Convert pKa to Ka and vice versa

[Melomare17]

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How do you calculate the pKa to Ka and vice versa in your head?? Any shortcuts/simple steps?

 

[STxHorn]

How do you calculate the pKa to Ka and vice versa in your head?? Any shortcuts/simple steps?

If you can picture logs in your head, then yes.

It's not hard if you've never tried it. It's easy to approximate if you think of everything in scientific notation.

 

[MedPR]

How do you calculate the pKa to Ka and vice versa in your head?? Any shortcuts/simple steps?

log2=0.3
log3=0.477

logAB = logA + logB

So for Ka = 2*10^-7:
pKa=-logKa
pKa=-log(2*10^-7)
pKa=-log(10^-7) - log2
pKa=7-0.3
pKa=6.7.

 

[nico vazquez]

10^-ka

 

[Melomare17]

log2=0.3
log3=0.477

logAB = logA + logB

just something you memorize?

 

[MedPR]

just something you memorize?

Yeah. It makes calculating logs a lot simpler. You can get to just about any number from 2 and 3. For example, log8 is just log2+log2+log2 = 0.3*3.

log9 = log3+log3

etc.

 

[folktale]

Yeah. It makes calculating logs a lot simpler. You can get to just about any number from 2 and 3. For example, log8 is just log2+log2+log2 = 0.3*3.

log9 = log3+log3

etc.

Converting for example [H+] to pH seems pretty simple, but I always have a hard time doing vice versa, especially with non-whole number pH values.

For example, a pH of 9.6, to find the [H+], it'd be:

9.6 = -log[H+]

and I'd end up with [H+] = 10^-9.6

I understand that it'd be some value of [H+] = # x 10^-10, but I don't know how to get that # value. From the looks of it, it's going to be the antilog of 0.4, since that's the difference between 10-9.6, but it's just giving me a hard time. Any shortcuts to this? What would be that # value on this case?

 

[SaintJude]

Good question! Would like to know myself.

 

[MedPR]

Converting for example [H+] to pH seems pretty simple, but I always have a hard time doing vice versa, especially with non-whole number pH values.

For example, a pH of 9.6, to find the [H+], it'd be:

9.6 = -log[H+]

and I'd end up with [H+] = 10^-9.6

I understand that it'd be some value of [H+] = # x 10^-10, but I don't know how to get that # value. From the looks of it, it's going to be the antilog of 0.4, since that's the difference between 10-9.6, but it's just giving me a hard time. Any shortcuts to this? What would be that # value on this case?

Good question! Would like to know myself.

Yes, it is the antilog of 0.4. log2=0.3, log3=0.48, so the number value is about halfway between 2 and 3 but a little closer to 3. So probably like 2.6*10^-10.

 

[folktale]

Yes, it is the antilog of 0.4. log2=0.3, log3=0.48, so the number value is about halfway between 2 and 3 but a little closer to 3. So probably like 2.6*10^-10.

So another example would be 10^0.6

since log 2 = 0.3, 2*log2 = 0.6, so it would be 4?

log2 + log2 = log (2*2) = log 4 = 0.6

 

[MedPR]

Yup!