Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: A5HLEY on March 09, 2008, 03:16:41 PM
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Calculate [OH-] and pH for the following:
a). .210M KClO
b). .451M Ba(C2H3O2)2
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Try by yourself first. That's in forum rules (http://www.chemicalforums.com/index.php?page=forumrules).
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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...
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Put information from ICE table into dissociation constant and solve for x. You will need Ka/Kb values, these should be in your book.
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So do I use this formula:
Kb = x^2/(.210-x)
and will x be [OH-]?
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Looks like.
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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!
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Sorry, I meant to edit, not quote. :)
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First OK, second - wrong. What is initial concentration of acetate in solution?
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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
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DOH! I hate careless mistakes. :(
So stop repeating them over and over ;)
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DOH! I hate careless mistakes. :(
So stop repeating them over and over ;)
I try. Does what I redid look better?
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No. Worse.
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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.
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However, I have figured out what my mistakes were, and have acquired the correct answer.
That's the point :P