How many mL of 0.500M KOH are required to completely neutralize 25.00mL of 0.100 M H3PO4

Answer 1

Write the balanced equation:

3KOH + H₃PO₄ ==> K₃PO₄ + 3H₂O

moles H₃PO₄ present = 25.00 ml x 1 L/1000 ml x 0.100 mol/L = 0.0025 moles

mols KOH needed = 0.0025 mol H₃PO₄ x 3 mol KOH/mol H₃PO₄ = 0.0075 moles

volume KOH needed: 0.500 mole/L (x L) = 0.0075 moles

x = 0.015 L x 1000 ml/L = 15 mls