[jennielynn_1980]

A solution of hydrofluoric acid contains 2.0g of HF per litre and has a pH of 2.2.  What is the dissociation constant for HF?


I don't even know where to start here.  The only things I can think of are:
1) find the moles of HF
2) possibly find the [H+] and/or [H₃O+]

The equation would be :

HF(aq) <--> H+(aq) + F-(aq)

If anyone could point me in the right direction that would be great.  Thanks

[Dan]

K_a = [H⁺][F^-]/[HF]

Where [ x ] = concentration of x in mol/dm³
                         
You can probably get away with treating HF as a weak acid, ie. that [HF] at equilibrium is 2 g/L (although you will have to convert to appropriate units for your calculation)

[chiralic]

Read this:

http://www.mpcfaculty.net/mark_bishop/weak_acid_equilibrium.htm

regards,

chiralic

[jennielynn_1980]

So if K_a = [H+][F-]/[HF]

and there is 0.1 mol/L of HF which I got from this:

20.g/L of HF
mol = g        g/mol

=2.0/20 = 0.10 mol/L of HF

and I know that
[H+] = [F-] = 10^-pH
[H+] =[F-] = 10^-2.2

So [H+] = [F-] = 0.0063 mol/L
and HF = 0.1 mol/L

Then
K_a = (0.0063)2/(0.10)
    = 4.0 x 10-4

Is this even close to being right?

[Borek]

Is this even close to being right?

Close, but [HF] <> 0.1.

[HF] + [F^-] = 0.1

[jennielynn_1980]

Why? If HF is 0.1mol/L at equilibrium then why did you add the F to get 0.1?

[Dan]

Why? If HF is 0.1mol/L at equilibrium then why did you add the F to get 0.1?

Your answer uses an approximation, the equlibrium concentration of HF is not 0.1 mol/L because some of the HF has dissociated, so it is in fact a bit less than 0.1 mol/L.

Now the approximation you have made is that barely any of the HF has dissociated, so the equilibrium concentration of HF is close enough to 0.1 mol/L for you to ignore this. This approximation is used for weak acids, and HF is pretty weak, and you are probably expected to use it - in which case your answer is the one your teacher is looking for. However, you will probably impress your teacher if you do it the full way as well, and it is useful to be able to do it without having to make the aforementioned approximation.

Ok, so where does Borek's expression come from?

We have 0.1 mol of HF, which partially dissociates in water

HF   <----------> H⁺ + F^-

Before putting it in water,

moles of HF = 0.1
mol H⁺ = 0
mol F^- = 0

At equilibrium, "x" moles of H⁺ (and F^-) are present, and they must have come from the HF, so the conc of HF must have lowered, so,

moles of HF = 0.1 - x
mol H⁺ = x
mol F^- = x

So it follows that at equilibrium, [HF] = (0.1 - x) mol/L, not 0.1 mol/L

(and also that [HF] + [F^-] = (0.1 - x) + x = 0.1 , Borek's expression)

[jennielynn_1980]

Thanks!  You know, I actually did it Boreks way the first attempt I made and then I switched it to the other way because I was assuming I was supposed to use the pH given in the question to determine the concentrations.