some pH calculations

[Addallat]

First Question


Molar Concentration of HCl: 5*10^-5 M
What is the pH?



Since it's a strong acid, then pH will = the negative log of the initial concentration of the acid

so my question is, since 5 is greater than the square root of 10 (3.16) then is the pH closer to 4 than it is to 5?

will the pH be roughly 4.3ish?



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[Addallat]

This is what I've been taught in dealing with molar concentrations of a strong acid that:



you have a molar concentration

3.16 * 10^-10

instead of saying pH is exactly 10 ( the -log of M) , you have to account for the 3.16

since 3.16 is approximately the square root of 10 then it must follow that pH must be about 9.5

 

[Synapsis]

TBR has a fantastic strategy for this. It takes just a little getting used to.

Memorize:
log10= 1
log2= 0.3
log3= 0.477
-logx * 10^-y = y-logx

Everything you get, transform into those memorized numbers. So if we apply that to your example...

-log(5 x 10^-5)

Using -logx * 10^-y = y-logx, you end up with pH= 5 - log(5)

Now log(5) you need to convert to some combination of log2, log10, and/or log3, which we memorized. So log(5) can be rewritten as log(10/2)...log10 - log2 = 1 - 0.3

Substitute that back above for pH = 5 - (1 - 0.3)...pH = 5 - (0.7)...pH = 4.3

Hopefully I didn't make a math error there. There's a couple ways to either estimate or calculate. I know a lot of people prefer estimation, but this is the method for me. Takes about 10-15 seconds, and it's remarkably accurate.

 

[HesitantManatee]

TBR has a fantastic strategy for this. It takes just a little getting used to.

Memorize:
log10= 1
log2= 0.3
log3= 0.477
-logx * 10^-y = y-logx

Everything you get, transform into those memorized numbers. So if we apply that to your example...

-log(5 x 10^-5)

Using -logx * 10^-y = y-logx, you end up with pH= 5 - log(5)

Now log(5) you need to convert to some combination of log2, log10, and/or log3, which we memorized. So log(5) can be rewritten as log(10/2)...log10 - log2 = 1 - 0.3

Substitute that back above for pH = 5 - (1 - 0.3)...pH = 5 - (0.7)...pH = 4.3

Hopefully I didn't make a math error there. There's a couple ways to either estimate or calculate. I know a lot of people prefer estimation, but this is the method for me. Takes about 10-15 seconds, and it's remarkably accurate.

+1


pH calculations questions are free points on the MCAT