Question 16

[Guest]

if Ka2 = [H+][SO42-] / [HSO4-] = 1.2 x 10-2 and [HSO4-] = 1.0 M, then wouldn't you solve [H+] but squar rooting 1.2 x 10^-2? When you squar root it, the answer is approximately .1. So [H+]total = 1 + 0.1 = 1.1M. Answer D

[Guest]

Sorry.... let me say that again
if Ka2 = [H+][SO42-] / [HSO4-] = 1.2 x 10-2 and [HSO4-] = 1.0 M, then wouldn't you solve [H+] by squar rooting 1.2 x 10^-2? When you squar root it, the [H+] is approximately .1. So [H+]total = 1 + 0.1 = 1.1M. Answer D

[admin]

There is no need to take the square root of Ka2.

Consider this: sulfuric acid is among the strongest acids (CHM 6.1). When you start with 1 M sulfuric acid, you end up with 1 M hydrogen ions and 1 M sulfate ions.

Therefore:

Ka2 = [H+][SO42-] / [HSO4-]

Simplifies to:

Ka2 = [SO42-]

[rahman.405365]

how does Ka2 = [H+][SO42-] / [HSO4-] ??

shouldn't Ka2 = [H+][SO42-] where [H+] = [SO42-], so that [H+][H+] = Ka2, resulting in sqrt Ka2 = [H+]??

[admin]

Ka2 is simply the acid dissociation constant for the second dissociable proton. It is not a good idea to try to memorize the equation, but rather to know where it comes from. Let's begin with the following equilibrium:

HSO4- <--> H+ + SO4--

Now it's just like a regular equilibrium requiring calculation of an equilibrium constant meaning the product of the products divided by the product of the reactants. In this case the EQ constant K is called Ka2 indicating that it refers to the second (2) dissociable acid (a) equivalent (also know as a proton!).

Thus:

Ka2 = [H+][SO42-] / [HSO4-]

Here are some references if you wish to see more examples of the use of Ka2:

http://members.aceweb.com/patrussell/approximations/Borate.htm

The following is a Power Point presentation of acids and bases. Some of it is very, very elementary but it deals with many useful issues including a clear presentation of Ka2:

http://library.tedankara.k12.tr/IB/mustafa/Ch16-%20Notes.ppt

[anon3543]

This discussion confuses me. What I am thinking:

Ka1 = [H+][HSO4-] / [H2SO4] = 10^3

This obviously a strong acid, so it's pretty much a complete dissociation of the first proton. If that were true, you have 1 M H+ (dissociated from the 1 M H2SO4), and -log(1M H+) = 0 pH!! That number doesn't scare me or anything, it just means it's very acidic, and it can only get more acidic with the dissociation of the second proton. It's probably not really that low because of the approximation made when I said all of the first protons dissociated... Either way, it's much lower of a pH than any of the answer choices. PLEASE respond, I don't want to miss easy pH questions, and I want to get this straight. Thank you.

[admin]

[anon3543]

that was silly of me. thanks! : )

[rs53]

I'm honestly really confused about #16, even after reading the explanation online and on the forum. I'm having the same problem--I understand the concept of Ka's but if the second dissociation is exactly like the first, then why is the Ka2 just equal to the [SO4 2-] ?

If we were to treat it just like the other case of the first proton dissociating, then we should also consider the [H+] due to dissociation of the 2nd proton, which is equal to the [SO4 2-]...I really don't understand why we didn't set each equal to x, and then solve using the square root of Ka2. If the concentrations of sulfate ion and protons are the same, then we can't set the Ka2=[SO4]!! please help

[admin]

[rs536324]

WOW haha this makes so much more sense now! thanks a lot, that really helped.

one last thing--can we generally use this approach for polyprotic acids, as long as the 2nd dissociation is small?

[admin]

Exactly!

[hulk475390]

I'm still a bit confused. First you say that [H+]=1 from the first equation, and then later you state that [H+]=[S04-2] (which is .012). How can [H+] equal too different amounts. I understand how you get the first, but once you assume it equals 1 why do you add to it? The concept makes sense but the math doesn't.

[GraemeMatt8033]

bump confused on the same concept.

So H2SO4 dissociates completely (assuming) to 1M [H+] and 1M HSO4

Than HSO4 dissociates to H and SO4.
So initially for this reaction, we have: 1M HSO4; 1M H+ and 0M SO4.

So we will decrease HSO4 by an amount x while concomitantly increasing H+ by the amount x, and also increasing SO4 by x.

So our equation should look like:

Ka2= [h][so4]/[hs04] which from the previous dissociation equals

1.2x10^-2= [1+x][x]/[1-x]

So in your explanation when stating that the two values cancelled each other out [HSO4 and H] did not take into account the dissociation of HSO4 and the production of more H, making HSO4 amounts smaller and H amounts greater, not allowing them to cancel out.

The math just doesnt seem to make sense, sure you can cancel things out, but they are in the right order of time. We do have 1 M HSO4 and 1M H at the same time, but as soon as HSO4 begins to dissociate (via Ka2) they are no longer the same, so how could you treat them the same in the Ka2 equation???

[emilie.maz5603]

The initial answer of B is CORRECT!
Clarification: Sulfuric acid is a strong polyprotic acid in which the first proton ionization is the most important ionization in an aqueous solution. Moreover, the first ionization step is strong so the complete ionization of the first proton will give a [H+] = 1.0 M. Now, consider the second ionization step in which the proton now acts as a weak acid due to the much smaller Ka₂ value in comparison to the large Ka ₁ value and so use Ka₂ and ionization of proton 2 as follows:
HSO₄⁻(aq) + H₂O (l) ↔ H₃O+(aq)SO₄⁻(aq)
I                  1.0                      1.0            0.0

C                 -x                      +x                +x

E                1.0 – x                 +x             +x


and so, [H+] = [HSO₄⁻] = approximately 1.0 M, since Ka1 is so high and the solution is 1 M H₂SO₄ .
However, HSO₄⁻ is also an acid and will dissociate as a weak acid with the following (from above table):

So, Ka= [H+][SO₄²⁻] / [HSO₄⁻] = (x)(x) / (1.0-x) = 0.012
X² + 1.01x² – 0.012 = 0
Using the quadratic equation to solve,
X = +0.01169 or x = -1.024
So, x must = +0.01169 and thus, [H+] = 1.0M from first ionization + 0.01169 from second ionization and thus, [H+] = 1.01169 M and the answer is B (1.01 M)
Thus, THE EXPLANATION IS CORRECT! Answer is B !




Though the AAMC lists the quadratic equation is required knowledge, it is highly unlikely that you would ever use it on the real test.  And, even in this problem, an estimation would have easily given you the correct answer.