Question

in a titration, 60 mL of Phosphoric acid (H3PO4) completely nutralizes 750 mL of sodium carbonate solution containing 3.18 gm of NA2CO3 in one litre solution. calculate the molarity of phosphoric acid.

Answer 1

Dear Student,

We first need to calculate the moles of Na₂CO₃ in a litre of solution, which is = 3.18 g / 106 g mol^-1 = 0.03 moles

Since 0.03 moles of Na₂CO₃ are present in 1000 ml of solution, then in 750 ml of solution the moles of of Na₂CO₃ = $\frac{0.03}{1000}\times 750=0.0225moles$

The chemical reaction is : $2H_{3}P{O}_4+3N{a}_{2}C{O}_3\to 2N{a}_{3}P{O}_4+3C{O}_2+3H_{2}O$

Since, 3 moles of Na₂CO₃​ require 2 moles of H₃PO₄, then, 0.0225 moles of Na₂CO₃​ will need = $\frac{2}{3}\times 0.0225=0.015moles$

Molarity of phosphoric acid = $\frac{0.015}{60}\times 1000=0.25M$

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