Question

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).

(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

Answer 1

a) only HClO is present

we know that

HClO is a weak acid

and for weak acids

[H+] = sqrt ( Ka x C)

so

[H+] = sqrt ( 2.9512 x 10-8 x 0.21)

[H+] = 7.872 x 10-5

now

pH = -log [H+]

so

pH = -log 7.872 x 10-5

pH = 4.104

b)

now

moles = molarity x volume (L)

so

moles of HClO = 0.21 x 50 x 10-3 = 10.5 x 10-3

moles of KOH = 0.21 x 25 x 10-3 = 5.25 x 10-3

now

the reaction

HClO + KOH ----> KClO + H20

we can see that

moles of HClO reacted = moles of KOH added = 5.25 x 10-3

so

moles of HClO left = 10.5 x 10-3 - 5.25 x 10-3 = 5.25 x 10-3

also

moles of KClO formed = moles of KOH added = 5.25 x 10-3

now

HClO and KClO are a buffer solution

we know that

for buffers

pH = pKa + log [salt / acid]

so

pH = pKa + log [ KClO / HClO]

pH = 7.53 + log [ 5.25 x 10-3 / 5.25 x 10-3 ]

pH = 7.53

c)

now

moles = molarity x volume (L)

so

moles of HClO = 0.21 x 50 x 10-3 = 10.5 x 10-3

moles of KOH = 0.21 x 30 x 10-3 = 6.3 x 10-3

now

the reaction

HClO + KOH ----> KClO + H20

we can see that

moles of HClO reacted = moles of KOH added = 6.3 x 10-3

so

moles of HClO left = 10.5 x 10-3 - 6.3 x 10-3 = 4.2 x 10-3

also

moles of KClO formed = moles of KOH added = 6.3 x 10-3

now

HClO and KClO are a buffer solution

we know that

for buffers

pH = pKa + log [salt / acid]

so

pH = pKa + log [ KClO / HClO]

pH = 7.53 + log [ 6.3 x 10-3 / 4.2 x 10-3 ]

pH = 7.7061

d)

now

moles = molarity x volume (L)

so

moles of HClO = 0.21 x 50 x 10-3 = 10.5 x 10-3

moles of KOH = 0.21 x 50 x 10-3 = 10.5 x 10-3

now

the reaction

HClO + KOH ----> KClO + H20

we can see that

moles of HClO reacted = moles of KOH added = 10.5 x 10-3

so

moles of HClO left = 10.5 x 10-3 - 10.5 x 10-3 = O

also

moles of KClO formed = moles of KOH added = 10.5 x 10-3

final volume = 50 + 50 = 100 ml

[KClO] = 10.5 x 10-3 x 1000 / 100

[KClO] = 0.105

now

KClO is a weak base

we know that

for weak bases

[OH-] = sqrt ( Kb x C)

also

Kb = 10-14 / Ka

so

[OH-] = sqrt ( 10-14 x 0.105 / 2.9512 x 10-8)

[OH-] = 1.886 x 10-4

now

pOH = -log [OH-]

pOH = - log 1.886 x 10-4

pOH = 3.724

pH = 14 - pOH

pH = 14 - 3.724

pH = 10.276

e)

now

moles = molarity x volume (L)

so

moles of HClO = 0.21 x 50 x 10-3 = 10.5 x 10-3

moles of KOH = 0.21 x 60 x 10-3 = 12.6 x 10-3

now

the reaction

HClO + KOH ----> KClO + H20

we can see that

moles of KOH reacted = moles of HClO added = 10.5 x 10-3

so

moles of KOH left = 12.6 x 10-3 - 10.5 x 10-3 = 2.1 x 10-3

also

moles of KClO formed = moles of HClO added = 10.5 x 10-3

final volume = 50 + 60 = 110 ml

[KOH] = 2.1 x 10-3 x 1000 / 110

[KOH] = 0.01909

now

KOH and KClO are present

we know that

KOH is a very strong base

so

[OH-] = [KOH] = 0.01909

poH = -log 0.01909

pOH = 1.72

now

pH = 14 - 1.72

pH = 12.28