Question

1.00L of 1.45M solution of weak acid, HA (Ka = 3.21E-4) is titrated with 1.45M NaOH

a) 200ml of NaOH is added
b) 500ml of NaOH is added
c) 1000ml of NaOH is added

Answer 1

we know that

moles = conc x volume (L)

so

moles of HA = 1.45 x 1 = 1.45

a) now 0.2 L of NaOH is added

so

moles of NaOH added = 1.45 x 0.2 = 0.29

now

NaOH + HA ---> NaA + H20

moles of HA reacted = moles of NaoH added = 0.29

moles of HA remaining = 1.45 - 0.29 = 1.16

moles of NaA formed = 0.29

now

HA and NaA form a buffer solution

we know that

for buffers

pH = pKa + log [ salt / acid ]

also

pKa = -log Ka

so

pH = -log Ka + log [ NaA / HA]

pH = -log 3.21 x 10-4 + log [ 0.29 / 1.16]

pH = 2.89

2)

now 0.5 L of NaOH is added

so

moles of NaOH added = 1.45 x 0.5 = 0.725

now

NaOH + HA ---> NaA + H20

moles of HA reacted = moles of NaoH added = 0.725

moles of HA remaining = 1.45 - 0.725 = 0.725

moles of NaA formed = 0.725

now

HA and NaA form a buffer solution

we know that

for buffers

pH = pKa + log [ salt / acid ]

also

pKa = -log Ka

so

pH = -log Ka + log [ NaA / HA]

pH = -log 3.21 x 10-4 + log [ 0.725 / 0.725]

pH = 3.49

3)

now 1 L of NaOH is added

so

moles of NaOH added = 1.45 x 1 = 1/45

now

NaOH + HA ---> NaA + H20

moles of HA reacted = moles of NaoH added = 1.45

moles of HA remaining = 1.45 - 1.45 = 0

moles of NaA formed = 1.45

final volume = 1 + 1 = 2 L

[NaA] = 1.45 / 2 = 0.725 M

now

only NaA remains in the solution

it undergoes hydrolysis

A- + H20 ---> HA + OH-

now

NaA is a weak base

so

for weak bases

[OH-] = sqrt ( Kb x C)

also

Kb = Kw / Ka

Kb = 10-14 / 3.21 x 10-4

Kb = 3.115 x 10-11

so

[OH-] = sqrt ( 3.115 x 10-11 x 0.725)

[OH-] = 4.75 x 10-6

now

pOH = -log [OH-]

so

pOH = -log 4.75 x 10-6

pOH = 5.323

now

pH = 14 - pOH

so

pH = 14 - 5.323

pH = 8.677