## pKa and pKb

Artin Allahverdian 2H
Posts: 76
Joined: Fri Sep 28, 2018 12:26 am

### pKa and pKb

If we are given the acid and it’s pKa, but we want to solve for the pH of its base, do we need to convert pKa to pKb?

Chem_Mod
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### Re: pKa and pKb

Yes, you would convert pKa to pKb using pKa+pKb=14 and then convert pKb to Kb using pKb= -logKb.

Briana Lopez 4K
Posts: 44
Joined: Wed Nov 22, 2017 3:01 am

### Re: pKa and pKb

when given pKa and asked for ph, solve for Ka with
pKa= -logKa
which would mean
10^(-pKa)= Ka

Then, if the reaction is acidic, find the Ka equilibrium equation for the reaction
Usually, an initial Molarity will be given for the Reactant
Once the Ka equation is written, set it equal to the Ka value calculated previously ^

When you solve for x, it should be the concentration of [H3O+]
Plug that into
pH= -log[H3O+] to obtain pH

The only way I see you needing pKb is if it is a basic reaction

Claudia Luong 4K
Posts: 59
Joined: Fri Sep 28, 2018 12:25 am

### Re: pKa and pKb

Yes. To calculate pH for base, we would need to use Kb and not Ka.

MichaelMoreno2G
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

### Re: pKa and pKb

Briana Lopez 4K wrote:when given pKa and asked for ph, solve for Ka with
pKa= -logKa
which would mean
10^(-pKa)= Ka

Then, if the reaction is acidic, find the Ka equilibrium equation for the reaction
Usually, an initial Molarity will be given for the Reactant
Once the Ka equation is written, set it equal to the Ka value calculated previously ^

When you solve for x, it should be the concentration of [H3O+]
Plug that into
pH= -log[H3O+] to obtain pH

The only way I see you needing pKb is if it is a basic reaction

Thank you so much. I was confused on how to do this problem but this clarified it well.

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