What quantity of 6.00 M aqueous H3PO4 (in mL) would be required to neutralize 589 mL of 2.60 M aqueous NaOH?

Answer 1

The molarity of NaOH times its volume in liters will give moles of NaOH. Based on the balanced equation

3NaOH + H3PO4 <--> 3H2O + Na3PO4

the moles of NaOH are converted to moles of H3PO4 using the stoichiometric ratio. Once the moles of H3PO4 are obtained, the volume is determined by dividing by the molarity of the H3PO4.

2.60 mol NaOH/1 L * 589 mL * 1L/1000mL * 1 mol H3PO4/ 3mol NaOH *1 L/ 6.00 mol H3PO4 = 0.0851 L or 85.1 mL