Question

The solubility of Ca(OH)$_2$ in water is:
[Given: The solubility product of Ca(OH)$_2$ in water = 5.5 $\times$ 10⁻⁶]

• A. 1.11 $\times$ 10⁻²
• B. 1.11 $\times$ 10⁻⁶
• C. 1.77 $\times$ 10⁻²
• D. 1.77 $\times$ 10⁻⁶

Hint:

Be very careful with the stoichiometry when writing the K$_{sp}$ expression. For a salt of the form A$_x$B$_y$, the equilibrium is $x$A$^{y+}$ + $y$B$^{x-}$, and the K$_{sp}$ expression is $[xS]^x[yS]^y$. For Ca(OH)$_2$, this becomes (S)(2S)$^2 = 4S^3$.

Answer 1

The Correct Option is A

The dissolution equilibrium for calcium hydroxide is:
Ca(OH)$_2$(s) $\rightleftharpoons$ Ca$^{2+}$(aq) + 2OH$^-$(aq)
Let 'S' be the molar solubility of Ca(OH)$_2$ in moles per liter.
From the stoichiometry of the dissolution, if 'S' moles of Ca(OH)$_2$ dissolve, it will produce 'S' moles of Ca$^{2+}$ ions and '2S' moles of OH$^-$ ions.
[Ca$^{2+}$] = S
[OH$^-$] = 2S
The solubility product constant, K$_{sp}$, is given by the expression:
K$_{sp}$ = [Ca$^{2+}$][OH$^-$]$^2$.
Substituting the concentrations in terms of S:
K$_{sp}$ = (S)(2S)$^2$ = 4S$^3$.
We are given K$_{sp}$ = 5.5 $\times$ 10⁻⁶.
5.5 $\times$ 10⁻⁶ = 4S$^3$.
S$^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6}$.
S = $\sqrt[3]{1.375 \times 10^{-6}} = \sqrt[3]{1.375} \times 10^{-2}$.
Calculating the cube root: $\sqrt[3]{1.375} \approx 1.112$.
So, the solubility S is approximately $1.11 \times 10^{-2}$ M.
This matches option (A). (Note: The provided answer key may have listed a different option, but this is the chemically correct calculation).