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Topic: Dihydrogen Phosphate's amphiprotic behavior  (Read 5276 times)

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Offline GoldShadow

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Dihydrogen Phosphate's amphiprotic behavior
« on: April 30, 2007, 02:15:33 PM »
This question appeared as a bonus on an exam:

The Ka of H3PO4 is 7.6 x 10-3.  Calculate the equilibrium constant for the formation of H3PO4 From H2PO4- and explain in words why we consider H2PO4- to be an acid despite its amphiprotic nature.

Using Kw=Ka*Kb I calculated that for H2PO4- + H2O<--> H3PO4 + OH-, Kb= 1.316 x 10-12.  But other than that I don't know what to do using the information given in this problem.  I looked up the Ka2 for H3PO4 and found that it was 6.2 x 10-8, which is bigger than the Kb of H2PO4-, but is there any way to determine using only the information given in the problem?

Offline english

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Re: Dihydrogen Phosphate's amphiprotic behavior
« Reply #1 on: April 30, 2007, 02:42:54 PM »
If you know the relative Ka and Kb of H2PO4-, then you can determine the average behavior of the compound in solution.  More prone to acidic character or basic?

What do high Ka and Kb values indicate?  What do the size of these constants say about the progress of their respective reactions? 

Strong acidic behavior is determined by favorable deprotonation of the acid, and strong basic character is determined by favorable protonation of the base.  This is the Bronsted-Lowry concept.
« Last Edit: April 30, 2007, 02:49:07 PM by g_english »

Offline Borek

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Re: Dihydrogen Phosphate's amphiprotic behavior
« Reply #2 on: April 30, 2007, 03:47:14 PM »
What is pH of NaH2PO4 solution?
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