350.ml of .25M KOH is reacted with 250.ml of .30M H3PO4. Determine which is the limiting reagent, the molarity of the salt formed, and the final molarity of the reactant that was in excess.

Answer 1

350 mL KOH x 0.25M= 87.5 mmoles
250 mL H3PO4 x 0.30M = 75 mmoles.

..........KOH + H3PO4 ==> KH2PO4 + H2O
initial...87.5...75.0......0........0
change...-75.0..-75.0.....+75.....+75
equil.....12.5....0.......75........75

BUT the KH2PO4 (75 mmoles of it) can react with what is left of the KOH.
.......KOH + KH2PO4 ==> K2HPO4 + H2O
init...12.5...75.........0........0
change -12.5..-12.5......12.5.....12.5
equil...0......62.5......12.5....12.5

So KOH is the limiting reagent since it ran out first. The problem is stated as if only one of the H ions on H3PO4 reacts. The salt concn (you pick out the one the question wants because there are two of them) is M = moles/L soln. The molarity of the one in excess is moles/L also. NOte the volume is 250 + 350 mL and convert to L.