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Thetis

Confusion Over Acid Base Log Rules

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Hoping someone can help me understand logarithm rules better when it comes to acid-base chemistry. I understand how to approximate pH, pOH, pKa and pKB given a concentration or a Ka/Kb. For example:

[H3O+]= 8.89 x 10^(-4) so pH = 4-0.9, pH = 3.1

 

I don't understand how to do the reverse and approximate [H3O+] or [OH-] from a given pOH or pH. For example, how do I find [OH-] given a pH of 3.1?

 

Or how would I find [H3O+] given a pH of 8.2??

 

Thanks so much for your help, this is really bugging me!

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So to convert pH to [H], or pOH to [OH], compute 10 to the exponent of the negative pH or pOH

 

i.e., if the pH of your standard premed student is 1, what is their [H] concentration?

 

pH = 1, therefore [H] = 10^-1 = 0.1

 

Note you can also convert [H] to [OH] and vice versa by the formula Kw=[OH][H], where Kw= 10^-14. so even with pH, you can calculate OH

 

Hope this helps!

 

-A

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Hey thanks for the help. Those Leah4Sci MCAT math videos are really good. The reason I was asking was because in my Kaplan 2015 prep book there is a fill in the blank question (General Chem, p. 347) where they give you a pOH, for example 5.19 and you are supposed to find the [OH-].

Usually in the book it says when approximating something is necessary (i.e. earlier on it says the quadratic formula doesn't need to be used on the MCAT) but in this case in the answers they have the table completely filled out with specific values, such as the [OH-] for a pOH of 5.19 is "6.46x10^-6". I was wondering how you would arrive at that answer just by approximating.

I guessed that the [OH-] would be approximately 10^-5 but they have a much more specific value.

 

I guess according to the Leah4Sci video I could estimate the number will be between 10^-5 and 10^-6, and then use a memorized range where .1=8*10^-..., .3=5*10^-..., .5=3*10^... so then I know it is some number between 8*10 and 5*10, and if I choose for example 6*10^-6 I would get a pOH of 6-0.6 = 5.4 which is close enough to the given pOH of 5.19. But if I use 6*10^-5 I get a pOH of 5-0.6 = 4.4 which undershoots th given pOH. So I go with [OH-]= 6*10^-6.

 

Is that the trick everyone else is using? The memorized range of .1 = 8*10^-x, .3 = 5*10^-x, .5 = 3*10^-x, .7 = 2*10^-x?

And does the same range work for positive logs or not?

 

Thanks!!

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