Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.170 M HClO(aq) with 0.170 M KOH(aq).

a) after addition of 40 mL of KOH

b) after addition of 50 mL of KOH

Answer 1

Ka of HClO = 2.9*10^-8

a)

when 40.0 mL of KOH is added

we have:

Molarity of HClO = 0.17 M

Volume of HClO = 50 mL

Molarity of KOH = 0.17 M

Volume of KOH = 40 mL

mol of HClO = Molarity of HClO * Volume of HClO

mol of HClO = 0.17 M * 50 mL = 8.5 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.17 M * 40 mL = 6.8 mmol

We have:

mol of HClO = 8.5 mmol

mol of KOH = 6.8 mmol

6.8 mmol of both will react

excess HClO remaining = 1.7 mmol

Volume of Solution = 50 + 40 = 90 mL

[HClO] = 1.7 mmol/90 mL = 0.0189M

[ClO-] = 6.8/90 = 0.0756M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 2.9*10^-8

pKa = - log (Ka)

= - log(2.9*10^-8)

= 7.538

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 7.538+ log {7.556*10^-2/1.889*10^-2}

= 8.14

Answer: 8.14

b)

when 50.0 mL of KOH is added

we have:

Molarity of HClO = 0.17 M

Volume of HClO = 50 mL

Molarity of KOH = 0.17 M

Volume of KOH = 50 mL

mol of HClO = Molarity of HClO * Volume of HClO

mol of HClO = 0.17 M * 50 mL = 8.5 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.17 M * 50 mL = 8.5 mmol

We have:

mol of HClO = 8.5 mmol

mol of KOH = 8.5 mmol

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

ClO- here is strong base

ClO- formed = 8.5 mmol

Volume of Solution = 50 + 50 = 100 mL

Kb of ClO- = Kw/Ka = 1*10^-14/2.9*10^-8 = 3.448*10^-7

concentration ofClO-,c = 8.5 mmol/100 mL = 0.085M

ClO- dissociates as

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

0.085 0 0

0.085-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((3.448*10^-7)*8.5*10^-2) = 1.712*10^-4

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

so, x = 1.712*10^-4 M

[OH-] = x = 1.712*10^-4 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.712*10^-4)

= 3.77

we have below equation to be used:

PH = 14 - pOH

= 14 - 3.77

= 10.23

Answer: 10.23