Question

A 0.025 mol sample of a weak acid, HA, is dissolved in 467 mL of water ant titrated with 0.41 M NaOH. After 33 mL of the NaOH solution has been added, the overall pH=5.580. Calculate the Ka value for HA.

Answer 1

we know that

moles = molarity x volume (L)

so

moles of NaOH added = 0.41 x 33 x 10-3

moles of NaoH added = 13.53 x 10-3

now

the reaction is

HA + NaOH --> NaA + H20

we can see that

moles of HA reacted = moles of NaOH added = 13.53 x 10-3

so

moles of HA remaining = 0.025 - 13.5 x 10-3 = 0.01147

now

moles of NaA formed = moles of NaOH added = 13.53 x 10-3

now

HA and NaA form a buffer

we know that

for buffers

pH = pKa + log [salt / acid ]

so

pH = pka + log[ NaA / HA]

so

5.58 = pKa + log [ 13.53 x 10-3 / 0.01147]

pKa = 5.508

now

we know that

pKa = -log Ka

so

5.508 = -log Ka

Ka = 3.102 x 10-6

so

Ka value for HA is 3.102 x 10-6