Respuesta :
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
The concentration of sodium ions will be "0.350 M".
Volume and concentration
Sodium sulphide's initial volume, v₁ = 50 mL
Sodium sulphide's initial concentration, s₁ = 0.874 M
Sodium sulphide's final volume, v₂ = 250 mL
By using Acidimetry- alkalimetry,
→ v₁ × s₁ = v₂ × s₂
or,
Final concentration will be:
s₂ = [tex]\frac{v_1\times s_1}{v_2}[/tex]
By substituting the values,
= [tex]\frac{50\times 0.874}{250}[/tex]
= [tex]\frac{43.7}{250}[/tex]
= 0.1748 M
Since,
One mole Na₂S ionised to give 2 moles of Na⁺, then
Concentration = 2 × 0.1748
= 0.3496 M
Thus the above answer is correct.
Find out more information about concentration here:
https://brainly.com/question/26756988
