Question

How can I calculate solubility of calcium hydroxide?

Answer 1

Hint: The solubility of calcium hydroxide is based on the dissociation of calcium hydroxide. The solubility product of calcium hydroxide will give the solubility of calcium hydroxide.

Complete step by step answer:
Calcium hydroxide is an inorganic compound. It consists of calcium (\[C{a^{2 + }}\]) and hydroxide (\[O{H^ - }\]) ions. The chemical formula of calcium hydroxide is \[Ca{\left( {OH} \right)_2}\]. It appears as a white powder. The synonyms of calcium hydroxide are hydrated lime, caustic lime, slack lime, or pickling lime. It is generated by the reaction of quicklime with water.
$CaO + {H_2}O \to Ca{(OH)_2}$
\[Ca{\left( {OH} \right)_2}\] is a very strong base with a pH value of \[12.8\] . Calcium hydroxide is sparingly soluble in water. So the compound remains in equilibrium on dissolution in water. But the amount of calcium hydroxide which dissolves in water completely dissociates into cations and anions. Thus the compound is assumed as a strong base due to \[100\% \] ionization. The dissolution of calcium hydroxide is shown as
$Ca{(OH)_2} \rightleftharpoons C{a^{2 + }} + 2O{H^ - }$
The solubility of calcium hydroxide is the property by virtue of which we get to know the amount of calcium hydroxide which dissolves in water. When a solute is in equilibrium with the solution the product of the concentration of ions gives an equilibrium constant which is known as solubility product. So the solubility product of calcium hydroxide is
${K_{sp}} = [C{a^{2 + }}]{[O{H^ - }]^2}$
The solubility product constant for \[Ca{\left( {OH} \right)_2}\] is reported as \[5.5 \times {10^{ - 6}}\].
$Ca{(OH)_2} \rightleftharpoons C{a^{2 + }} + 2O{H^ - }$
From the balanced equation, if \[x\] is the amount of dissociation of calcium hydroxide at equilibrium, then the concentration of calcium ion \[\left[ {C{a^{2 + }}} \right] = x\] and that of hydroxide ion \[\left[ {O{H^ - }} \right] = 2x\].
Thus \[{K_{sp}}\; = [C{a^{2 + }}]{[O{H^ - }]^2}\]
\[{K_{sp}} = x \times {(2x)^2} = 4{x^3}\]
So $4{x^3} = 5.5 \times {10^{ - 6}}$
${x^3} = 1.375 \times {10^{ - 6}}$
$x = 0.011$.
Hence the solubility of calcium hydroxide is \[0.011mol/L\].

Note:
The value of solubility products depends on temperature variance. In general, the solubility product (\[{K_{sp}}\]) of a solute increases with increase in temperature as the solubility is increased.