Question

You determine that it takes 19.26 ml of base to neutralize a 25 ml sample of your unknown acid solution. The pH of the solution when about 9.61 ml of base had been added was 2.90. You notice the concentration of the base was 0.1232 M.

a) What are the Ka and pKa of this unknown acid? (Show work)

b) What is the possible identity of this unknown acid?

c) What is the concentration of the unknown acid?

Answer 1

we know that

at equivalence point

Ma x Va = Mb x Vb

so

Ma x 25 = 0.1232 x 19.26

Ma = 0.0949

so

the concentration of acid is 0.0949

now

we know that

moles = conc x volume (ml) / 1000

so

moles of base added = 0.1232 x 9.61 x 10-3 = 1.183952 x 10-3

now

moles of acid = 0.0949 x 25 x 10-3 = 2.372832 x 10-3

now

acid + base ---> salt + water

we can see that

moles of acid reacted = moles of base added = 1.183952 x 10-3

moles of salt formed = moles of base added = 1.183952 x 10-3

now

finally

moles of acid left = 2.372832 x 10-3 - ( 1.183952 x 10-3)

moles of acid left = 1.18888 x 10-3

now

pH = pKa + log [salt /acid]

so

2.9 = pKa + log [ 1.183952 x 10-3 / 1.18888 x 10-3 ]

pKa = 2.9018

so

pKa of the acid is 2.9018

now

pKa = -log Ka

so

2.9018 = -log Ka

Ka = 1.2537 x 10-3

so

Ka of this acid is 1.2537 x 10-3

b)

the acid may be chloroacetic acid

c)

concentration of unknown acid is 0.0949 M