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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: bucket on August 04, 2008, 10:32:52 pm

Title: Acidity Constants
Post by: bucket on August 04, 2008, 10:32:52 pm: I dont get it.
How would you find the K_a of NH₄⁺
Title: Re: Acidity Constants
Post by: Mao on August 04, 2008, 10:36:05 pm: $NH_4^+(aq)\leftrightharpoons H_3O^+(aq)+NH_3(aq)$

$K_a=\frac{\left[NH_3\right]\left[H_3O^+\right]}{\left[NH_4^+\right]}$
Title: Re: Acidity Constants
Post by: bucket on August 04, 2008, 10:38:55 pm: why is this defined as an acidity constant, how is it showing acidity :s
Title: Re: Acidity Constants
Post by: Collin Li on August 04, 2008, 10:41:43 pm: It shows the extent that ammonium will donate its proton. The higher the $K_a$ , the more inclined the acid is to ionise (move to the right), and hence donate its proton.
Title: Re: Acidity Constants
Post by: bucket on August 04, 2008, 10:49:52 pm: thanks both of you.
Title: Re: Acidity Constants
Post by: bucket on August 04, 2008, 11:08:21 pm: ok well theres this question:

"Chloroacetic acid is a weak monoprotic acid with K_a of $1.3 \times 10^{-3}$ . 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 $K_a=\frac{[H_3O^+]^2}{1.0}$

and yeah wth. lol.
Title: Re: Acidity Constants
Post by: bturville on August 04, 2008, 11:29:53 pm: because the conc. of the conjugate base and $H_3O^+$ are the same, ie $x^2$
Title: Re: Acidity Constants
Post by: bucket on August 04, 2008, 11:31:20 pm: ooookay thanks man
Title: Re: Acidity Constants
Post by: Collin Li on August 04, 2008, 11:31:41 pm: Two mathematical equations we use:

1. $\mbox{[H_3O^+]} = \mbox{[conjugate base]}$

This is how we conclude the numerator is equal to $\mbox{[H_3O^+]}^2$

2. We make an assumption that $\mbox{[acid]}_{\mbox{initial}} = \mbox{[acid]}_{\mbox{equilibrium}}$

This lets us substitute in: $\mbox{[acid]}_{\mbox{equilibrium}} = 1.0$

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!
Title: Re: Acidity Constants
Post by: Ahmad on August 05, 2008, 12:07:33 am: The 2nd point coblin raised is an approximation that is used to avoid more difficulty calculation. If you instead used,

$\mbox{final acid} = \mbox{initial acid} - \mbox{H_3O^{+}}$

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. :)
Title: Re: Acidity Constants
Post by: Mao on August 05, 2008, 03:26:24 pm: For general interest, the final expression derived from the quadratic equation looks something like:

$[\mbox{H}_3\mbox{O}^+]^2+\mbox{K}_{\mbox{a}}[\mbox{H}_3\mbox{O}^+]-\mbox{K}_{\mbox{a}} \cdot [\mbox{HX}]_\mbox{init}=0$

$\implies [\mbox{H}_3\mbox{O}^+]=\frac{-\mbox{K}_{\mbox{a}} +\sqrt{\left(\mbox{K}_{\mbox{a}}\right)^2+4\cdot \mbox{K}_{\mbox{a}} \cdot [\mbox{HX}]_\mbox{init}}}{2}$

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 $HX+H_2O\leftrightharpoons H_3O^+ + X^-$ . This is also counterintuitive, as $[HX]$ 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? :P thanks =]
Title: Re: Acidity Constants
Post by: Collin Li on August 05, 2008, 05:39:18 pm: 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.
Title: Re: Acidity Constants
Post by: Mao on August 05, 2008, 05:50:25 pm: very nice! I shall accept that ionisation is $HX\leftrightharpoons H^+ + X^-$ , where Le Chatelier's principle works.
Title: Re: Acidity Constants
Post by: Collin Li on August 05, 2008, 08:16:32 pm: 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: $\frac{b}{a}$

Now, the percentage ionisation is: $\frac{b+x}{a+c-x}$

$a$ , $b$ and $c$ are positive numbers.

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

Let's suppose $\frac{b+x}{a+c-x} < \frac{b}{a}$ :

(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, $x \approx 0$ , and hence $\frac{b+x}{a+c-x} \approx \frac{b}{a+c} < \frac{b}{a}$ . 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:

$a(b+x) < b(a+c-x)$

$\implies ab + ax < ab + bc -bx$

$\implies x(a+b) < bc$

$\implies x < \frac{bc}{a+b}$

Can this ever be false?

Here's some intuition. We know that $x < c$ (Le Chatelier's principle), but that doesn't tell us anything. $\frac{bc}{a+b} < c$ since $\frac{b}{a+b} < 1$ . However, it should make us think: if I decrease $\frac{b}{a+b}$ , then I could possibly make $x > \frac{bc}{a+b}$ (i.e.: contradiction)

If we try to analyse the size of $x$ in terms of $a$ , $b$ and $c$ :

$K = \frac{b^2}{a}$

Hence, the new equilibrium is equal to $K$ :

$\implies \frac{(b+x)^2}{a+c-x} = \frac{b^2}{a}$

So the answer lies within this maths problem:

Prove, or prove incorrect, that $\frac{b+x}{a+c-x} < \frac{b}{a}$ , using the fact that: $\frac{(b+x)^2}{a+c-x} = \frac{b^2}{a}$ and that $a$ , $b$ and $c$ are positive constants.
Title: Re: Acidity Constants
Post by: Collin Li on August 05, 2008, 08:24:13 pm: Guys, that post was stupid! Here is the much easier reason why the above is true:

Percentage ionisation: $\frac{[A^-]}{[HA]} = \frac{K}{[H^+]}$

An addition of solid acid (i.e: virtually holding volume constant), will increase $[H^+]$ , and will not affect $K$ , hence percentage ionisation drops.

Oh well, at least the maths works as well.
Title: Re: Acidity Constants
Post by: Mao on August 05, 2008, 09:22:12 pm: Quote from: coblin on August 05, 2008, 08:16:32 pm
Hence, the new equilibrium is equal to $K$ :

$\implies \frac{(b+x)^2}{a+c-x} = \frac{b^2}{a}$

Prove, or prove incorrect, that $\frac{b+x}{a+c-x} < \frac{b}{a}$ , using the fact that: $\frac{(b+x)^2}{a+c-x} = \frac{b^2}{a}$ and that $a$ , $b$ and $c$ are positive constants.

Leave this for the maths guys.

lol, you might as well already done it yourself:

$\frac{(b+x)^2}{a+c-x} = \frac{b^2}{a}$

$\frac{b+x}{a+c-x}=\frac{b}{a}\cdot \frac{b}{b+x}$

since $b+x>b$ , $\frac{b}{b+x}<1$

$\therefore \frac{b+x}{a+c-x} < \frac{b}{a}$

yay, math works.

:. God exists
Title: Re: Acidity Constants
Post by: Pandemonium on August 05, 2008, 10:08:04 pm: Jesus, such a non sequitur statement to make.
Title: Re: Acidity Constants
Post by: bucket on September 24, 2008, 12:11:28 am: I should have listened in class :S.

When using an example with HCl, the book stated this:
"Since water is the solvent in aqueous solution and its concentration is virtually constant we write the equilibrium expression as
$K[H_2O]=\frac{[H_3O^+][Cl^-]}{[HCl]}$ "

Why is the concentration of water constant?? Is this related to K_W?
Title: Re: Acidity Constants
Post by: Collin Li on September 24, 2008, 12:16:19 am: It's constant because things like liquids and solids have a set type of structure (roughly, for liquids), and hence a set density. Water, the solvent, is a liquid. Adding another molecule of water is not going to change the concentration (much).

Another way to think of it is because we have heaps of water compared to any other species. In the reaction I'm consuming 1 molecule of water and 1 molecule of hydrochloric acid. Will we experience much change to the concentration of water? No, it'll still be high as, while the change in the hydrochloric acid concentration will be significant. It's sort of like taking 1 from 5, versus taking 1 from a 1 million.

On the other hand, aqueous substances and gases, have very variable concentrations.
Title: Re: Acidity Constants
Post by: bucket on September 24, 2008, 12:21:47 am: ahhhh thanks a million coblin =]