Question

Calculate the pH for each case in the titration of 50.0 mL of 0.240 M HClO(aq)0.240 M HClO(aq) with 0.240 M KOH(aq). Use the ionization constant for HClO.

What is the pH before addition of any KOH?

What is the pH after addition of 25.0 mL KOH?

What is the pH after addition of 30.0 mL KOH?

What is the pH after addition of 50.0 mL KOH?

What is the pH after addition of 60.0 mL KOH?

Answer 1

1.

Before addition of KOH , pH is due to ionisation of weak acid HOCl.

pKa of HOCl = 7.53

now,

pH of weak acid is,

pH = ( pKa - logC)

Or, pH = ( 7.53 - log 0.240)

Or, pH = 4.075.

2.

When 25 mL 0.240 M KOH is added reaction reaches at half neutralization point.

ICE table is

mmoles	HOCl	KOH	KOCl
I	(50×0.24) = 12	25×0.240= 6	0
C	-6	-6	+6
E	(12-6) = 6	0	6

So,

at half neutralization point , the solution is buffer where

[ HOCl] =[KOCl]

Using Hendersson -Hasselbalch equation

pH = pKa + log

So,

pH = pKa = 7.53.

3.

Now, when 30.0 mL KOH is added

Mmoles of Base = mmoles of salt (NaOCl) formed = 30× 0.24

= 7.2.

Mmoles of excess base remains = (50×0.24) - (30×0.24)

= 4.8.

The solution is now a buffer of HOCl and NaOCl.

So, by Hendersson -Hasselbalch equation

pH = pKa + log

Or, pH = 7.53 + log(7.2/4.8)

Or, pH = 7.53 +0.18

Or, pH = 7.71.

4.

When, 50.0 mL KOH is added , the reaction reaches to neutralization point . At this point pH is due to dissociation of salt NaOCl.

[Salt] = (mmoles of base added/total volume)

= (50× 0.240)/(50+50)

= 0.120 M

Now, pH of salt of weak acid strong base is

pH = 7 + (pKa + logC_salt)

= 7 + (7.53 + log0.12)

= 7 + ( 7.53 - 0.92)

= 7 + 3.31

= 10.31

5.

After addition of excess base ( 60.0 mL) KOH, pH is due to dissociation of strong base KOH.

Now, excess mmoles of KOH

= (mmoles of KOH - mmoles of acid )

= (60×0.240) - (50× 0.240)

= 2.4

Total volume = (50+60) = 110 mL.

Then,

[OH^-] = ( mmoles of KOH present/total volume)

= ( 2.4/110)

= 0.022 M.

Now, pOH = - log[OH^-] = - log(0.022) = 1.66.

Then, pH = 14 - 1.66 = 12.34