arc5005

7+ Year Member
Oct 5, 2011
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Medical Student (Accepted)
Please forgive me for struggling with this concept. It's been a long time since I've taken gen chem (~7 years), so i'm struggling understanding the relationship between pka values.

Question:

Which of the following relationships must be TRUE?

Reaction 1:
2 NH3 (aq) + CO2 (g) + H2O (l) -> (NH4)2CO3 (aq)

Reaction 2:
(NH4)2CO3 (aq) + CaSO4 (aq) -> (NH4)2SO4 (aq) + CaCO3 (s)

Reaction 3:
2 Ca5(PO4)3F (s) + 7 H2SO4 (aq) -> 3 Ca(H2PO4)2 (aq) + 7 CaSO4 (aq) + 2 HF (g)

I. pKa1(H3PO4) is less than pKa2(HF)
II. pKa2(H3PO4) is less than pKa2(H2SO4)
III. pKa1(HF) is less than pKa2(H2SO4)

Possible Answers:


A. I only
B. II only
C. I and II only
D. II and III only









Correct answer: A

Reasoning:

"First, the lower the pKa value is, the stronger the acid is. Second, when the pH of the solution exceeds the pKa of an acid, that site exists predominantly in the deprotonated state. Statement III can immediately be eliminated, because the first proton of sulfuric acid is strong, meaning it has a very low (negative) pKa value associated with it. From the products of the last equation in the passage, it can be seen that H2SO4-, SO42-, and HF all simulateneously exist in solution. Sulfuric acid has lost both of its protons, phosphoric acid has lost its first proton, and hydrofluoric acid has yet to lose a proton. The first proton of phosphoric acid is lost more readily than the proton of HF, which means that pKa1 of H3PO4 is lower than the pKa value of HF. Statement I is thus true. Because the second proton of sulfuric acid has been lost, while the second proton of phosphoric acid has not been lost, the second proton of sulfuric acid is more acidic than the second proton of phosphoric acid. The result numerically is that pKa2 of H2SO4 is less than pKa2 of H3PO4. Statement II is thus false."


Why I'm confused:
So, my understanding right now is that the lower a pka value is = the stronger the acid. Because the first proton in H2SO4 is strong it has a very low pKa value, making it's pka value less than the pka of HF. However statement III is comparing the pka1 of HF and the pka2 of H2SO4. Isn't the pKa2 of HS204 talking about the deprotonation of the second hydrogen in sulrfuric acid?

I'm confused about statement I. It's comparing the pka1 of H3PO4 which would be a low pKa value, because the first hydrogen would be readily available, making it a strong acid? But it then compares it to the pka2 of HF, HF only had 1 proton to provide, so i'm confused how it can have a pka2 value?

For statement II, the reaction formula shows that H2PO4 the second hydrogen was not deprotonated, whereas H2SO4 was deprotonated to SO4, meaning pka2 H2SO4 is lower than pka2 of H2PO4, thus making it a false statement. This one I understand.


Thank you in advance for any help!
 

NextStepTutor_1

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Hi @arc5005 ,

Great questions! Let's start with the quickly answerable ones and move on to the more involved ones from there.

First off, you're correct that HF cannot have a pKa2. It's interesting that the explanation just talks about the "pKa" of HF, so the reference to "pKa2(HF)" in statement I must be a typo.

W/ regard to statement III, you're correct that the pKa2 of H2So4 refers to the deprotonation of the second proton. What this question is getting at is that reaction 3 tells you that at some pH (we don't know what pH exactly), we have a mix of: H2PO4- (which has given up one proton), SO42- (which has given up both protons), and HF (which has not given up a proton). This means that sulfuric acid must give up both of its protons before HF gives up its first. (This is actually true, BTW -- the pKa1 of sulfuric acid is -3 and the pKa2 is 1.99, while the pKa of HF is 3.17). A key point to note here, though, is that we can't securely conclude that statement III is incorrect just based on the knowledge that sulfuric acid is strong and HF is weak. We have to have some source of external information about how the second proton of sulfuric acid behaves, which is provided to us by reaction 3.

Hope this clarifies things :)!
 
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arc5005

7+ Year Member
Oct 5, 2011
1,010
426
Status
Medical Student (Accepted)
Hi @arc5005 ,

Great questions! Let's start with the quickly answerable ones and move on to the more involved ones from there.

First off, you're correct that HF cannot have a pKa2. It's interesting that the explanation just talks about the "pKa" of HF, so the reference to "pKa2(HF)" in statement I must be a typo.

W/ regard to statement III, you're correct that the pKa2 of H2So4 refers to the deprotonation of the second proton. What this question is getting at is that reaction 3 tells you that at some pH (we don't know what pH exactly), we have a mix of: H2PO4- (which has given up one proton), SO42- (which has given up both protons), and HF (which has not given up a proton). This means that sulfuric acid must give up both of its protons before HF gives up its first. (This is actually true, BTW -- the pKa1 of sulfuric acid is -3 and the pKa2 is 1.99, while the pKa of HF is 3.17). A key point to note here, though, is that we can't securely conclude that statement III is incorrect just based on the knowledge that sulfuric acid is strong and HF is weak. We have to have some source of external information about how the second proton of sulfuric acid behaves, which is provided to us by reaction 3.

Hope this clarifies things :)!

Thank you thank you. This clarifies a lot.
 
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