[o1ocups]

When 2.65g of an unknown weak acid (HA) with a molar mass of 85.0 g/mol is dissolved in 250.0g of water, the freezing point of the resulting solution is -0.255 C. Calculate Ka for the unknown weak acid.

Hint from my professor: find the Van't Hoff factor then make an ICE table (or something like that). So I used the freezing point depression formula and found the Van't Hoff factor (i). But what do I do with it? Van't Hoff factor is the ratio of the moles of solute actually in solution and the moles of solid solute added...So do I just set it equal to x/(mole of HA) or something?

I guess what I am really asking is: if you have the Van't Hoff factor of the resulting solution at equilibrium and the initial concentration of the reactant (HA), how do you find the equilibrium concentrations of each reactant and product (i.e. HA, H+ and A-)? I know how to use the ICE table, but I don't know how to use the Van't Hoff factor to find ONE equilibrium concentration so that I can proceed with the table.

Can someone please give me a hint? Thanks in advance!!

[Borek]

Think about stoichiometry of dissociation. If 1 is initial amount of acid and x is amount of dissociated acid, what will be value of Van't Hoff factor?

[o1ocups]

Think about stoichiometry of dissociation. If 1 is initial amount of acid and x is amount of dissociated acid, what will be value of Van't Hoff factor?

I am not sure. Wouldn't it just be x? because x/1=x?
But isn't there supposed to be more molecules at the end of a dissociation? So is it more like 1+x?
I don't know 

I actually still don't understand what the van't Hoff factor is 

[Borek]

Freezing point depression depends on the molality of dissolved molecules/ions, right?

If you dissolve something that doesn't dissociatie, its molality is just that of the subtance.

If you dissolve something that dissociates, molality of dissolved ions is higher than the milality of original substance, as each molecule of original substance gives several ions in the solution.

That's where the Van't Hoff factor comes in - it described how many ions/molecules appear in the solution per each dissolved molecule.

[o1ocups]

That's where the Van't Hoff factor comes in - it described how many ions/molecules appear in the solution per each dissolved molecule.

OK I think now I understand the concept. Thanks.

So, if 1 is the initial amount of acid and x is the amount of dissociated acid,
HA H⁺+A^-
Would the van't Hoff factor be 2/x ? Because if 1=HA, then 2=amount of ions appearing on the other side?
But that doesn't make sense because I already have the real van't Hoff factor, which is less than 2 

[Borek]

If 1 is initial amount of acid and x dissociated, x of HA disappeared, but x of H⁺ and x af A^- appeared in the solution. How many ions/molecues do you have in solution now?

[o1ocups]

If 1 is initial amount of acid and x dissociated, x of HA disappeared, but x of H⁺ and x af A^- appeared in the solution. How many ions/molecues do you have in solution now?

I have the molecules of the rest of the initial acid (1-x) and the H⁺ and A^-  ions (x+x=2x), so I have (1-x)+x+x ions/molecules, so does that equal to the van't Hoff factor?!! 

So after all i=1+x

But of course the initial amount of acid doesn't equal to 1 in this problem, right? So I just solve for x and get all the equilibrium concentrations and get Ka, right? Thank you so much!! 

[o1ocups]

Did I do it wrong if I got a negative Ka value? 

[Borek]

Show what and how you did.

i=1+x is OK.

[o1ocups]

So I got the moles of the acid (HA) by dividing the mass (2.65g) by the molar mss (85.0g), which turned out to be 0.031176471 moles, then I divided that number by 0.250L of water (I assumed that 1g of water = 1 mL of water) to get the molarity or concentration of HA, which is 0.125 M (3 sig figs right?). Then, I used the same number of moles and divided it by the kg of water (250g=0.250kg) to find the molality, and that is 0.125 m. From here, I used the Freezing Point Depression equation ΔTf=Kf*m(i) to find i, so i=ΔTf/Kf*m=-0.255/(1.86*0.125)=-0.255/-0.2325=1.097 (here I suspect that I did something wrong because our professor actually gave us the value of i in lecture and he said it was 1.151.).

Then I used the ICE table and the van't Hoff equation that you helped me derive:

	HA	H+	A-
I	0.125	0	0
C	-x	+x	+x
E	0.125-x	x	x

You said that i=1+x is OK, but I changed 1 to 0.125 and so I had i=0.125+x which probably caused the problem..?
When I did 1.151=0.125+x, I got x=1.026
Then when I used Ka=[H+][A-]/[HA], I got Ka=(x)²/(0.125-x)=(1.026)²/(0.125-1.026)
And since 0.125-1.026 is a negative number my Ka is also negative, which just doesn't make sense.
Where did I do wrong?

[Borek]

You said that i=1+x is OK, but I changed 1 to 0.125 and so I had i=0.125+x which probably caused the problem..?

That's it.

x is not an absolute value, it is fraction of the acid concentration - that's because Van't Hoff factor is relative to the concentration. So your initial [HA] is 1*0.125, and [H⁺]/[OH^-] are x*0.125.

You may also think about x as dissociation fraction.

[o1ocups]

I finally got it! Thanks. 
(Even though MC counted my first two answers wrong b/c of rounding errors!!)