Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as colligative properties. For this problem, we use boiling point elevation concept.
ΔT(boiling point) = (Kb)mi
ΔT(boiling point) = (0.51 C-kg / mol )(4.0 mol / 2.05 kg ) (2)
ΔT(boiling point) = 1.99 C
T(boiling point) = 101.99 C
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Answer:
Ksp = 2.74 x 10⁻⁵
Explanation:
The solubility equilibrium for Ca(OH)₂ is the following:
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)
I 0 0
C + s + 2s
E s 2s
According to the ICE table, the expression for the solubility product constant (Kps) is:
Ksp = [Ca²⁺] x ([OH⁻])² = s x (2s)² = 4s³
Then, we calculate Ksp from the solubility value (s):
s = 0.019 M
⇒ Ksp = 4s³ = 4 x (0.019)³ = 2.74 x 10⁻⁵
