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
The Ksp value of Ca(OH)2 is [tex]2.74 \times 10^-^5[/tex].
The Ksp is the solubility product constant.
The solubility equilibrium is
[tex]Ca(OH)_2(s) = Ca^2^+(aq) + 2 OH^-(aq)[/tex]
I 0 0
C s +2s
E s 2s
By the ICE table, the expression for Ksp is
[tex]\rm Ksp = [Ca^2^+] \times ([OH^-])^2 = s \times (2s)^2 = 4s^3[/tex]
Now, calculate the Ksp
s, solubility value is 0.019 M
then, [tex]\rm Ksp = 4s^3= 4 \times (0.019)^3 = 2.74 \times 10^-^5[/tex]
Thus, the Ksp value is [tex]2.74 \times 10^-^5[/tex]
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