Finding Kb from pOH

[the_fella]

I have a practice problem that gives me only the pOH (5.19) and wants me to find pH, Ka, Kb, and tell if it's an acid or base. I know pH is 8.81, and it's thus a base. But beyond this, I'm stuck. Any suggestions?

EDIT: Ok. I'm stupid. They want the Hydroinum and Hydroxide ion concentrations, which I still don't know how to find.

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[Labrat07]

What's the pH and pOH?? you need to understand the formula not just plug and chug. Hydronium = H +, Hydroxide = OH-. We can give you the answers but i think it's more beneficial for you to find the method

 

[amy_k]

This page gives you the formulas you need: http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm

Your method would be to calculate the [OH-] from pOH, then find pH from pOH which you have done, and finally calculate [H+] from pH.

 

[the_fella]

This page gives you the formulas you need: http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm

Your method would be to calculate the [OH-] from pOH, then find pH from pOH which you have done, and finally calculate [H+] from pH.

Thank you. I suspected as much. Now, my question, is how does one do that without a calculator? How do I get from 10^-8.81 to 6.46*10^-6 (which is the answer given), without a calculator?

 

[NextStepTutor_1]

pOH + pH = 14
You knew this and were able to find the pH which equals 8.8

When you see a "p" think "-log(...)" where the "..." is whatever comes after the "p"
So in using the pH, we know it is also equal to the -log[H]
8.8 = -log[H]
so [H] = 10^-8.8, round this if you need to to 10^-9
As you can see, we have a very low conc. of acid, which, for verification purposes, confirms your initial assessment that we're dealing with a basic solution

Now do the same to find the [OH] using the pOH.

 

[the_fella]

What's the pH and pOH?? you need to understand the formula not just plug and chug. Hydronium = H +, Hydroxide = OH-. We can give you the answers but i think it's more beneficial for you to find the method

I stated the pH and pOH in my OP. The pOH was given and is 5.19 From there, I calculated the pH as 8.81, making it basic. I understand that pH is the -log of [H3O] and pOH = -log [OH]. I know I need to take the antilog, but Idk how to arrive at the answers they give by doing so. For instance 10^-5.19 would be the [OH], but the answer says that equals 6.46*10^-6. How do I get that without a calculator?

 

[krukshanks]

I stated the pH and pOH in my OP. The pOH was given and is 5.19 From there, I calculated the pH as 8.81, making it basic. I understand that pH is the -log of [H3O] and pOH = -log [OH]. I know I need to take the antilog, but Idk how to arrive at the answers they give by doing so. For instance 10^-5.19 would be the [OH], but the answer says that equals 6.46*10^-6. How do I get that without a calculator?

Usually the answer choices are very different in magnitude. To estimate the value of 10^-5.19, you really just need to know that the value is between 10^-5 and 10^-6. 6.46*10^-6 is probably the only answer within that range.

 

[the_fella]

Usually the answer choices are very different in magnitude. To estimate the value of 10^-5.19, you really just need to know that the value is between 10^-5 and 10^-6. 6.46*10^-6 is probably the only answer within that range.

This particular problem is a problem in the Kaplan review. It doesn't give choices, just asks us to calculate the answer. Your answer is right. But how do you know it's not closer to 5?

 

[krukshanks]

This particular problem is a problem in the Kaplan review. It doesn't give choices, just asks us to calculate the answer. Your answer is right. But how do you know it's not closer to 5?

I could assure you this is not required for MCAT. But if you really want to know how to estimate this, here is my strategy:
We know the answer is between 10^-5 and 10^-6, aka 10*10^-6 and 1*10^-6, and thus the answer should be in the form of "something"*10^-6, with "something" being a number between 1 and 10. We want to estimate "something".

"something"*10^-6 = 10^-5.19 ~ (10^0.8)*(10^-6)

Thus "something" ~ 10^0.8 = (10^0.5)*(10^0.3)
10^0.5 is just the root of 10, which is around 3 (because 3*3=9~10)
10^0.3 ~ 10^(1/3), which is a tiny bit more than 2 (because 2^3=8~10)
So "something" should be around 2*3 = 6, which is around the exact answer, 6.46.

That would be how you would estimate stuff like this, but you needn't and shouldn't learn this: MCAT doesn't test this, and this is a complete waste of time for the sake of MCAT prep. This might be useful if you do highly quantitative job in the future and need to give something a rough estimation right away, but I highly doubt that this thing would ever be useful as you could almost always access a calculator/wolfram alpha.