Question 16

[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?