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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: A5HLEY on March 09, 2008, 03:16:41 PM

Title: Calculate [OH-] and pH
Post by: A5HLEY on March 09, 2008, 03:16:41 PM: Calculate [OH-] and pH for the following:

a). .210M KClO
b). .451M Ba(C2H3O2)2
Title: Re: Calculate [OH-] and pH
Post by: Borek on March 09, 2008, 03:25:59 PM: Try by yourself first. That's in forum rules (http://www.chemicalforums.com/index.php?page=forumrules).
Title: Re: Calculate [OH-] and pH
Post by: A5HLEY on March 09, 2008, 08:47:32 PM: Ok, well I know that

Kb = Kw/Ka where Kw = 1E-14

I set up an ICE table:
base- + H20 <==> HBase + OH-
Initial .210 0 0
Change -x +x +x
Equilibrium .210-x +x +x

And I also know that pH = 14 - pOH

But I don't know what to do from there...
Title: Re: Calculate [OH-] and pH
Post by: Borek on March 10, 2008, 04:47:11 AM: Put information from ICE table into dissociation constant and solve for x. You will need Ka/Kb values, these should be in your book.
Title: Re: Calculate [OH-] and pH
Post by: A5HLEY on March 10, 2008, 02:36:23 PM: So do I use this formula:

Kb = x^2/(.210-x)

and will x be [OH-]?
Title: Re: Calculate [OH-] and pH
Post by: Borek on March 10, 2008, 02:48:44 PM: Looks like.
Title: Re: Calculate [OH-] and pH
Post by: A5HLEY on March 10, 2008, 03:42:44 PM: Ok, can someone PLEASE PLEASE check this for me. I did all the work and everything... I just need you to check it. I have to submit it to WebAssign tonight, and I only have one submission. I've bolded my actual answers

Ok, so for a).
.210 M KOH
base- + H20 <==> HBase + OH-
Initial .210 0 0
Change -x +x +x
Equilibrium .210-x +x +x

Kb = Kw/Ka = (1E-14)/(3.0E-8)
Kb = 3.33E-7

So Kb = (x^2)(.210-x), thus

3.33E-7 = (x^2)/(.210-x)
x = [OH-] = 2.6428E-4

pOH = -log[OH-] = -log(2.6428E-4) = 3.57
pH = 14 - pOH = 10.43

b). .415M Ba(C2H3O2)2

base- + H20 <==> HBase + OH-
Initial .415 0 0
Change -x +x +x
Equilibrium .415-x +x +x

Kb = Kw/Ka = (1E-14)/(1.8E-5) = 5.56E-10

5.56E-10 = (x^2)/(.415-x)
x= [OH-] = 1.519012837E-5

pOH = -log(1.519012837E-5) = 4.82
pH = 14 - pOH = 9.18

If you check this for me, I will be FOREVER greatful.. thank you!
Title: s
Post by: A5HLEY on March 10, 2008, 03:43:04 PM: Sorry, I meant to edit, not quote. :)
Title: Re: Calculate [OH-] and pH
Post by: Borek on March 10, 2008, 04:20:32 PM: First OK, second - wrong. What is initial concentration of acetate in solution?
Title: Re: Calculate [OH-] and pH
Post by: A5HLEY on March 10, 2008, 04:30:58 PM: DOH! I hate careless mistakes. :(

Ok, so

5.56E-10 = (x^2)/(.451-x)
x= 1.583527707E-5

pOH = -log(x) = 4.80
pH = 9.20
Title: Re: Calculate [OH-] and pH
Post by: Borek on March 10, 2008, 05:04:02 PM: Quote from: A5HLEY on March 10, 2008, 04:30:58 PM
DOH! I hate careless mistakes. :(

So stop repeating them over and over ;)
Title: Re: Calculate [OH-] and pH
Post by: A5HLEY on March 10, 2008, 05:13:56 PM: Quote from: Borek on March 10, 2008, 05:04:02 PM
Quote from: A5HLEY on March 10, 2008, 04:30:58 PM
DOH! I hate careless mistakes. :(

So stop repeating them over and over ;)

I try. Does what I redid look better?
Title: Re: Calculate [OH-] and pH
Post by: Borek on March 10, 2008, 05:48:04 PM: No. Worse.
Title: Re: Calculate [OH-] and pH
Post by: A5HLEY on March 10, 2008, 06:04:24 PM: I find your attitude towards my work to be quite rude. I apologize that I am a lowly high school student struggling with chemistry. At least I'm trying to work it out rather than being like "OMGZZ PLZ I HAV HW DUE TONITE AND I NEED U 2 DO IT!!!111." I don't find it very admirable that you ridicule my mistakes rather than help me. Perhaps you're in the wrong place, considering that this is essentially a chemistry help forum.

I bow to you, oh great forum administrator. Such an important position on such an important internet forum, oh my, I am feeling quite small compared to your almighty greatness.

However, I have figured out what my mistakes were, and have acquired the correct answer. So I will no longer burden you with my careless mistakes.
Title: Re: Calculate [OH-] and pH
Post by: Borek on March 10, 2008, 06:25:24 PM: Quote from: A5HLEY on March 10, 2008, 06:04:24 PM
However, I have figured out what my mistakes were, and have acquired the correct answer.

That's the point :P