A.) a solution of 0.205M KOH is used to neutralize 25.0mL H3PO4, if 45.6mL KOH is required to reach neutralization, what is the balanced equation? what is molarity of H3PO4 solution?
not sure how to do this any work would be much appreciated!!!
not sure how to do this any work would be much appreciated!!!
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3KOH + H3PO4 -------> 3H2O + K3PO4
molarity = no. of moles of solute/volume of solution in litres
so no. of moles of soute = molarity X volume of solution in litres
so no. of moles of KOH = 0.205 X 45.6/1000 = 9.348 X 10^-3
now 3 moles of KOH neutralizes 1 moles of H3PO4
so 1 mole of KOH will neutralize 1/3 moles of H3PO4
so 9.348 X 10^-3 moles will neutralize 1/3 X 9.348 X 10^-3 = 3.116 X 10^-3 moles of H3PO4
now number of moles of H3PO4 = 3.116X10^-3
volume = 25/1000 = 0.025 L
molairty = 3.116 X 10^-3/0.025 = 0.125 M
so molarity of H3PO4 = 0.125
molarity = no. of moles of solute/volume of solution in litres
so no. of moles of soute = molarity X volume of solution in litres
so no. of moles of KOH = 0.205 X 45.6/1000 = 9.348 X 10^-3
now 3 moles of KOH neutralizes 1 moles of H3PO4
so 1 mole of KOH will neutralize 1/3 moles of H3PO4
so 9.348 X 10^-3 moles will neutralize 1/3 X 9.348 X 10^-3 = 3.116 X 10^-3 moles of H3PO4
now number of moles of H3PO4 = 3.116X10^-3
volume = 25/1000 = 0.025 L
molairty = 3.116 X 10^-3/0.025 = 0.125 M
so molarity of H3PO4 = 0.125