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The pH of a saturated solution of $Ca(OH)_2$ is $9.$ The solubility product $(K_{sp})$ of $Ca(OH)_2$ is:

medium

Equilibrium

2019

chemistry

$0.5 \times 10^{-15}$

$0.25 \times 10^{-10}$

$0.125 \times 10^{-15}$

Explanation

To solve this problem, we need to find the solubility product $(K_{sp})$ of $Ca(OH)_2. \\$ Given: $\\$

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$0.5 \times 10^{-10}$

• The pH of the saturated solution is

9. \\

First, calculate the concentration of

OH^-

ions in the solution:

\\

Since pH + pOH = 14, we have:

\\ pOH = 14 - 9 = 5 \\

The concentration of

OH^-

ions is given by:

\\ [OH^-] = 10^{-pOH} = 10^{-5} \\

The dissociation of

Ca(OH)_2

in water is represented by the equation:

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

Let

s

be the solubility of

Ca(OH)_2

in mol/L. Then:

\\ [Ca^{2+}] = s \\ [OH^-] = 2s \\

Since we know

[OH^-] = 10^{-5},

we can write:

\\ 2s = 10^{-5} \\ s = \frac{10^{-5}}{2} = 5 \times 10^{-6} \\

Now, calculate the solubility product

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

Substitute the value of

s: \\ K_{sp} = 4 \times (5 \times 10^{-6})^3 \\ K_{sp} = 4 \times 125 \times 10^{-18} \\ K_{sp} = 500 \times 10^{-18} \\ K_{sp} = 0.5 \times 10^{-15} \\

Therefore, the solubility product

K_{sp}

of

Ca(OH)_2

is

0.5 \times 10^{-15}. \\

This corresponds to Option 1.