Question

Calculate the concentration of 45.0 mL of a phosphoric acid solution that neutralized 20.8 mL of 0.532 M sodium hydroxide. Assume complete ionization of the acid.

Answer 1

0.0820M H₃PO₄

Step-by-step explanation:

THe phosphoric acid, H₃PO₄ reacts with sodium hydroxide, NaOH, thus:

H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

The moles of NaOH used to neutralize the acid are:

Moles NaOH:

20.8mL = 0.0208L * (0.532mol / L) = 0.0111 moles NaOH.

That means moles of phosphoric acid in the solutionare:

0.0111 moles NaOH * (1mole H₃PO₄ / 3 moles NaOH) =

0.00369 moles H₃PO₄

These moles are in 45.0mL = 0.045L. The molar concentration of the solution are:

0.00369 moles H₃PO₄ / 0.045L =

0.0820M H₃PO₄