Question

If 100.0 mL of a 0.500 M NaOH(aq) solution is required to completely neutralize 125.0 mL of H3PO4(aq) according to the balanced equation below, then what was the molarity of the H3PO4(aq)?
H3PO4(aq) + 3 NaOH(aq) à 3 H2O(l) + Na3PO4(aq)

Answer 1

V ( NaOH ) = 100.0 mL => 100.0 / 1000 = 0.1 L
M(NaOH) = 0.500 M

Number of moles NaOH :

n = M x V

n = 0.500 x 0.1 =>0.05 moles of NaOH

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

1 mole H₃PO₄ ------- 3 moles NaOH
? mole H₃PO₄ -------0.05 moles NaOH

moles H₃PO₄ = 0.05 x 1 / 3

=> 0.0166 moles of H₃PO₄

V( H₃PO₄) = 125.0 mL => 125.0 / 1000 = 0.125 L

M = n / V

M ( H₃PO₄) = 0.0166 / 0.125

 => 0.1328 M

hope this helps!