What is the molarity of an aqueous phosphoric acid solution if 13.13 mL is completely neutralized by 19.64 mL of 0.09291 M NaOH? Collapse question part Part 1 Incorrect answer. Did you write the correct molecular formula for phosphoric acid? Write a balanced molecular equation for the reaction. Be sure to include the appropriate phases for all reagents and products.

The molarity of this solution is 0.0695 M

Explanation:

Step 1: The balanced equation

H3PO4 (aq) + 2NaOH (aq) → Na2HPO4 (aq) + 2H2O (l)

Step 2: Given data

Volume of phosphoric acid (H3PO4) = 13.13 mL

volume of NaOH = 19.64 mL

Molarity of NaOH = 0.09291 M

Step 3: Calculate moles of NaOH

Number of moles = Molarity * volume

Number of moles = 0.09291 M * 19.64 mL * 10^-3 L

Number of moles = 0.001825 moles

Step 4: Calculate moles of H3PO4

For 2 moles of NaOH there is 1 mole of H3PO4 consumed to produce 1 mole Na2HPO4

So for 0.001825 moles NaOH, there is consumed 0.001825/2 = 0.0009125 moles of H3PO4

Step 5: Calculate molarity of H3PO4

Molarity = number of moles / Volume

Molarity = 0.0009125 moles / (13.13 * 10^-3L) = 0.0695 M

The molarity of this solution is 0.0695 M