Calculating Ka from pKa

[Mrudula Akkinepally]

Hello everyone!

I know that pka is the negative log of ka, but how do we go from pka to ka?

Thank you :)

[Liam Bertrand 3]

When you have log10(a)=c, to get to exponential form it would be 10^c= a. Thus, since the pka is c in this example, raise 10 to whatever the pka is and it should give you ka. Someone might want to double check me on that one though because its been a while since I've done that kind of math

[Emily Vu 1L]

Hi! So the log rules are exactly the same if we were to go from pH to [H₃O+], and vice versa. So to convert from pKa to Ka, you take 10^-pKa.

Hope this helps!

[Tessa House 3A]

Because -log(Ka)=pKa, you can rearrange this to get 10^-pKa=Ka. Plug in the pKa and you can get the Ka. You can always just rearrange the original equation to solve for Ka. I hope this helps!

[Ziyan Peng 3A]

Ka=10^(-pKa), just like how [H3O+]=10^(-pH) and how [OH-]=10^(-pOH). I find grouping these together easier to remember, especially if you don't want to run the math through your head every time like me :)

[Nishka Vipul 1J]

Hello everyone!

I know that pka is the negative log of ka, but how do we go from pka to ka?

Thank you :)

Hi! To calculate Ka from pKa you would use this: Ka=10^(-pKa)! Hope this helps. I would recommend reviewing log rules, it really helped me understand this so that it's not just another equation to memorize.

[Stephen Min 1I]

To derive Ka, you would 10^-(pKa). This is because pKa is in the form of -log[Ka]. Its the same format as converting between pH and [H3O+].

[Nayra Gharpetian 3F]

to find Ka from pKa you would do 10^(-pKa)