Question

When you titrate 100 mL of 0.10 M HF with 0.10 M KOH

a) When will this titration reach the equivalence point?

b) Calculate pH when you add 5 mL of KOH.

c) Calculate pH when you add 50 mL of KOH.

d) Calculate pH at the equivalence point.

e) Calculate pH when you add 101 mL of KOH.

(show each step)

Answer 1

a) we know that

at half equivalence point

Ma x Va = Mb x Vb

so

100 x 0.1 = 0.1 x Vb

Vb = 100

so

after adding 100 ml of KOH

equivalence point is reached
b)

now

moles = molarity x volume (L)

so

moles of HF = 0.1 x 100 x 10-3

moles of HF = 100 x 10-4

moles of KOH added = 0.1 x 5 x 10-3

moles of KOH added = 5 x 10-4

the reaction is

HF + KOH ---> KF + H20

moles of HF reacted = moles of KOH added = 5 x 10-4

moles of HF remaining = 95 x 10-4

moles of KF formed = 5 x 10-4

now

we know that

pH = pKa + log [salt / acid ]

so

pH = 3.17 + log [ 5 x 10-4 / 95 x 10-4 ]

pH = 1.89

c)

now

moles of KOH added = 0.1 x 50 x 10-3 = 50 x 10-4

now

KOH + HF ---> KF + H20

moles of HF reacted = moles of KOh added = 50 x 10-4

moles of HF remaining = 50 x 10-4

moles of KF formed = 50 x 10-4

now

pH = pKa + log [ KF / HF]

pH = 3.17 + log [ 50 x 10-4 / 50 x 10-4 ]

pH = 3.17

d)

at equivalence point

moles of KOH = moles of HF = 100 x 10-4

now

the reaction is

KOH + HF ---> KF + H20

moles of KF formed = 100 x 10-4

now

KF is a weak base

so

[OH-] = sqrt ( kb x C)

also

Kb = kw/ ka

so

kb = 10-14 / 6.76 x 10-4

Kb = 1.48 x 10-11

so

[OH-] = sqrt ( 1.48 x 10-11 x 100 x 10-4 )

[OH-] = 3..846 x 10-7

pOH = -log [OH-]

so

pOH = -log 3.846 x 10-7

pOH = 6.415

now

pH = 14 - pOH

so

pH = 14 - 6.415

pH = 7.585

so

pH at equivalence point is 7.585

e)

now

moles of KOH added = 101 x 10-4

moles of HF = 100 x 10-4

the reaction is

HF + KOH ----> KF + H20

moles of KOH remainign = 1 x 10-4

now

[OH-]= [KOH ] = 10-4

pOH = -log [OH-]

pOH = -log 10-4

pOH = 4

so

pH = 14 - 4

pH = 10

so

the pH is 10