Question

In a saturated solution of the sparingly soluble strong electrolyte $AgI{O_3}$(Molecular mass = 283) the equilibrium which sets in is
$AgI{O_3}(s) \rightleftharpoons A{g^ + }(aq) + IO_3^ - (aq)$
If the solubility product constant ${K_{sp}}$of $AgI{O_3}$ at a given temperature is 1.0$ \times {10^{ - 8}}$, what is the mass of $AgI{O_3}$ contained in 100 ml of its saturated solution ?
a) 28.3$ \times {10^{ - 2}}$g
b) 2.83$ \times {10^{ - 3}}$g
c) 1.0\[ \times {10^{ - 7}}\]g
d) 1.0$ \times {10^{ - 4}}$g

Answer 1

Hint: The solubility product constant gives us the product of the solubility of all components in a solution. In case of $AgI{O_3}$, it can be given by the formula -
${K_{sp}}$=$[A{g^ + }][IO_3^ - ]$

Complete answer:
First, let us write what is given to us and what we need to find out.
Given :
Solubility product constant (${K_{sp}}$) of $AgI{O_3}$= 1.0$ \times {10^{ - 8}}$
Volume of saturated solution = 100 ml
Molecular mass = 283
To find :
mass of $AgI{O_3}$
We know that ${K_{sp}}$ of solution is given by -
${K_{sp}}$=$[A{g^ + }][IO_3^ - ]$
If we say that ‘s’ is the solubility of electrolyte
So, $[A{g^ + }]$=$[IO_3^ - ]$=$[AgI{O_3}]$= s
Thus, ${K_{sp}}$= (s) (s)
${K_{sp}}$=${s^2}$
${s^2}$= 1.0$ \times {10^{ - 8}}$
‘s’ = 1.0$ \times {10^{ - 4}}$M
So, 1000 ml of solution will have electrolyte = 1.0$ \times {10^{ - 4}}$moles
And we have a solution of 100 ml.
Thus, 100 ml of solution will have electrolyte =1.0$ \times {10^{ - 5}}$moles
We have molar mass = 283 g/mol
So, the mass of electrolyte in 100 ml solution = 1.0$ \times {10^{ - 5}} \times 283$
mass of electrolyte in 100 ml solution = 2.83$ \times {10^{ - 3}}$g
So, this is our answer.

Thus, the correct option is option b.).

Note:
The saturated solution is the one which has dissolved the maximum amount of solute. It can not dissolve more solute. Thus, it has the maximum number of moles of an electrolyte that can be present. While the unsaturated solution is one which can dissolve more solute.