# Dissociation constant

1. May 24, 2005

### ChemRookie

I understand this tells us how strongly binded a solution is.
So I hve hydroflouric acid, it says it contains 2.0g of HF per litre and has a PH of 2.2. How do I figure what the dissociation constant for HF is?

Thanks

2. May 24, 2005

### JFo

use the ph to find the concentration of H+

remember pH = -log[H+] => [H+] = 10^-pH

use the info they gave you about the amount of HF to get molarity of HF solution.

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

[F-] = [H+] (neglecting dissociation of H20)

then use the defintion for the dissociation constant

Last edited: May 25, 2005
3. May 24, 2005

### ChemRookie

oh boy, I have hated this equilibrium unit.

I dont follow exactly. can you define some of the things? and is molarity (moles)? that be 0.1.

4. May 24, 2005

### maverick280857

A good idea is to list all knowns first.

You've been given the strength of the solution in grams per litre. You have to convert it into moles per litre first (Molarity). Molarity = moles of solute per litre of solution. Then, consider the ionization of HF and account for how much HF has reacted and what concentration H+ and F- ions have been formed as a result.

Cheers
Vivek

5. May 24, 2005

### maverick280857

Molarity = number of moles of solute per litre of solution

Molarity of HF before ionization = moles of HF given to you/volume of solution in litres. Let this equal a.

Now, consider the ionization of HF:

:::::::::: HF----------------->H+ + F-
t = 0_____a_______________ ~0____0
t = teq___a-x_______________x_____x

Here x is the concentration of HF that has been consumed during the ionization so that at equilibrium, H+ and F- are formed in equal amounts, i.e. x.

The crucial step: Note that I have put a ~ sign in the first line below H+. This is to indicate that the solution contains some amount of hydrogen ions due to the autoionization of water which we have neglected due to its small value. It is conceptually incorrect to say that H+ in such a solution is zero before HF ionizes. However, once HF has ionized then you can make an order of magnitude approximation and say that almost all the hydrogen ions (or more precisely hydronium ions) in solution are due to ionization of HF. Its like this: if x = 4 then $4 + 10^{-7}$ (example) is approximately 4.

Now,

$$K_{c} = \frac{[H^+][F^-]}{[HF]}$$

REMEMBER: These are molarities in the square brackets.

Now you should be able to use the definition of pH and solve the problem. Stated differently this is precisely what JFo told you:

"All the H+ and F- come from HF so the initial concentration of HF = concentration of HF at any time t + concentration of H+ (or F-) at that time."

Please try all this on paper and do many equilibrium problems after this to come to terms with the methods which look complex initially. You should look for constraints like charge balance, mass balance, etc.

Hope that helps...

cheers
Vivek

PS--the underscores are for spacing only....I couldn't figure out how to make that table well spaced :-D

Last edited: May 24, 2005
6. May 24, 2005

### ChemRookie

thanks for your effort. I get it more, but I am still not comfortable with this.
I'm going to need to give it a look more later, and keep looking back at it more and more.

7. May 25, 2005

### GCT

$$K=[H30+][F-]/[HF]$$

In the case of a simple dissociation of a monoprotic acid

$$K=[x][x]/[initial~conc.~acid~-x]$$

note that all you'll need to solve for K is x, and the initial concentration.

note that x is the hydronium concentration, which can be deduced using the pH.

note that the initial concentration can be found also, find the moles of HF, the volume is 1 L.