Chemistry
Lia134
17

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)

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(1) Answers
michaelaroman

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!

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