13.
An acidic solution containing 0.01 M La3+ is treated with NaOH until La(OH)3
precipitates. At what pH does this occur? (Ksp=2×10-21)
a. 6.8
b. 7
c. 7.8
d. 8
La(OH)₃ precipitates when the ion product exceeds its Ksp of 2×10-21 in a 0.01 M La³⁺ solution. The correct pH is 7.77, closest to option c (7.8). This occurs as NaOH addition raises [OH⁻] to the precipitation threshold.
Precipitation Calculation
For La(OH)3(s) ⇌ La³⁺ + 3OH⁻, Ksp = [La³⁺][OH⁻]³ = 2×10-21.
At precipitation onset, [La³⁺] = 0.01 M (initial concentration).
- [OH⁻]³ = Ksp / 0.01 = 2×10-21 / 0.01 = 2×10-19.
- [OH⁻] = (2×10-19)1/3 = 5.85×10-7 M.
- pOH = -log(5.85×10-7) = 6.23; pH = 14 – 6.23 = 7.77.
Option Analysis
| pH | [OH⁻] (M) | Q = [La³⁺][OH⁻]³ | Result vs Ksp (2×10-21) |
|---|---|---|---|
| 6.8 | 1.58×10-8 | 2.51×10-24 | Q < Ksp (no precipitation) |
| 7.0 | 1.00×10-7 | 1.00×10-23 | Q < Ksp (no precipitation) |
| 7.8 | 5.01×10-7 | 2.51×10-21 | Q > Ksp (precipitates) |
| 8.0 | 1.00×10-6 | 1.00×10-20 | Q > Ksp (precipitates) |
Option c (7.8) matches as precipitation begins near pH 7.77. Lower pH values keep Q below Ksp, preventing precipitation.
CSIR NET Application
This calculation tests solubility equilibria understanding for rare earth hydroxides. Precipitation pH depends on Ksp and cation concentration; here, La³⁺’s low solubility (pKsp ≈21) requires near-neutral pH despite 3 OH⁻ stoichiometry. Practice similar problems with varying [Mⁿ⁺] and Ksp for exam success.
