[Dan]

For NaF, a base, work out the pOH using K_b, then convert to pH.

Can you show your answer to the first part so I can check it's right?

Can you also post all dissociation constants you have access to?

[specialk08]

So is this the correct procedure for Part B?

NaF -> Na+ + F-

Then set up the equilibrium using Fluoride?

F- + H2O   HF + OH-
.10 - x                 +x    +x

Then Kb = Kw/Ka
= 1.10x10^-14 / 6x10^-4

Kb = 6.6x10^-18

Kb = x^2
       0.10 - x
Once again use the approximation...
Kb = x^2
       .10

So solve for x, and x = [OH-], get your pOH and calculate pH?

Part A
I calculated the pH as 2.83!

I have a very,very very large handout that shows Ka values, pKa values and Kb/pKb values for the conjugate bases. So basically I have access to every value I will need for these questions. Did you want me to post the actual values?

[specialk08]

If this is correct, I still need assistance with the last one, having a fair bit of difficulty

[Dan]

I agree with your value (2.83) for part A.

For part B, what is your value for pOH? How is pOH related to pH? Hint: read about the dissociation constant of water...

[specialk08]

Did I use the correct value for Ka for part B? I used the Ka value for fluoride and calculated Kb from that? Is that the correct process?

If this is correct, I determined x to equal 8.12x10^-10.

So [OH-] = 8.12x10^-10 and pOH therefore is -log[OH-] = -log[8.12x10^-10] = 9.09 and so pH = 4.91?

[Dan]

Your method is OK, but your numbers have come out wrong - post you working, including the value of K_b you are using.

[specialk08]

NaF -> Na+ + F-

F- + H2O    HF + OH-
.10 - x                 +x    +x

So to determine Kb, I have the value of Kw and Ka (Ka for Fluoride = 6.0 x 10^-4)

Kb = Kw/Ka
Kb= 1.10x10^-14 / 6x10^-4
Kb = 6.6x10^-18

Kb = x^2
       0.10 - x
Approximation:
Kb = x^2
       .10

6.6x10^-18 = x^2
                    0.10
So cross multiply:
6.6x10^-19 = x^2
Take sq rt of both sides
x = 8.12x10^-10

[OH-] = 8.12x10^-10 and using that pOH = -log[OH]:

pOH = -log[8.12x10^-10]
pOH = -(-9.09)
pOH = 9.09

pOH + pH = 14
9.09 + pH = 14
pH = 14 -9.09
pH = 4.91

[specialk08]

Not sure how to edit a post, but when I said Ka for Fluoride, I meant Ka for Hydrofluoric acid. It's the same value, I just typed fluoride by mistake.

[Dan]

Kb = Kw/Ka
Kb= 1.10x10^-14 / 6x10^-4 - correct
Kb = 6.6x10^-18 - incorrect

This is wrong. You have written (1.1 x 10^-14)/(6.6 x 10^-4), which is fine but you have calculated (1.1 x 10^-14)x(6.6 x 10^-4) instead.

[specialk08]

Oh rats. Ok great. So once I fix that error the answer should be correct?

And what about the third one...? Not sure on that

[specialk08]

So I think I successfully corrected my mistake and the correct pH for this question is 8.11.

[Dan]

8.11 is what I get too, good.

For the last part you need to combine the information you have got from the first two parts.

Now I'm not actually 100% on this next bit, if I'm wrong then our resident pH guru will probably correct me...

You know [H⁺] generated by 0.1 M anilinium, and you know [^-OH] generated by 0.1 M fluoride. Combine them and what happens? What is left?

[specialk08]

You know [H⁺] generated by 0.1 M anilinium, and you know [^-OH] generated by 0.1 M fluoride. Combine them and what happens? What is left?

Okay. so H+ generated by anilinium is 0.00148 and OH- generated by fluoride is 1.29x10^-6. So multipyling these together I get 1.909x10^-9. I'm not sure what I'm supposed to do now..I'm not really sure what you are getting at haha

[fledarmus]

LOL - I think he meant "combine them" chemically, as in, "what do you get when you mix your H⁺ with your OH^-"?

[specialk08]

awkward....
so water haha...