You prepare exactly 1.00 L of a solution containing 0.02 mol of. Na_2HPO_4, 0.02 mol of...

Question

Question

science chemistry

Answer 1

Answer #1

Na2HPO4 = 2Na+ + HPO4-2

NaH2PO4 ---> Na+ + H2PO4-

Na3PO4 --> 3Na+ + PO4-3

then

reactions:

PO4-3 + H+ =HPO4-2

HPO4-2 + H+ = H2PO4-

charge balance:

[ChargE] = [Na+](0.02*2 +0.02*1 + 0.01*3) + [HPO4-](0.02) + [H2PO4-2](0.02) + [PO4-3](0.01)

Answer 2

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