A 25.0 mL sample taken from a saturated solution of Ca(OH)2 required 36.0 mL of 0.100 M HCl to titrate.
a) Calculate the concentration in moles/L of hydroxide ions in the original sample.
b) Calculate the concentration of calcium ions in the original solution
c) Calculate the Ksp of Calcium Hydroxide for these conditions
Thanks!
Chemistry, Titration quesiton + Equilibrium?
2016-03-25 5:12 pm
回答 (2)
2016-03-25 6:25 pm
✔ 最佳答案
A. moles HCl used = 0.0360 L X 0.100 mol/L = 3.6X10^-3 mol HCl. This will equal moles OH- in the original solution, so [OH-] = 3.6X10^-3 mol / 0.0250 L = 0.144 M OH-B. Since each Ca2+ has 2 OH-, [Ca2+] = 0.144 M / 2 = 0.072 M Ca2+
C. Ksp = [Ca2+][OH-]^2
Ksp = 0.072 (0.144)^2 = 1.5X10^-3
(This is quite far from the value of Ksp given in my textbook.)
2016-03-25 6:31 pm
(a)
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
No. of moles of HCl = 0.100 × 36.0 = 3.60 mmol
1 mole of Ca(OH)₂ needs 2 moles of HCl for complete neutralization.
No. of moles of Ca(OH)₂ = 3.60 × (1/2) = 1.80 mmol
Concentration of Ca(OH)₂ = 1.80 / 25.0 = 0.072 mol/L
Dissociation of 1 mole of Ca(OH)₂ gives 2 moles of OH⁻ ions.
[OH⁻] in the original sample = 0.072 × 2 = 0.144 mol/L
(b)
Dissociation of 1 mole of Ca(OH)₂ gives 1 mole of Ca²⁺ ions.
[Ca²⁺] in the original sample = 0.072 mol/L
(c)
Ksp = [Ca²⁺] [OH]² = 0.072 × 0.144² = 1.49 × 10⁻³ (mol³/L³)
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
No. of moles of HCl = 0.100 × 36.0 = 3.60 mmol
1 mole of Ca(OH)₂ needs 2 moles of HCl for complete neutralization.
No. of moles of Ca(OH)₂ = 3.60 × (1/2) = 1.80 mmol
Concentration of Ca(OH)₂ = 1.80 / 25.0 = 0.072 mol/L
Dissociation of 1 mole of Ca(OH)₂ gives 2 moles of OH⁻ ions.
[OH⁻] in the original sample = 0.072 × 2 = 0.144 mol/L
(b)
Dissociation of 1 mole of Ca(OH)₂ gives 1 mole of Ca²⁺ ions.
[Ca²⁺] in the original sample = 0.072 mol/L
(c)
Ksp = [Ca²⁺] [OH]² = 0.072 × 0.144² = 1.49 × 10⁻³ (mol³/L³)
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