Question

A sample containing Na2CO3 & NaOH is dissolved in 100 ml solution. 10 ml of this solution requires 25 ml of 0.1 N HCl when phenolphthalein is used as an indicator. If methyl orange is used as indicator 10 ml of same solution requires 30 ml of same HCl. Calculate % of Na2CO3 in the sample. [Nearest Integer]

Answer 1

40

Answer 2

Let “a” = moles of Na₂CO₃ and “b” = moles of NaOH in 10 mL aliquot.

• With phenolphthalein:  NaOH:      NaOH + HCl → NaCl + H₂O  (1:1)  Na₂CO₃:    Na₂CO₃ + HCl → NaHCO₃ + NaCl  (1:1) Thus,  b + a = (25 mL × 0.1 N)/1000 = 0.0025 eq                                        (1)

• With methyl orange:  NaOH:      NaOH + HCl → NaCl + H₂O  (1:1)  Na₂CO₃:    Na₂CO₃ + 2HCl → 2NaCl + H₂CO₃  (1:2) Thus,  b + 2a = (30 mL × 0.1 N)/1000 = 0.003 eq                               (2)

Subtract (1) from (2):  (b + 2a) – (b + a) = 0.003 – 0.0025  ⇒ a = 0.0005 moles

Substitute into (1):  b = 0.0025 – 0.0005 = 0.0020 moles

Calculate mass in 10 mL aliquot:

• Na₂CO₃: 0.0005 moles × 106 g/mol = 0.053 g
• NaOH: 0.0020 moles × 40 g/mol = 0.080 g

Since these values are for 10 mL, in 100 mL solution they’ll be 10× higher:

• Na₂CO₃ = 0.053 g × 10 = 0.53 g
• NaOH = 0.080 g × 10 = 0.80 g

Total sample mass = 0.53 g + 0.80 g = 1.33 g

Percentage of Na₂CO₃ in the sample = (0.53/1.33) × 100 ≈ 39.85% ≈ 40%