[specialk08]

Calculate the pH of the following solutions:

0.10M Aniliunium Chloride
0.10M Sodium Fluoride
0.10M Anilinium Fluoride

Kind of confused on the start up of this!

[Dan]

You don't have enough information to complete the question, you are going to need some dissociation constants.

[specialk08]

Well I have a table of Ka and Kb values, that's what you mean right?
Did you want me to post them here?

[Dan]

Yes, it would be useful to know what information you have access to.

Have a look at this page:

http://www.science.uwaterloo.ca/~cchieh/cact/c123/wkacids.html

and attempt the first question.

[specialk08]

I just have a question before I make an attempt.. I can't neglect the Cl can I? We did a few examples in class where it was possible to neglect the Na ion, but I don't think it is the same case for Cl, am I correct?
So for the first one I have a Ka value for the Anilinium Ion as 2.2x10^-5, and then it's conjugate base of Aniline is 4.6x10^-10. I'm not sure which one is appropriate to use etc. We have reading assignments that we have to complete before the stuff is thoroughly covered in class and I'm just not too certain on it. Thanks!

[Dan]

So for the first one I have a Ka value for the Anilinium Ion as 2.2x10^-5, and then it's conjugate base of Aniline is 4.6x10^-10. I'm not sure which one is appropriate to use etc.

K_a for anilinium is 2.2 x 10^-5 (pK_a = 4.66)

I think 4.6 x 10^-10 is the k_b (not K_a) of aniline (the conjugate base of anilinium). pk_b of aniline is 9.34.

Start by writing out the equilibrium for the dissociation of anilinium.

You can ignore chloride based on the assumption that it is a very very weak base and does not participate. Note that this is not the case for fluoride, but we'll get to that later.

[specialk08]

The table I have states that the pKa for anilinium is 4.66! So I imagine I should go by that. Anyway!!

So the dissociation of Anilinium is as follows?

C6H5NH3+ + H2O  C6H5NH2- + H3O+?

[Dan]

My apologies, I wrote pK_a and pK_b where I meant K_a and K_b. I have edited my previous post accordingly.

Your equilibrium looks good, except that aniline is neutral - remember to balance charges.

C₆H₅NH₃⁺ + H₂O C₆H₅NH₂ + H₃O⁺

Now, have a look at that link I gave you, paying particular attention to Example 2 on the webpage. You may also want to read into the acid dissociation constant and the Henderson-Hasselbalch equation (in your books or online).

[specialk08]

Ok so I have:

C6H5NH3+ = C6H5NH2 + H3O+
.10 - x              -x              -x

Ka = x^2
        .10 - x

Then given that Ka = 2.2x10^-5

2.2x10^-5 = x^2
                    .10-x

Then when I try to solve that quadratic I get a negative under the square root -_-
   

[Dan]

You should have a +ve under the square root:

if K_a = x²/(0.1 - x)

then x² + xK_a - 0.1K_a = 0

and x = [-K_a + (K_a² + 0.4K_a)^1/2]/2

However, for a weak acid, such as anilinium, x << Concentration (0.1 in this case).

You can use the approximation: 0.1 - x ~ 0.1, which makes the calculation much simpler.

[specialk08]

Ok. So how do I go about doing the approximation? I just neglect the x value? Sorry I haven't done chemistry in a long time and the fact that we haven't gone over this in lecture yet is not great.

[Dan]

Ok, so since it is a weak acid, most of the acid does not dissociate - so the value of x is very small in comparison to the concentration of anilinium. You can sustitute (0.1 - x) for 0.1 because x << 0.1.

So K_a = x²/(0.1 - x)

is approximately: K_a = x²/0.1

Now solve: x = (0.1K_a)^1/2

The quadratic solution is more accurate, but this approximation gives a pretty accurate solution and is much easier.

[specialk08]

Alright makes sense. Great. So then I solve that value and use that value to calculate pH? Is there an equation that relates these two? Would it be the value for x and take the negative log of that value?

[Dan]

Yes, x = [H⁺]; pH = -log[H⁺]

[specialk08]

Ok! Great!! Thanks a lot! So now are the other two done in the same way or is there a different procedure to follow since we can't neglect the Fluoride?