Question

A .025 mol sample of a weak acid, HA, is dissolved in 485 mL of water and titrated with .41M NaOH. After 29 mL of the NaOH solution has been added the overall pH = 5.414. Calculate the Ka value for HA.

Answer 1

we know that

moles = molarity x volume (L)

so

moles of NaOH added = 0.41 x 29 x 10-3

moles of NaOH added = 11.89 x 10-3

now

the reaction is

HA + NaOH ---> NaA + H20

we can see that

moles of HA reacted = moles of NAOh added = 11.89 x 10-3

so

moles of HA left = 0.025 - 11.89 x 10-3 = 0.01311

now

moles of NaA formed = moles of NaOH added = 11.89 x 10-3

all the NaOH is consumed

and

HA and NaA are left in the solution

HA and NaA form a buffer

and for buffers

pH = pKa + log [salt / acid ]

so

pH = pKa + log [ NaA / HA]

5.414 = pKa + log [ 11.89 x 10-3 / 0.01311]

pKa = 5.456

now

pKa = -log Ka

so

5.456 = -log Ka

Ka = 3.496 x 10-6

so

the value of Ka is 3.496 x 10-6