更新1:
Also what is the experimental value of Ksp?
How to find OH-?
2017-11-01 3:47 pm
Suppose a student titrates a 10.00-mL aliquot of saturated Ca(OH)2 solution to the equivalence point with 16.37 mL of 0.0196 M HCl. What was the initial [OH − ]?
回答 (1)
2017-11-02 1:17 am
Ca(OH)₂ completely dissociates in aqueous solution to form OH⁻ions, while HCl completely dissociates in aqueous solution to form H⁺ ions.
Consider the titration of saturated Ca(OH)₂ solution against HCl solution :
OH⁻ + H⁺ → H₂O
Mole ratio OH⁻ : HCl = OH⁻ : H⁺ = 1 : 1
No. of moles of HCl reacted in the titration = (0.0196 mol/L) × (16.37/1000 L) = 0.000321 mol
No. of moles of OH⁻ ions reacted in the saturated solution of Ca(OH)₂ = 0.000321 mol
[OH⁻] in the saturated Ca(OH)₂ solution = (0.000321 mol) / (10.00/1000 L) = 0.0321 M
Consider the saturation solution of Ca(OH)₂ :
Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq) …… Ksp
At equilibrium, each mole of Ca(OH)₂ dissociates to form 1 mole of Ca²⁺ ions and 2 moles of OH⁻ ions.
At equilibrium :
[OH⁻] = 0.0321 M
[Ca²⁺] = 0.0321/2 M
Ksp = [Ca²⁺] [OH⁻]² = (0.0321/2) × (0.0321)² = 1.65 × 10⁻⁵
Consider the titration of saturated Ca(OH)₂ solution against HCl solution :
OH⁻ + H⁺ → H₂O
Mole ratio OH⁻ : HCl = OH⁻ : H⁺ = 1 : 1
No. of moles of HCl reacted in the titration = (0.0196 mol/L) × (16.37/1000 L) = 0.000321 mol
No. of moles of OH⁻ ions reacted in the saturated solution of Ca(OH)₂ = 0.000321 mol
[OH⁻] in the saturated Ca(OH)₂ solution = (0.000321 mol) / (10.00/1000 L) = 0.0321 M
Consider the saturation solution of Ca(OH)₂ :
Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq) …… Ksp
At equilibrium, each mole of Ca(OH)₂ dissociates to form 1 mole of Ca²⁺ ions and 2 moles of OH⁻ ions.
At equilibrium :
[OH⁻] = 0.0321 M
[Ca²⁺] = 0.0321/2 M
Ksp = [Ca²⁺] [OH⁻]² = (0.0321/2) × (0.0321)² = 1.65 × 10⁻⁵
收錄日期: 2021-04-18 17:54:48
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