Question

#1. Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HClO(aq) with 0.200 M KOH(aq). The ionization constant of HClO is Ka = 4.0×10^–8.

(a) Before addition of any KOH

(b) After addition of 25.0 mL of KOH

(c) After addition of 30.0 mL of KOH

(d) After addition of 50.0 mL of KOH

(e) After addition of 60.0 mL of KOH

#2 Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.200 M pyridine, C5H5N(aq) with 0.200 M HBr(aq). The ionization constant for pyridine is Kb = 1.7×10^–9.

(a) Before addition of any HBr

(b) After addition of 12.5 mL of HBr

(c) After addition of 20.0 mL of HBr

(d) After addition of 25.0 mL of HBr

(e) After addition of 29.0 mL of HBr

Answer 1

Unfortunately this is a lengthy problem because it is not possible to use one method and plug in different values ( except for the second and third , b and c, volumes )
Equation:
HClO + KOH → KClO + H2O
1 mol HClO reacts with 1 mol KOH to produce 1 mol KClO plus water

You will notice that throughout most of the problem you have a solution containing a weak acid ( HClO) and the conjugate base of that acid , from the salt KClO. This is a buffer solution .

a) Before any addition of KOH
What you are asked to do is calculate the pH of the acid solution: You do this by using the Ka equation:
Ka = [H+] / [acid]
4.0*10^-8 = [H+]² / 0.200
[H+]² = (4.0*10^-8 ) * 0.200
[H+]² = 8.0*10^-9
[H+] = √( 8.0*10^-9)
[H+] = 8.944*10^-5 M
pH = -log [H+]
pH = -log(8.944*10^-5)
pH = 4.05

b) After addition of 25.0mL of KOH. This produces a mixture of unreacted HClO and KClO in solution . This is a buffer solution . You determine the pH of a buffer solution using the Henderson - Hasselbalch equation.
This particular problem is easily solved - You react 50mL of 0.200M HClO with 25mL of 0.200M KClO. You have added exactly half the KOH required to neutralise all the acid . You are at the half equivalence point - the concentration of unreacted acid and salt are the same . The pH of this solution is equal to the pKa of the acid:
pKa = - log Ka = -log ( 4.0*10^-8) = 7.40
Therefore the pH of the buffer is 7.40.
But for completion let us do this using the H-H equation - this will help us with question c) next

Mol HClO in 50.0mL of 0.200M HClO solution = 50/1000*0.2 = 0.01 moles HClO
Mol KOH in 25.0ml of 0.200M solution = 25/1000*0.2 = 0.005 mol KOH
These react to produce 0.005 mol KClO , and there is 0.01-0.005 = 0.005 mol unreacted HClO remaining.
Calculate the molarity of each compound in solution: Final volume = 50+25 = 75mL = 0.075L
HClO = 0.005/0.075 = 0.0666M
KClO = 0.005/0.075 = 0.0666M

Now apply the H-H equation:
pH = pKa + log ( [salt] /[acid])
pH = 7.40 + log ( 0.0666/0.0666)
pH = 7.40 + log 1.0
pH = 7.40 + 0.00
pH = 7.40.

c) After addition of 30mL of KOH
We will use the above working out , but shortened:
Mol HClO in 50.0mL of 0.20M HClO solution = 50/1000*0.20 = 0.01 moles HClO
Mol KOH in 30.0ml of 0.20M solution = 30/1000*0.20 = 0.006 mol KOH
These react to produce 0.006 mol KClO , and there is 0.01-0.006 = 0.004 mol unreacted HClO remaining.
Calculate the molarity of each compound in solution: Final volume = 50+30 = 80mL = 0.080L
HClO = 0.004/0.080 = 0.05M
KClO = 0.006/0.080 = 0.075M

Now apply the H-H equation:
pH = pKa + log ( [salt] /[acid])
pH = 7.40 + log ( 0.075/0.05)
pH = 7.40 + log 1.5
pH = 7.40 + 0.1761
pH = 7.58

d) After addition of 50mL of KOH. What is significant about this titration volume : You are reacting 50mL of 0.20M HClO with 50mL of 0.20 KOH. You are neutralising all the HClO and producing a solution that contains only KClO . What you are doing in effect is calculating the pH of a KClO solution ( no longer a buffer solution - so you do not use the H-H equation)
Fro our above work , you are reacting 0.01 mol HClO with 0.01 mol KOH to produce 0.01 mol KClO dissolved in 100mL of solution = 0.1L
Molarity of the KClO solution = 0.01/0.1 = 0.1M KClO solution.

Calculate pH of a 0.1M solution of KClO
KClO dissociates:
KClO ↔ K+ + ClO-
The CLO- reacts with water :
ClO- + H2O → HClO + OH-

You require the Kb of HClO
Ka of HClO = 4.0*10^-8
Kb = 10^-14 / (Ka)
Kb = 10^-14 / ( 4.0*10^-8)
Kb = 2.5*10^-7

Use the Kb equation:
Kb = [ClO-] [ OH-] / [KClO]
Because [ClO- ] = [OH-] , and [KClO] = 0.1
Kb = [OH-]² / 0.1
2.5*10^-7 = [OH-]² / 0.1
[OH-]² = (2.5*10^-7) * 0.1
[OH-] ² = 2.5*10^-8
[OH-] = 1.58*10^-4

To calculate pH you require [H+]
[H+] = 10^-14 / [OH-]
[H+] = 10^-14 / (1.58*10^-4)
[H+] = 6.32*10^-11
pH = -log [H+]
pH = -log (6.32*10^-11)
pH = 10.19

e) After addition of 60mL of KOH. You saw in question d) above that 50mL of KOH had neutralised all the HClO. What you are now doing is adding 10mL excess of the 0.200M KOH . The final solution volume is 50+60 = 110mL
Molarity of the KOH solution:
M1V1 = M2V2
M1*110 + 0.2*10
M1 = 0.018M KOH solution . The KClO in solution does not affect the pH of the solution.
Because [KOH] = 0.018M , then [OH-] = 0.018M
[H+] = 10^-14 / 0.018
[H+] = 5.5*10^-13M
pH = -log 5.5*10^-13
pH = 12.26