Antilog

[kmp0410]

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Similar threads are all very old so I'm starting a new one. Let me see if I'm understanding how to calculate this correctly.

I'm using BR shortcut log method, not sure if other places are different, but I like this one so I'd rather not learn something else.

So I'm finding the [H+] and I have pH. So [H]= 10^-pH.

Say my pH is 3.7 so my [H]= 10^-3.7. So now my goal from here should be to break this into two parts, one being a number I know the log to, correct?

So I would break it into 10^0.3 * 10^-4.

Since Log 2 = 0.3 then 10^0.3 = 2

Giving me [H] = 2 * 10^-4 I understand this, when the number is easily broken down into a whole number and a log I know.

But say my pH = 4 so I'm at 10^-4, I am unsure on how to proceed with this.

Maybe 10^1 * 10^-5

Since Log 10 = 1 then 10^1 = 10

So then I'm at 10 x 10^-5 = 1.0 x 10-4?

Seems simple, is there a short cut/concept for whole numbers that I'm missing?

Thanks

 

[kmp0410]

Well I think I was overthinking the whole number pH thing.

If pH = x then [H+] = 1.0 * 10^-x

Is this correct (for strong acids)?

 

[mehc012]

Well I think I was overthinking the whole number pH thing.

If pH = x then [H+] = 1.0 * 10^-x

Is this correct (for strong acids)?

Yes, this is correct at all times.

-log([H+]) = pH

log([H+]) = -pH

10^(log([H+])) = 10^-pH

[H+] = 10^-pH