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

To calculate the solubility of calcium hydroxide, Ca(OH)₂, in water using its solubility product, we need to apply the concept of solubility product constant (K_sp). The solubility product constant is the product of the molar concentrations of the dissociated ions each raised to the power of their respective coefficients from the balanced equation.

The dissociation of Ca(OH)₂ in water can be represented as:

Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^{-}(aq)

Let the solubility of Ca(OH)₂ be S mol/L. At equilibrium, the concentrations of the ions will be as follows:

• Concentration of Ca^{2+} = S mol/L
• Concentration of OH^{-} = 2S mol/L

The expression for the solubility product (K_sp) of Ca(OH)₂ is given by:

K_{sp} = [Ca^{2+}][OH^{-}]^2 = S \cdot (2S)^2 = 4S^3

Substituting the given value of K_sp:

4S^3 = 5.5 \times 10^{-6}

Solving for S:

S^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6}

S = \sqrt[3]{1.375 \times 10^{-6}}

Upon calculating the cube root, we get:

S \approx 1.11 \times 10^{-2} mol/L

Therefore, the solubility of Ca(OH)₂ in water is 1.11 \times 10^{-2} mol/L.

The correct answer is: 1.11 × 10^-2

This step-by-step calculation shows how the solubility product relates to the solubility of a compound, exemplified by the calculation of the solubility of Ca(OH)₂ in water.