A little help solving log problems

[UFRalo]

So, I know this is an important thing to know...I just can't seem to wrap my head around it. Anybody know a good tool to figure these problems out or can just give me a simple explanation?

For example, what is -log 8.6x10^-3

Simple right lol

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[Avery07]

These are really simple.

8.6 x 10^-3

First, you write the 10^-3 part as 3.0.

Now you look at the 8.6 part. If it is lower (0-4) it will be about 2.5-3.0 (closer to your number). However if it is a high number (6-9), it will be closer to the number below the 10^-x part or in this case, 2 (2.0-2.5)

So for -log 8.6x10^-3, we start with 3, but 8.6 is closer to 2 than 3 so our pH will be somewhere between 2-2.5.

The actual pH here is 2.06.



Another example--

-log 2.4 x 10^-11

1. 11.0
2. 2 is closer to 0 than it is to 10
3. The antilog must be close to 11 so let's say our pH is 10.50-11.00
4. Actual pH or antilog value is 10.61

 

[lex1489]

Good explanation by Avery. Just another thing to help with the thought process-- Assuming we're working with H+ concentration here and trying to get a pH, the 8.6 means the concentration of H+ is LARGER than if it were 1 x 10^-3.

A larger concentration gives a LOWER pH value. When I think about it in terms of the how the change in concentration directly affects acidity/basicity, I tend to avoid moving the pValue in the wrong direction.

 

[UFRalo]

thanks guys