What volume of 0.100 M sodium hydroxide is required to completely neutralize 50.0 mL of 0.100 M phosphoric acid?
The balanced equation for the reaction between sodium hydroxide (NaOH) and phosphoric acid (H₃PO₄) is:
3NaOH + H₃PO₄ -> Na₃PO₄ + 3H₂O
From the equation, we can see that 3 moles of sodium hydroxide react with 1 mole of phosphoric acid.
To determine the volume of sodium hydroxide required, we can use the equation:
moles of phosphoric acid = volume of phosphoric acid x concentration of phosphoric acid
moles of sodium hydroxide = moles of phosphoric acid / (3 moles NaOH / 1 mole H₃PO₄)
volume of sodium hydroxide = moles of sodium hydroxide / concentration of sodium hydroxide
Given:
Volume of phosphoric acid = 50.0 mL = 0.050 L
Concentration of phosphoric acid = 0.100 M
Concentration of sodium hydroxide = 0.100 M
Calculations:
moles of phosphoric acid = 0.050 L x 0.100 M = 0.005 moles
moles of sodium hydroxide = 0.005 moles / (3 moles NaOH / 1 mole H₃PO₄) = 0.0016667 moles
volume of sodium hydroxide = 0.0016667 moles / 0.100 M = 0.01667 L
Therefore, the volume of 0.100 M sodium hydroxide required to completely neutralize 50.0 mL of 0.100 M phosphoric acid is 0.01667 liters or 16.7 mL.