Answer:
Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M
≈0.350 M
Explanation:
From the question it is clear that,
Initial volume of sodium sulphide solution is (v₁) = 50mL
Initial concentration of sodium sulphide solution is (s₁) =0.874 M
Final volume of sodium sulphide solution is (v₂) = 250mL
Let, the final concentration of sodium sulphide solution is s₂, then according to acidimetry-alkalimetry,
v₁ × s₁ = v₂ × s₂
Or, s₂ = v₁ × s₁/v₂
= 50 × 0.874 / 250
= 0.1748 M
Therefore, concentration of 250mL sodium sulphide solution is 0.1748 M
Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M
≈0.350 M
