To calculate the volume of 0.200 M phosphoric acid that will react completely with 50 mL of 0.100 M sodium hydroxide, we need to determine the balanced chemical equation for the reaction between phosphoric acid and sodium hydroxide.

The balanced chemical equation is:

H3PO4 + 3NaOH → Na3PO4 + 3H2O

From the balanced chemical equation, we can see that 1 mole of phosphoric acid reacts with 3 moles of sodium hydroxide. Therefore, we need to calculate the number of moles of sodium hydroxide present in 50 mL of 0.100 M sodium hydroxide:

Moles of NaOH = Molarity of NaOH × Volume of NaOH in liters

Moles of NaOH = 0.100 M × 50 mL / 1000 mL/L

Moles of NaOH = 0.00500 moles

Since 1 mole of phosphoric acid reacts with 3 moles of sodium hydroxide, we need 1/3 of the moles of sodium hydroxide to determine the moles of phosphoric acid required.

Moles of H3PO4 = Moles of NaOH / 3

Moles of H3PO4 = 0.00500 moles / 3

Moles of H3PO4 = 0.00167 moles

Now we can calculate the volume of 0.200 M phosphoric acid required to provide 0.00167 moles of phosphoric acid:

Volume of H3PO4 = Moles of H3PO4 / Molarity of H3PO4

Volume of H3PO4 = 0.00167 moles / 0.200 M

Volume of H3PO4 = 8.35 mL

Therefore, 8.35 mL of 0.200 M phosphoric acid will react completely with 50 mL of 0.100 M sodium hydroxide.