Question

An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?

Answer 1

`pKa=3.58`

Step-by-step explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

`pH=-log([H^+])`

`[H^+]=10^(-pH)=10^(-3.4)=3.98x10^(-4)`

And the percent ionization is:

`\% \ ionization=([H^+])/([HA])*100\%`

We compute the concentration of the acid, HA:

`[HA]=([H^+])/(\% \ ionization)*100\%=(3.98x10^(-4))/(66\%) *100\%\\\\`

`[HA]=6.03x10^(-4)`

Thus, the Ka is:

`Ka=([H^+][A^-])/([HA])=(3.98x10^(-4)*3.98x10^(-4))/(6.03x10^(-4))\\ \\Ka=2.63x10^(-4)`

So the pKa is:

`pKa=-log(Ka)=-log(2.63x10^(-4))\\\\pKa=3.58`

Regards.