Calculate the concentration (M) of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to a total volume of 250.0 mL. 4.37 0.175 0.525 0.350 0.874

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

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