Q1(e). Specific conductance of saturated BaSO₄
The solubility product of BaSO₄ is 1×10⁻¹⁰ md² dm⁻⁶. If the ionic conductance at infinite dilution at 25°C of Ba⁺² =127.28×10⁻⁴ m² mho mol⁻¹ and SO₄⁻² =160.0×10⁻⁴ m² mho mol⁻¹, calculate the specific conductance of BaSO₄.
Answer in brief
κ=2.87×10⁻⁴ S m⁻¹ (=2.87×10⁻⁶ S cm⁻¹).
Key points
- Solubility. BaSO₄(s) ⇌ Ba²⁺ + SO₄²⁻, so Kₛₚ=s²: 3pt] s=1×10⁻¹⁰=1.0×10⁻⁵ mol dm⁻³ =1.0×10⁻⁵×10³ mol m⁻³ =1.0×10⁻² mol m⁻³
- Molar conductivity. The saturated solution is so dilute that Λₘ≈Λ°. By Kohlrausch's law of independent migration: 3pt] Λ°(BaSO₄)=λ°(Ba²⁺)+λ°(SO₄²⁻)=(127.28+160.0)×10⁻⁴=287.28×10⁻⁴ S m² mol⁻¹
- Specific conductance. Λₘ=κ/c, so κ=Λ° c with c in mol m⁻³: 3pt] κ=287.28×10⁻⁴×1.0×10⁻², κ=2.87×10⁻⁴S m⁻¹=2.87×10⁻⁶S cm⁻¹
- S m² mol⁻¹ × mol m⁻³ = S m⁻¹. The common slip is to use c in mol dm⁻³, which gives an answer 10³ times too small.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Reading: the unit “md² dm⁻⁶” is mol² dm⁻⁶; mho = S.
- Solubility. BaSO₄(s) ⇌ Ba²⁺ + SO₄²⁻, so Kₛₚ=s²: 3pt] s=1×10⁻¹⁰=1.0×10⁻⁵ mol dm⁻³ =1.0×10⁻⁵×10³ mol m⁻³ =1.0×10⁻² mol m⁻³
- Molar conductivity. The saturated solution is so dilute that Λₘ≈Λ°. By Kohlrausch's law of independent migration: 3pt] Λ°(BaSO₄)=λ°(Ba²⁺)+λ°(SO₄²⁻)=(127.28+160.0)×10⁻⁴=287.28×10⁻⁴ S m² mol⁻¹
- Specific conductance. Λₘ=κ/c, so κ=Λ° c with c in mol m⁻³: 3pt] κ=287.28×10⁻⁴×1.0×10⁻², κ=2.87×10⁻⁴S m⁻¹=2.87×10⁻⁶S cm⁻¹
- Units check: S m² mol⁻¹ × mol m⁻³ = S m⁻¹. The common slip is to use c in mol dm⁻³, which gives an answer 10³ times too small.
- In practice the conductance of the water itself is subtracted: even good conductivity water conducts about as well as this saturated solution.
Application. run backwards, s=κ/Λ° gives the solubility and Kₛₚ of a sparingly soluble salt (BaSO₄, AgCl, PbSO₄) from one conductance measurement.
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