[A5HLEY]

Calculate [OH-] and pH for the following:

a). .210M KClO
b). .451M Ba(C2H3O2)2

[Borek]

Try by yourself first. That's in forum rules.

[A5HLEY]

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...

[Borek]

Put information from ICE table into dissociation constant and solve for x. You will need Ka/Kb values, these should be in your book.

[A5HLEY]

So do I use this formula:

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

and will x be [OH-]?

[Borek]

Looks like.

[A5HLEY]

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!

[A5HLEY]

Sorry, I meant to edit, not quote.

[Borek]

First OK, second - wrong. What is initial concentration of acetate in solution?

[A5HLEY]

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

[Borek]

DOH! I hate careless mistakes.

So stop repeating them over and over

[A5HLEY]

DOH! I hate careless mistakes.

So stop repeating them over and over

I try. Does what I redid look better?

[Borek]

No. Worse.

[A5HLEY]

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.

[Borek]

However, I have figured out what my mistakes were, and have acquired the correct answer.

That's the point