Question

In a saturated of the sparingly soluble strong electrolyte \[Agl{O_3}\]​ (molecular mass = 283), the equilibrium which sets in is
\[AgI{O_3}\left( s \right)\underset {} \leftrightarrows A{g^ + }\left( {aq} \right) + I{O^ - }\left( {aq} \right)\]
If the solubility product constant ${K_{sp}}$​ of\[Agl{O_3}\]​ at a given temperature is$1.0 \times {10^{ - 8}}$, what is the mass of \[Agl{O_3}\] contained in 100 mL of its saturated solution?
A.$1.0 \times {10^{ - 7}}$
B.$1.0 \times {10^{ - 4}}$
C.$28.3 \times {10^{ - 2}}$
D.$2.83 \times {10^{ - 3}}$

Answer 1

Hint: To solve these types of questions you should know the basic principles and the basic formula of the solubility product constant i.e. ${\left[ {{M^{y + }}} \right]^x}{\left[ {{A^{x - }}} \right]^y}$to find the correct option.

Complete answer:
According to the question we are given,
The solubility product constant ${K_{sp}}$ of \[Agl{O_3}\] = $1.0 \times {10^{ - 8}}$
And we know the molecular mass of \[Agl{O_3}\] which is 283.
We have to calculate the mass of \[Agl{O_3}\] contained in 100ml of its saturated solution.
We know that ${K_{sp}}$ equation is written as: ${\left[ {{M^{y + }}} \right]^x}{\left[ {{A^{x - }}} \right]^y}$
So,
Let us consider the solubility of \[Agl{O_3}\] as “s”.
Then, ${K_{sp}}$ =${\left[ {Ag} \right]^ + }{\left[ {L{O_3}} \right]^ - }$
${K_{sp}}$=$1.0 \times {10^{ - 8}}$
So, $1.0 \times {10^{ - 8}} = {s^2}$
Then s =${10^{ - 4}}$ mol/liter
Now, formula of solubility is “\[Solubility{\text{ }}in{\text{ }}gram{\text{ }} \times 100\]”
Substituting the value in the formula of solubility
$ \Rightarrow $\[{10^{ - 4}} \times \dfrac{{283}}{{1000}} \times 100\]
$ \Rightarrow $$283 \times {10^{ - 5}}$
$ \Rightarrow $$2.83 \times {10^{ - 3}}$g/100 ml
Hence, the correct option is D.

NOTE: In the above formulation we used a term the solubility of product constant Ksp that can be defined as it is the constant of equilibrium for a solid substance that dissolves in an aqueous solution. The more soluble a substance is, the higher its Ksp value will become. Its equation is: ${\left[ {{M^{y + }}} \right]^x}{\left[ {{A^{x - }}} \right]^y}$