Question

What is the molar solubility of $\text{Al(OH}{{\text{)}}_{\text{3}}}$in $\text{0}\text{.2 M NaOH}$solution?
Given that, the solubility product of $\text{Al}{{\left( \text{OH} \right)}_{\text{3}}}\text{ = 2}\text{.4 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-24}}}$
A)$12\times {{10}^{-23}}$
B) $12\times {{10}^{-21}}$
C) $3\text{ }\times {{10}^{-19}}$
D) $3\times {{10}^{-22}}$

Answer 1

Solubility is the property of solute to get dissolved in the solvent to form a solution. The solubility product is the measure of the number of solute dissolves. the equilibrium constant of the solubility ${{\text{K}}_{\text{sp}}}$ is proportional to the concentration of ions of solute in a solution.

Complete step by step answer:
Solubility is the characteristic feature of solute. It determines the dissolution of a solute in the solvent to form a solution. The solubility product ${{\text{K}}_{\text{sp}}}$ is the measure of the concentration of ions that goes into the solution.
The solubility product for general solute $\text{A}{{\text{B}}_{\text{2}}}$ as shown below.
$\text{A}{{\text{B}}_{\text{2}}}\rightleftarrows {{\text{A}}^{\text{+}}}\text{+2}{{\text{B}}^{\text{-}}}$
Therefore, ${{\text{K}}_{\text{sp}}}=\left[ {{\text{A}}^{\text{+}}} \right]{{\left[ {{\text{B}}^{\text{-}}} \right]}^{\text{2}}}$
Where $\left[ {{\text{A}}^{\text{+}}} \right]$ and $\left[ {{\text{B}}^{\text{-}}} \right]$ represent the molar solubility of the solute $\text{A}{{\text{B}}_{\text{2}}}$.
We are given an $\text{Al(OH}{{\text{)}}_{\text{3}}}$ in $\text{NaOH}$ solution. The data given is
Molarity of $\text{NaOH}$solution$\text{=0}\text{.2M}$
Solubility product $\text{Al(OH}{{\text{)}}_{\text{3}}}$is $2.4\times {{10}^{-24}}\text{M}$
To find the molar solubility (S) of $\text{Al(OH}{{\text{)}}_{\text{3}}}$
The $\text{Al(OH}{{\text{)}}_{\text{3}}}$dissociates itself into $\text{A}{{\text{l}}^{\text{3+}}}$and$\text{O}{{\text{H}}^{\text{-}}}$. One mole of $\text{Al(OH}{{\text{)}}_{\text{3}}}$dissociates into one mole of $\text{A}{{\text{l}}^{\text{3+}}}$and three moles of $\text{O}{{\text{H}}^{\text{-}}}$-.The reaction of $\text{Al(OH}{{\text{)}}_{\text{3}}}$as follows:
$\text{Al(OH}{{\text{)}}_{\text{3}}}\rightleftarrows \text{A}{{\text{l}}^{3+}}+3\text{O}{{\text{H}}^{\text{-}}}$
Thus the solubility product ${{\text{K}}_{\text{sp}}}$for $\text{Al(OH}{{\text{)}}_{\text{3}}}$ is
${{\text{K}}_{\text{sp}}}\text{=}\left[ \text{A}{{\text{l}}^{\text{3+}}} \right]{{\left[ \text{O}{{\text{H}}^{\text{-}}} \right]}^{\text{3}}}$
Let the concentration of ions formed by the dissociation of $\text{Al(OH}{{\text{)}}_{\text{3}}}$ as ‘S’.Let the concentration of $\text{Al(OH}{{\text{)}}_{\text{3}}}$ as ‘S’ before dissociation in solution.
$\begin{aligned} & \text{ Al(OH}{{\text{)}}_{\text{3}}}\rightleftarrows \text{A}{{\text{l}}^{3+}}+3\text{O}{{\text{H}}^{\text{-}}} \\\ & Before\text{ S 0 0} \\\ & After\text{ 0 S 3S} \\\ \end{aligned}$
${{\text{K}}_{\text{sp}}}\text{=}\left[ \text{A}{{\text{l}}^{\text{3+}}} \right]{{\left[ \text{O}{{\text{H}}^{\text{-}}} \right]}^{\text{3}}}$
Since we know that the $\text{Al(OH}{{\text{)}}_{\text{3}}}$ is prepared in $\text{0}\text{.2 M NaOH}$solution. We know that the$\text{NaOH}$ dissociates in solution.
$\text{NaOH}\to \text{N}{{\text{a}}^{+}}+\text{O}{{\text{H}}^{\text{-}}}$
one mole $\text{NaOH}$of gives one mole of $\text{N}{{\text{a}}^{+}}$ the one mole of $\text{O}{{\text{H}}^{\text{-}}}$oh as shown below;

& \text{ NaOH}\to \text{N}{{\text{a}}^{+}}+\text{O}{{\text{H}}^{\text{-}}} \\\ & Before\text{ 0}\text{.2M 0 0} \\\ & After\text{ 0 0}\text{.2M 0}\text{.2M} \\\ \end{aligned}$$ Thus we can say that$$\text{0}\text{.2 M NaOH}$$results in the formation of $$\text{0}\text{.2M}$$ $$\text{N}{{\text{a}}^{+}}$$ and $\text{O}{{\text{H}}^{\text{-}}}$ Therefore, the concentration $\text{O}{{\text{H}}^{\text{-}}}$ for$$\text{NaOH}$$is $$\text{0}\text{.2 M}$$. Now, let us find out the total concentration of $\text{O}{{\text{H}}^{\text{-}}}$ in the solution. $$\text{Total con}\text{.of O}{{\text{H}}^{\text{-}}}\text{=concentration of O}{{\text{H}}^{\text{-}}}\text{ from Al(OH}{{\text{)}}_{\text{3}}}\text{+concentration of O}{{\text{H}}^{\text{-}}}\text{ in NaOH}$$ $\left[ \text{O}{{\text{H}}^{\text{-}}} \right]=3S+0.2$ The molar solubility of a solute is equal to the number of moles of solute dissolved per liter of the solution after the solution has reached the point of saturation. It is obtained from the solubility product.its unit is mol/L or M. Let's substitute the value for the concentration of hydroxide in the equation of $${{\text{K}}_{\text{sp}}}$$$${{\text{K}}_{\text{sp}}}\text{=}\left[ \text{A}{{\text{l}}^{\text{3+}}} \right]{{\left[ \text{O}{{\text{H}}^{\text{-}}} \right]}^{\text{3}}}$$ $${{\text{K}}_{\text{sp}}}\text{=(S)(}3S+0.2{{)}^{3}}$$ Since, $\text{S0}\text{.2M}$, the S term can be neglected. Therefore we get, $$\begin{aligned} & \text{=}\left( S \right){{\left( 0.2 \right)}^{\text{3}}} \\\ & =(S)(0.08) \\\ & =0.08S \\\ \end{aligned}$$ Since we know that$${{\text{K}}_{\text{sp}}}=2.4\times {{10}^{-24}}$$ $$\begin{aligned} & 2.4\times {{10}^{-24}}=0.08S \\\ & S=\frac{2.4\times {{10}^{-24}}}{0.08} \\\ & S=3.0\times {{10}^{-22}} \\\ \end{aligned}$$ Therefore molar solubility(S) of $\text{Al(OH}{{\text{)}}_{\text{3}}}$is $$3.0\times {{10}^{-22}}$$. **Hence, the correct option is (D)** **Note:** Remember that the molar solubility (S) is very low as compared to the concentration of the solution. Therefore neglect the term S for simplification. The solubility product is not the same as that of molar solubility. Solubility is the amount dissolved in solution but molar solubility is no solute dissolved per liter of solution.