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2. Calcule la solubilidad molar de Hidróxido de Berilio en una solución que es 0.0150m naoh(ac). be(oh)2(s) → Be+2 (ac) + 2oh-(ac) kps = 2.00 x10-18
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The solubility of Be(OH)2 in a 0.0150M NaOH solution can be calculated using the following steps:
1. Write the equilibrium expression for the dissolution of Be(OH)2:
Be(OH)2(s) ↔ Be+2(ac) + 2OH-(ac)
Ksp = [Be+2][OH-]^2
2. Set up an ICE table (Initial, Change, Equilibrium) to determine the concentrations at equilibrium:
| | Be+2 | 2OH- |
|-----|--------|----------|
| I | 0 | 0.0150 |
| C | +s | +2s |
| E | s | 0.0150+2s|
3. Substitute the equilibrium concentrations into the Ksp expression:
2.00 x 10^-18 = (s)(0.0150 + 2s)^2
4. Since Ksp is very small, we can assume 2s is negligible compared to 0.0150:
2.00 x 10^-18 ≈ (s)(0.0150)^2
5. Solve for s, which represents the molar solubility of Be(OH)2:
s ≈ 8.89 x 10^-15 M
Therefore, the molar solubility of Be(OH)2 in a 0.0150M NaOH solution is approximately 8.89 x 10^-15 M.
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