Question

The solubility of calcium hydroxide is $s\,mol.liter{e^{ - 1}}$. The solubility product under the same condition will be:
A.$4{s^3}$
B.$2{s^3}$
C.$2{s^2}$
D.${s^3}$

Answer 1

Solubility is the maximum amount of solute that can dissolve in a known quantity of solvent at a certain temperature. The solubility product is calculated with help of the solubility of the
compound and mole of the compound.

Complete Step by step solution:
The solubility product is calculated with equilibrium constant and its value depends on temperature. ${K_{sp}}$ increases with an increase in temperature when temperature increase solubility is also increased.
For the reaction
${A_a}{B_b}(s) \rightleftharpoons a{A^ + }(aq) + b{B^ - }(aq)$
The solubility expression is ${K_{sp}} = {\left[ {{A^{b + }}} \right]^a}{\left[ {{B^{a - }}} \right]^b}$
For these equations:
1.A and B represent different ions and solids. In these equations, there are also referred to as “products”.
2.a and b represent coefficients used to balance the equation.
3.(aq) and (s) indicate which state the products are in (aqueous or solid, respectively)
4.Brackets stand for molar concentration.

With the use of the above example, we calculate the solubility product of calcium hydroxide,
Calcium hydroxide is dissolved into $C{a^{2 + }}$ and $2O{H^ - }$
$Ca{(OH)_2} \to [C{a^{2 + }}] + {[O{H^ - }]^2}$
$Ca{(OH)_2}$ Is a solid, its concentration before and after the reaction is the same. The equilibrium equation can be :
${K_{sp}} = \left[ {Ca{{\left( {OH} \right)}_2}} \right]$
${K_{sp}}$ is known as the solubility product of concentrations of ions of both barium and sulphate is equal to the solubility product constant. In the above reaction, calcium hydroxide is taken ‘s’ as the value of solubility.
$\left[ {C{a^{2 + }}} \right] = s\,mol\,li{t^{ - 1}}$
$\left[ {O{H^ - }} \right] = 2s\,mol\,li{t^{ - 1}}$
The solubility product ${K_{sp}} = \left[ {C{a^{2 + }}} \right]\left[ {O{H^ - }} \right] = s{\left( {2s} \right)^2} = 4{s^3}$

Thus the answer is (A)

Note: To write ${K_{sp}}$ expressions, correctly, you need to have a good knowledge of chemical names, polyatomic ions, and the charges associated with each ion. In these equations, each concentration(represented by square brackets) is raised to the power of its coefficient in the balanced ${K_{sp}}$ expression. Unlike ${K_a}$ and ${K_b}$ for acids and bases, the relative values ${K_{sp}}$ cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions. The solubility product is the product of the solubilities of the ions in units of molarity ($mol\,li{t^{ - 1}}$)