2-,c = 3 mmol/59.4828 mL = 0.0504M

ClO2- dissociates as

ClO2- + H2O -----> HClO2 + OH-

0.0504 0 0

0.0504-x x x

Kb = [HClO2][OH-]/[ClO2-]

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 ((8.319*10^-13)*5.043*10^-2) = 2.048*10^-7

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

so, x = 2.048*10^-7 M

[OH-] = x = 2.048*10^-7 M

use:

pOH = -log [OH-]

= -log (2.048*10^-7)

= 6.6886

use:

PH = 14 - pOH

= 14 - 6.6886

= 7.3114

Answer: 7.31

4)when 39.0 mL of KOH is added

Given:

M(HClO2) = 0.12 M

V(HClO2) = 25 mL

M(KOH) = 0.087 M

V(KOH) = 39 mL

mol(HClO2) = M(HClO2) * V(HClO2)

mol(HClO2) = 0.12 M * 25 mL = 3 mmol

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

mol(KOH) = 0.087 M * 39 mL = 3.393 mmol

We have:

mol(HClO2) = 3 mmol

mol(KOH) = 3.393 mmol

3 mmol of both will react

excess KOH remaining = 0.393 mmol

Volume of Solution = 25 + 39 = 64 mL

[OH-] = 0.393 mmol/64 mL = 0.0061 M

use:

pOH = -log [OH-]

= -log (6.141*10^-3)

= 2.2118

use:

PH = 14 - pOH

= 14 - 2.2118

= 11.7882

Answer: 11.79