Calculate Ka for an acid?

This is a very dilute acid solution - the [H+] from the auto-ionisation of the water near to pH 7.00 has to be taken into account .

The [H+] in the solution is:

[H+] = 10^-pH

[H+] = 10^-6.24

[H+] = 5.75*10^-7

The [H+] comes from 2 sources: 1) the H+ contributed by the acid and 2) the H+ from the dissociation of the water

But we cannot say that [H+] from the water = 10^-7 , so the balance has come from the acid. When you add the acid to the water , it produces H+ions. This causes the dissociation equilibrium of the water:

H2O ↔ H+ + OH- to be upset , and the equilibrium shifts to the left. The [H+] from the water is less than 10^-7.

You cannot say: [H+] for the acid = (5.75*10^-7) - (1*10^-7) = 4.75*10^-7

Let the [H+] from the dissociation of water = X . Then [OH-] from the water must = X

Kw = [H+] [OH-]

We know that [H+] in total = (5.75*10^-7)

Substitute and solve for [OH-]:

[OH-] = (10^-14) / ( 5.75*10^-7)

[OH-] = 1.74*10^-8

But [OH-] = [H+] from the dissociation of the water

Therefore [H+] from the acid must be: (5.75*10^-7) - ( 1.74*10^-8)

[H+] from the acid = 5.58*10^-7

We now substitute this into the Ka equation:

Ka = [H+]² / [acid]

Ka = (5.58*10^-7)² / (0.0195 - ( 5.58*10^-7) )

Ka = (5.58*10^-7)² / 0.0195

Ka = 3.114*10^-13 / 0.0195

Ka = 1.60*10^-11

That is not 1.7*10^-11 but hardly different. Do you have the correct Ka value?

Bro the question's asking for a pKa not a Ka...

Wanted to submit this question too this evening

No idea what to say