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-]

[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

[admin]

[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]

[admin]

Exactly!

[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.

[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+][SO₄#8322;⁻] / [HSO₄⁻] = (x)(x) / (1.0-x) = 0.012
X2 + 1.012x – 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.

[GraemeMatt8033]

[admin]

Just trying to follow your math and logic. So in your second ICE table you included the previous 1 M HSO4 and at the begining include the 1M H, but then you ignore it in the C and E portions and state H as x. Where did the 1M go?

If you look at the final answer in both explanations, the 1 M was then added to the second [H+] of ~ 0.1M giving a total of 1.01 M (thus, answer B).

My ultimate question is why isn't the 1M H+ from the first reaction taken into account when the 1M from HSO4 is. At the beginning it was taken into account, but the additional H that was generated from the second dissociation wasn't.

Either or those two things clarified would make this questions make sense, I just don't understand why they weren't taken into consideration (each respectively)

Essentially, the two concentrations are the concentrations from the two separate protons from H2SO4. Moreover, the first H+ is from the H2SO4 which is a strong dissociation (or a strong acid in which all of the first proton is dissociated to ~ 1.0 M). As for the second proton dissociation, it is released as a weak acid as it comes from HSO4- and not H2SO4 and thus, the second proton acts as a weak acid and is essentially sparingly released. So in a nut shell, the two separate concentrations are based on the release of the two separate protons each from H2SO4 initially for the first proton release and then HSO4- for the second proton release. However, at the end, you add both the concentrations in which in reality, the second proton concentration is insignificant in relation to the first proton concentration.

[jsfkt78927]

for some reason, i dont believe that 85% of people are getting this question right....

[prestonker2934]

I knew the first proton would dissociate all the way so That means my pH is 1. Then the second proton by ka2 is 100x less likely to dissociate than the 1st proton cuz ka=10^-2, which is .01. Is my thinking wrong here?