Question

. A sample of ammonium phosphate can be prepared by the reaction of aqueous ammonia and phosphoric acid. 3NH3 (aq) + H3PO4 (aq) (NH4),PO4 (aq) 25.0cm of 1.25 mol/dm phosphoric acid is neutralised by 45.3cm of aqueou ammonia (a) Calculate the concentration in mol/dm3 of the ammonia used.​

Answer 1

The concentration in mol/dm³ of the ammonia used is 2.07 mol/dm³

Balanced equation

3NH₃ (aq) + H₃PO₄ (aq) -> (NH₄)₃PO₄ (aq)

From the balanced equation above,

• The mole ratio of the acid, H₃PO₄ (nA) = 1
• The mole ratio of the base, NH₃ (nB) = 3

How to determine the molarity of NH₃

• Volume of acid, H₃PO₄ (Va) = 25 cm³
• Molarity of acid, H₃PO₄ (Ma) = 1.25 mol/dm³
• Volume of base, NH₃ (Vb) = 45.3 cm³
• Molarity of base, NH₃ (Mb) = ?

MaVa / MbVb = nA / nB

(1.25 × 25) / (Mb × 45.3) = 1 / 3

31.25 / (Mb × 45.3) = 1 / 3

Cross multiply

Mb × 45.3 = 31.25 × 3

Divide both side by 45.3

Mb = (31.25 × 3) / 45.3

Mb = 2.07 mol/dm³

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