The solubility product of a sparingly soluble salt A₂X₃ is 1.1 × 10⁻²³. If specific conductance of the solution is 3 × 10⁻⁵ S m⁻¹, the limiting molar conductivity of the solution is x × 10⁻³ S m² mol⁻¹. The value of x is ___________.

Solution

Chemistry ma’am · 4 steps

Get the solubility from Kₛₚ, turn it into mol m⁻³, then divide κ by it.

1. Write Ksp. A₂X₃ ⇌ 2A³⁺ + 3X²⁻. If the solubility is s, then [A³⁺] = 2s and [X²⁻] = 3s.

   Kₛₚ = (2s)²(3s)³ = 108 s⁵

2. Find s. s⁵ = (1.1 × 10⁻²³)/108 ≈ 1.0 × 10⁻²⁵, so s = 1 × 10⁻⁵ mol L⁻¹. In SI units that is 1 × 10⁻⁵ × 1000 = 1 × 10⁻² mol m⁻³.
3. Molar conductivity. The solution is very dilute, so its molar conductivity is the limiting value: Λₘ° = κ/c = (3 × 10⁻⁵ S m⁻¹)/(1 × 10⁻² mol m⁻³) = 3 × 10⁻³ S m² mol⁻¹.
4. So. Λₘ° = 3 × 10⁻³ S m² mol⁻¹, so x = 3.

The common slip: 3000. Dividing κ by s in mol L⁻¹ without converting to mol m⁻³ gives 3 S m² mol⁻¹, which is x = 3000, not 3.

The idea to keep. For a sparingly soluble salt, the saturated solution counts as infinitely dilute, so Λₘ° = κ / s. Match the units: S m⁻¹ needs s in mol m⁻³.

Answer from the NTA final answer key. Solution by PrepWiser.