Question

suppose that 100 mL of .150 M H3PO4 are mixed with 20 mL of .10 M Na3PO4; what is the composition (millimoles of each) and pH of the solution after mixing? for H3PO4, K1= 7.11*10^-3; K2= 6.32*10^-8; K3= 7.1*10^-13

Answer 1

mmol of H3PO4 = 100*0.15 = 15

mmol of NA3PO4 = 20*0.1 = 2

then

2 mmol of Na3PO4 + 2 mmol of H+ = 2 mmol of HPO4-2

mmol of H3PO4 = 15-2 = 13 mmol of H3PO4 left

note that there is stil more reaction

2 mmol of HPO4-2+ 2 mmol of H+, then = 2 mmol of H2PO4-

mmol of H3PO4 left = 13-2 = 11 mmol of H3PO4 left

this is a buffer

pH = pKa2 + log(H2PO3-2 / H3pO4)

pKa2 = -log(6.32*10^-8) = 7.20

so

pH = 7.2 + log(2/11) = 6.45