pKa and pKb
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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?
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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.
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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
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
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- Joined: Fri Sep 28, 2018 12:25 am
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- 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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