anti logs

[furyhecla]

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A question in a practice test gives pka = 6.4 (the ultimate question is to figure out pH, but I need to get through the first part of the answer, which I don't understand).

In the answer (in part of the process to discuss the answer to the question), it states to get the Ka, one has to take the anti log (or inverse log) of -6.4.

They say that Ka = 4 x 10^-7.

I am not sure how to do this....

I understand this logic (which is part of the answer as well):
If log x = -1, then x = 0.1
If log x = -2, then x = 0.01
If log x = -3, then x = 0.001...

So I understand if log x = 6.4, then x will be ~1 x 10^-6...

but why is it 10^-7, and where does the 4 come from?

Thanks for any tips

 

[indya]

to find the anti log, just take 10 to the (negative)pka.

ka= 10^-pka= 10^-6.4=4.0x10^-7


this works because pka= -log(Ka)

 

[BerkReviewTeach]

Thanks for any tips

Why you are converting from pKa to Ka on a question that is asking for pH? The biggest tip I can suggest is that it is highly unlikely you need to do this for such a question. Because the pKa is 6.4, it's a weak acid in solution. So it's either an aqueous solution of the weak acid or a buffer containing the weak acid and its conjugate base, which in either case can be solved in about ten to fifteen seconds using pKa (not Ka).

 

[furyhecla]

10^-6.4=4.0x10^-7

Thanks for replying... this is the exact part I don't get.... it must be some rule of logarithm manipulation that I don't understand....

Where does the 4 come from? In 4 x 10^-7... where does the '-7' come from? I guess my question is how does one solve a power that is a decimal?

Thanks...

 

[GreenRabbit]

I think you should spend some time reviewing logs outside of SDN...

If you have 10^-6.4, it's somewhere between 10^-6 and 10^-7, right?
Well 4 x 10^-7 is between 10^-6 and 10^-7, a more precise answer is 3.98 x 10^-7. Neither of those numbers "come" from anything.

 

[furyhecla]

@Greenrabbit.. thanks that makes a lot of sense. That was the logic that I had missed.

 

[gunj122]

Why you are converting from pKa to Ka on a question that is asking for pH? The biggest tip I can suggest is that it is highly unlikely you need to do this for such a question. Because the pKa is 6.4, it's a weak acid in solution. So it's either an aqueous solution of the weak acid or a buffer containing the weak acid and its conjugate base, which in either case can be solved in about ten to fifteen seconds using pKa (not Ka).

just to be clear.. pH is a measure of concentration, correct? and pKa a property of the acid itself, at a given temperature?

so in order to answer to OP's question, there must be a given concentration?