Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

2 answers:

8 0

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3