Question

What is the molarity of a solution of h3po4 if 10.2 ml is neutralized by 53.5 ml of 0.20m koh what is the molarity of a solution of h3po4 if 10.2 ml is neutralized by 53.5 ml of 0.20m koh 0.67m h3po4 0.15m h3po4 0.35m h3po4 0.37m h3po4

Answer 1

The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M

Balanced equation

H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O

From the balanced equation above,

• The mole ratio of the acid, H₃PO₄ (nA) = 1
• The mole ratio of the base, KOH (nB) = 3

How to determine the molarity of H₃PO₄

• Volume of acid, H₃PO₄ (Va) = 10.2 mL
• Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
• Volume of base, Ca(OH)₂ (Vb) = 53.5 mL

• Molarity of acid, H₃PO₄ (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 10.2) / (0.2 × 53.5) = 1 / 3

(Ma × 10.2) / 10.7 = 1 / 3

Cross multiply

Ma × 10.2 × 3 = 10.7

Ma × 30.6 = 10.7

Divide both side by 30.6

Ma = 10.7 / 30.6

Ma = 0.35 M

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