Question

The solubility of $AgBr{O_3}$ (formula weight $ = 236$) is $0.0072\,g$ in $1000\,mL$. What is the value of ${K_{sp}}$?
A. $2.2 \times {10^{ - 8}}$
B. $3 \times {10^{ - 10}}$
C. \[3 \times {10^{ - 5}}\]
D. $9.3 \times {10^{ - 10}}$

Answer 1

Hint: The solubility product of an ionic compound is defined as the equilibrium constant for dissociation of a solid (ionic) substance into an aqueous solution. It is represented by the symbol ${K_{sp}}$. As the solubility of a solid increases with an increase in temperature, so the value of solubility product increases on increasing the temperature.

Complete answer: As per question, the given data is as follows:
Given mass of $AgBr{O_3} = 0.0072g$
Molecular mass of $AgBr{O_3} = 236\,gmo{l^{ - 1}}$
Therefore, number of moles of $AgBr{O_3} = \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$
Substituting values:
$ \Rightarrow n = \dfrac{{0.0072}}{{236}}$
$ \Rightarrow n = 3.05 \times {10^{ - 5}}\;{\text{moles}}$
Given volume of solution$ = 1000\;mL$
$ \Rightarrow V = 1L$
We know that the concentration of a compound is the ratio of number of moles to the volume of solution. Therefore, concentration of $AgBr{O_3}$ in aqueous solution will be as follows:
Concentration $C = \dfrac{n}{V}$
Substituting values:
$ \Rightarrow C = \dfrac{{3.05 \times {{10}^{ - 5}}}}{1}$
$ \Rightarrow C = 3.05 \times {10^{ - 5}}mol{L^{ - 1}}$
Now, the dissociation of $AgBr{O_3}$ in aqueous solution takes place as follows:
$AgBr{O_3} \rightleftharpoons A{g^ + } + BrO_3^ - $
ICE table for the concentration of molecules in the given reaction is as follows:

	$\left[ {AgBr{O_3}} \right]$	$\left[ {A{g^ + }} \right]$	$\left[ {BrO_3^ - } \right]$
Initial concentration	$3.05 \times {10^{ - 5}}mol{L^{ - 1}}$	$0$	$0$
Change	$ - 3.05 \times {10^{ - 5}}$	$ + 3.05 \times {10^{ - 5}}$	$ + 3.05 \times {10^{ - 5}}$
Concentration at equilibrium	$0$	$3.05 \times {10^{ - 5}}mol{L^{ - 1}}$	$3.05 \times {10^{ - 5}}mol{L^{ - 1}}$

Therefore, the solubility product of $AgBr{O_3}$ can be expressed as follows:
${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {BrO_3^ - } \right]$
Substituting the values of final concentration from ICE table:
$ \Rightarrow {K_{sp}} = 3.05 \times {10^{ - 5}} \times 3.05 \times {10^{ - 5}}$
\[ \Rightarrow {K_{sp}} = 9.3 \times {10^{ - 10}}\]
Hence, the solubility product of $AgBr{O_3}$ under given conditions \[ = 9.3 \times {10^{ - 10}}\].
So, option (D) is the correct answer.

Note:
It is important to note that in ICE table, I stands for initial concentration of reactants and products, C stands for change in concentration i.e., the concentration of compounds required for a reaction to achieve equilibrium and E stands for equilibrium concentration i.e., the concentration of reactants and products after achieving equilibrium.