TBR GChem Question

[happyfellow]

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When 0.1 moles NaH2PO4 and 0.2 moles Na2HPO4 are mixed in 100mL, what is the pH of the solution?

A. pH < (pka1+pka2)/2
B. pka2 > pH > (pka1 + pka2)/2
C. (pka2 + pka3)/2 > pH > pka2
D. pH > (pka2 + pka3)/2

Here's my thought process. "Okay, I have more moles of HPO4 (corresponds to pka3) than I do of H2PO4 (corresponds to pka2). Therefore, my pH of my solution will be closer to pka3. My pka2 value is lower so if I average it with pka3 then it will be lower than my pH." Therefore, I am led to pick D, but the correct answer is C.

Would someone mind helping me out? Thanks

-happy

 

[IntelInside]

When 0.1 moles NaH2PO4 and 0.2 moles Na2HPO4 are mixed in 100mL, what is the pH of the solution?

A. pH < (pka1+pka2)/2
B. pka2 > pH > (pka1 + pka2)/2
C. (pka2 + pka3)/2 > pH > pka2
D. pH > (pka2 + pka3)/2

Here's my thought process. "Okay, I have more moles of HPO4 (corresponds to pka3) than I do of H2PO4 (corresponds to pka2). Therefore, my pH of my solution will be closer to pka3. My pka2 value is lower so if I average it with pka3 then it will be lower than my pH." Therefore, I am led to pick D, but the correct answer is C.

Would someone mind helping me out? Thanks

-happy

This is a buffer/titration problem. At 1/2 equivalence point we know that there is a 1:1 ratio between the acid and its conj base therefore pH = pKa. But in this scenario we see that we have more moles of the base and therefore pH is going to be a little more than pKa2 (this eliminates B and makes C a possible answer). Now based on a titration curve for this sample and what samples I have I know that at equivalence point (100% of species 3) the pH is approximately (pKa2 + pKa3)/2, but we dont have 100% of species 3 in this problem we only have a little bit more than what we should have at 1/2 equivalence. Therefore D is out.

pH = (pKa1+pKa2)/2

This formula only applies when the solution only contains amphoteric ions as the only major acid-base species!!​

 

[SoftwareKevin]

When 0.1 moles NaH2PO4 and 0.2 moles Na2HPO4 are mixed in 100mL, what is the pH of the solution?

A. pH < (pka1+pka2)/2
B. pka2 > pH > (pka1 + pka2)/2
C. (pka2 + pka3)/2 > pH > pka2
D. pH > (pka2 + pka3)/2

Here's my thought process. "Okay, I have more moles of HPO4 (corresponds to pka3) than I do of H2PO4 (corresponds to pka2). Therefore, my pH of my solution will be closer to pka3. My pka2 value is lower so if I average it with pka3 then it will be lower than my pH." Therefore, I am led to pick D, but the correct answer is C.

I think you're confusing where ph=pka2 and ph=pka3 show up on the graph. They are not the equivalence points (the area of graph where pH increases rapidly). They are in the middle of the buffer regions (the flat part of the graph where pH increases slowly). As the other poster pointed out, if you had equal quantities of HPO4-3 and H2PO4- you would have ph = pKa2. You're going to be just a little past this, but still in the same buffer region, so pH should be only slightly larger than pKa2, but definitely closer to pka2 than pka3.