How many ml of 0.100 M NaOH are required to completely neutralize 50.0 ml of 0.125 M H3PO4 solution?

Question

How many ml of 0.100 M NaOH are required to completely neutralize 50.0 ml of 0.125 M H3PO4 solution?

Answer 1

Hello Jana

 

The balanced chemical reaction for this neutralization reaction is:

 

3NaOH (aq) + H₃PO₄ (aq) → Na₃PO₄ (aq) + 3H₂O (l)

 

We need to first determine the # of moles of H₃PO₄ from its volume and molarity.  Then convert it to moles of NaOH using stoichiometry.  Finally, use the calculated moles of NaOH and its molarity (0.100 M) to figure out the volume needed.

 

Convert volume of H₃PO₄ from mL to L.

 

50.0 mL x (1 L / 1000 mL) = 0.0500 L

 

Molarity = mol / L

 

mol = Molarity x L

 

      = 0.125 mol/L x 0.0500 L = 0.00625 mol H₃PO₄

 

Moles of NaOH.....

 

0.00625 mol H3PO4 x (3 mol NaOH / 1 mol H3PO4) = 0.0188 mol NaOH (rounded to 3 significant figures)

 

Molarity = mol / L

 

L = mol / Molarity

 

   = 0.0188 / 0.100 = 0.188 L

 

Volume in mL = 0.188 L x (1000 mL / 1 L) = 188 mL of NaOH required