Question

A solution of HCl is standardized by titration against 0.242 g of primary standard Na2CO3. The titration requires 35.9 mL of HCl. A solid sample weighing 0.152 g and containing sodium oxide is dissolved in water (Na2O + H2O → 2NaOH) and titrated with HCl, requiring 34.7 mL. What is the percentage of Na2O in the sample?

Answer 1

Basis of Calculation : 100 mL of Na2CO3 solution

Molarity of primary standard (Na2CO3) = (0.242/105.9888)/0.1 = 0.02283 M₁

Let us assume that same quantity of Na2CO3 is rundown as of HCl. Then, V₁ = V₂ = 35.9 mL

Then according to molarity formulae, M₁V₁ = M₂V₂

M2 = M₁V₁/V₂ = 0.02283 X 0.0359/ 0.0359 = 0.02283.

Molarity of HCL solution = 0.02283.

According to Na2O + H2O ---> 2NaOH stoichiometry, 1 mole of Na2O dissolves in water to give 2 moles of NaOH and this NaOH solution is titrated against HCl of molarity 0.02283 M.

Given, sample weighs 0.152g and is Na2O, i.e., moles of Na2O = 0.152/ 61.9789 = 0.002452.

Then, 0.002452 moles of NaOH dissolves in water to give 2 X 0.002452/1 = 0.004904 moles of NaOH.

weight of NaOH = 0.004904 X 40 = 0.19616 gm

The question is not clear.

If the weight percentage of Na2O in sample is aked to calculate, then = weight of Na2O /Total weight (weight of NaOH) = 0.152 / 0.19616 = 0.7749 w%

If it is mole percentage of Na2O is asked to calculate, then = moles of Na2O/ Total moles

= 0.002452/ 0.004904 = 0.5 mole%.