pKa and pKb

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Bryce Ramirez 1J
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Joined: Sat Aug 24, 2019 12:16 am

pKa and pKb

Postby Bryce Ramirez 1J » Sun Feb 02, 2020 11:41 pm

Can someone explain the basics to pKa and pKb. I thought it was the product of the two equals pKw and that their product is equal to 14. And when calculating it, you just need to do the -log of the ja or kb. Is there anything else I should know fundamentally when applying this to a problem?

Aiden Metzner 2C
Posts: 104
Joined: Wed Sep 18, 2019 12:21 am

Re: pKa and pKb

Postby Aiden Metzner 2C » Sun Feb 02, 2020 11:44 pm

You know the most essential parts. Just know that for acidic reactions you should apply pKa. If you are only given pKb then you can calculate pKa because their products are equal to 14.

Posts: 102
Joined: Wed Sep 18, 2019 12:15 am

Re: pKa and pKb

Postby 005162902 » Mon Feb 03, 2020 11:02 pm

pKa is -logKa and pKb is -logKb. If you are familiar with the equation Kw= Ka x Kb and the equation Kw= 1.0x 10^-14, you can make the equation Ka x Kb= Kw= 1.0x 10^-14. Then, you can fill multiply the equation by -log and convert it to Kb and pKa. When you do this, you get the equation pKb + pKa = 14.

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Joined: Thu Jul 11, 2019 12:16 am

Re: pKa and pKb

Postby 805312064 » Thu Feb 06, 2020 10:00 pm

Also, Ka is calculated using the equation [H3o+][A-]/[H], while Kb is calculated using [OH-][BH+]/{B]

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