Acidity Constants

[bucket]

I dont get it.
How would you find the K_a of NH₄⁺

[Mao]

[bucket]

why is this defined as an acidity constant, how is it showing acidity :s

[Collin Li]

It shows the extent that ammonium will donate its proton. The higher the , the more inclined the acid is to ionise (move to the right), and hence donate its proton.

[bucket]

thanks both of you.

[bucket]

ok well theres this question:

"Chloroacetic acid is a weak monoprotic acid with K_a of . For a 1.0M solution of chloroacetic acid calculate the pH"

looking at the answers, they sub values into the acidity constant and end up with

and yeah wth. lol.

[bturville]

because the conc. of the conjugate base and are the same, ie

[bucket]

ooookay thanks man

[Collin Li]

Two mathematical equations we use:

1.

This is how we conclude the numerator is equal to

2. We make an assumption that

This lets us substitute in:

Note that step 2 seems 'invisible' for students who just plug in the numbers. You should actually try to think about why it's an issue to just substitute in 1.0 without making an assumption. The assumption is also a horribly contradictory one, but it also makes sense to do it for convenience, and I can explain that to you too (another day, if you're interested, or someone else can).

Also, you should write out the assumption in step 2 in your working, otherwise your working may not be accepted. You need to explain your rationale in working out.

Good night!

[Ahmad]

The 2nd point coblin raised is an approximation that is used to avoid more difficulty calculation. If you instead used,



you would encounter a quadratic equation, which although not impossible to solve, certainly involves more computation which is often just not worth it. For a weak acid, you'll find that the approximation usually yields results within 5% of the more exact method, which is typically acceptable. 

[Mao]

For general interest, the final expression derived from the quadratic equation looks something like:





and, I have a question:
When I dilute a weak acid, why does the percentage dissociation change?
using the above-provided K_a and the above quadratic equation, and initial concentration of 1M, 0.1M and 0.01M, the percentage dissociation were 3.5%, 11% and 30% respectively.
I tried to use Le Chatelier's principle to explain this increase in yield, and thought of lower concentrations as diluting a solution of higher concentration. But I'm stuck on the fact that both sides have the same number of particles . This is also counterintuitive, as increases, shouldn't the equilibrium be driven to the right (and yield/percentage dissociation should increase as molarity increases)?
at the same time, I don't see anything wrong with the above quadratic equation, the K_a it reproduced at the calculated concentrations were maticulous...

please explain? thanks =]

[Collin Li]

You are supposed to ignore solids and solvents when assessing the concentration fraction.

This is because we are only concerned about the interaction of dynamic concentrations of the reactants and products. Solvents and solids have a generally constant concentration.

Hence, dilution affects the products moreso (because there are two aqueous products and only one aqueous reactant).

This pushes the reaction to the right. Now, we have to be careful here. This means that the amount of protons in solution will increase, but the concentration will decrease overall because we can't forget we originally diluted the solution -- Le Chatelier's only states that the change will be partially opposed.

But for percentage ionisation, the formula is a 1:1 ratio of the conjugate base to the acid. The dilution factor cancels out because both the product and the reactant have been diluted, and so the only change that percentage ionisation captures is the change in the amount -- which was an increase in the amount of conjugate base.

To challenge your claim that it is counter-intuitive: your rationale seems to be that when [HX] increases, the equilibrium is driven to the right, so hence when [HX] decreases, the equilibrium should be driven to the left. Quite right, but also remember that [H+] decreases, and [A-] decreases too! Two effects are pulling the reaction to the right and only one effect is pulling the reaction to the left. This is exactly why the dilution rule works with Le Chatelier's principle.

[Mao]

very nice! I shall accept that ionisation is , where Le Chatelier's principle works.

[Collin Li]

Let's model a general equilibrium for a weak acid:

Code: [Select]

                     [HA]   [H+]   [A-]
initial equilibrium  a      b      b
addition of HA       a+c    b      b

change               -x     +x     +x

re-equilibriate      a+c-x  b+x    b+x

We are trying to investigate whether addition of acid has a definite effect on the percentage ionisation of an acid. At first, it may be tempting to say that addition of the acid will only have a partial decrease of the acid, so hence the denominator will remain large. However, it isn't really that obvious:

The percentage ionisation is initially:

Now, the percentage ionisation is:

, and are positive numbers.

By Le Chatelier's principle: as the change will only be partially opposed, however as well. So it really depends on how much the numerator has increased compared to the increase of the denominator.

Let's suppose :

(This is what is tempting to assume, but looking at it now, it is not clear, so hence we will try to contradict it)

Side-note: For a weak acid, , and hence . This means that the percentage ionisation will probably decrease for weak enough acids, but it is not clear if this is true for all acids.

The inequality can only be true if:









Can this ever be false?

Here's some intuition. We know that (Le Chatelier's principle), but that doesn't tell us anything. since . However, it should make us think: if I decrease , then I could possibly make (i.e.: contradiction)


If we try to analyse the size of in terms of , and :



Hence, the new equilibrium is equal to :



So the answer lies within this maths problem:

Prove, or prove incorrect, that , using the fact that: and that , and are positive constants.

[Collin Li]

Guys, that post was stupid! Here is the much easier reason why the above is true:

Percentage ionisation:

An addition of solid acid (i.e: virtually holding volume constant), will increase , and will not affect , hence percentage ionisation drops.

Oh well, at least the maths works as well.