Question

0.276 g of Na2CO3.xH2O was weighed out accurately and dissolved in water. Titration with 0.050 mol cm-3 sulphuric acid required 20.00 cm3 for neutralisation. Calculate the value of x.

Answer 1

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

C(H₂SO₄)= 0.05 mol/L

V(H₂SO₄) = 20 ml = 0.02 L

n(H₂SO₄) $= 0.05 \times 0.02 = 0.001 \;mol$

n(H₂SO₄) = n(Na₂CO₃) = 0.001 mol

M(Na₂CO₃) = 106 g/mol

m(Na₂CO₃) $= n \times M = 0.001 \times 106 = 0.106 \;g$

∆m = 0.276 – 0.106 = 0.17 g

M(H₂O) = 18 g/mol

n = m/M

n(H₂O) $= \frac{0.17}{18} = 0.0094 \;mol$

$x = \frac{n(H_2O)}{n(Na_2CO_3)} = \frac{0.0094}{0.001} ≈ 10$

Answer: 10.