Gen Chem math problem

[Murl]

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Is there an easy way to do the inverse log function with paper and pencil. I understand the chemistry in calculating pH or pOH with a calculator but can't do it without. Will you be expected to do it on the DAT?

How hard are the questions on the DAT compared to Destroyer?

 

[Kneecoal]

it's actually pretty easy with a little bit of practice. start with an easy one. if your acid concentration is 1x10^-3M...

pH = -log[H+]
pH = -log(1x10^-3)

because your "initial" number in the molarity is 1, your pH is simply the opposite sign of the -3 in the exponent. so here pH = 3.

but say if your acid concentration is something like 4x10^-3M, you need to get a little more creative. where above the pH was strictly 3 because your "initial" number was 1, here because your "initial" number is higher than 1, your pH is now going to be between 3 and 3-1, so between 2 and 3.

now go back to this problem's "initial" number: 4. because your "initial" number will pretty much always be between 1 and 9 with these problems, think on a scale of 1 to 9. the closer your initial number is to the "1" side, the closer your pH will be to the higher number (in our case, closer to 3, instead of 2), and the closer you are to the "9" side, the closer your pH will be to the lower end of our possible pH. off the top of my head, i would guess with our "initial" number of 4, our pH to be around 2.5. just checked it on the calculator, and it says 2.4.

hopefully that helps a little bit. if you can get the hang of it, you shouldn't have to worry too much, because i'm pretty sure they would never give you two choices very close to each other.

 

[Murl]

it's actually pretty easy with a little bit of practice. start with an easy one. if your acid concentration is 1x10^-3M...

pH = -log[H+]
pH = -log(1x10^-3)

because your "initial" number in the molarity is 1, your pH is simply the opposite sign of the -3 in the exponent. so here pH = 3.

but say if your acid concentration is something like 4x10^-3M, you need to get a little more creative. where above the pH was strictly 3 because your "initial" number was 1, here because your "initial" number is higher than 1, your pH is now going to be between 3 and 3-1, so between 2 and 3.

now go back to this problem's "initial" number: 4. because your "initial" number will pretty much always be between 1 and 9 with these problems, think on a scale of 1 to 9. the closer your initial number is to the "1" side, the closer your pH will be to the higher number (in our case, closer to 3, instead of 2), and the closer you are to the "9" side, the closer your pH will be to the lower end of our possible pH. off the top of my head, i would guess with our "initial" number of 4, our pH to be around 2.5. just checked it on the calculator, and it says 2.4.

hopefully that helps a little bit. if you can get the hang of it, you shouldn't have to worry too much, because i'm pretty sure they would never give you two choices very close to each other.

thanks I appreciate it, that makes more sense

 

[jay47]

Remember that -log(X) is the same as 1/(log[X]), that explains why as your number in front of the exponent gets larger, your pH actually drops. Thought I would throw that in just because I nearly forgot that when I was reading your explanation and thought, "waa?". Good explanation btw.

 

[nze82]

Here is a little short cut for you:

Let's say you've found [H+] = 4x10^-12

pH = -log (4x10^-12) = 12 - log(4)

Remember that log(1) = 0; therefore, log(4) must be a bit larger than 0, but still smaller than 1 (remember that log(10) = 1). In fact, log(4) ~ 0.6.
Therefore:

pH = 12 - 0.6 ~ 11.4

Check with your calculator and you'll find out that -log (4x10^-12) = 11.39
I'd say 11.4 is a very accurate estimate😉

In general:

-log(mx10^-n) = n - log(m)

 

[Murl]

Thanks for all of your help, these suggestions are already helping me.