Let\'s consider the dissociation of phosphoric acid H3PO4. All possible protonat
ID: 938211 • Letter: L
Question
Let's consider the dissociation of phosphoric acid H3PO4. All possible protonation states, with H number from zero to three, must co-exist in equilibrium. H3PO4 <---> H2PO4 (-1) (pK1 = 2) <---> HPO4 (-2) (pK2 = 7) <---> PO4 (-3) (pK3 = 12) 8.6: Let's consider the dissociation of phosphoric acid HsPO4. All four possible protonation states, with H number from zero to three, must co-exist in equilibrium. Find relative populations of all four states at the pH of the human blood. 7.4.Explanation / Answer
If: H3PO4 [1M]
Then, second state H2PO4-1:
pH = pKa - Log10([A-]/[HA])
7.4 = 2 - Log10([A-]/[1M])
7.4 - 2 = - Log10([A-]/[1M])
5.4 = -Log10([A-]/[1M])
10-5.4 = 10Log10([A-]/[1M])
10-5.4 = ([A-]/[1M])
10-5.4x [1M] = [H2PO4-1]
[3.98x10-6M] = [H2PO4-1]
Then, third state H1PO4-2:
pH = pKa - Log10([A-]/[HA])
7.4 = 7 - Log10([A-]/[1M])
7.4 - 7 = - Log10([A-]/[1M])
0.4 = -Log10([A-]/[1M])
10-0.4 = 10Log10([A-]/[1M])
10-0.4 = ([A-]/[1M])
10-0.4x [1M] = [H1PO4-2]
[0.398 M] = [H1PO4-2]
Then, four state PO4-3:
pH = pKa - Log10([A-]/[HA])
7.4 = 12 - Log10([A-]/[1M])
7.4 - 12 = - Log10([A-]/[1M])
(-4.6) = -Log10([A-]/[1M])
104.6 = 10Log10([A-]/[1M])
104.6 = ([A-]/[1M])
104.6x [1M] = [PO4-3]
[3.98x104 M] = [PO4-3]
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