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Question: The solubility of \(AgBr{O_3}\) (formula weight \( = 236\)) is \(0.0072\,g\) in \(1000\,mL\). What i...

The solubility of AgBrO3AgBr{O_3} (formula weight =236 = 236) is 0.0072 g0.0072\,g in 1000 mL1000\,mL. What is the value of Ksp{K_{sp}}?
A. 2.2×10−82.2 \times {10^{ - 8}}
B. 3×10−103 \times {10^{ - 10}}
C. 3×10−53 \times {10^{ - 5}}
D. 9.3×10−109.3 \times {10^{ - 10}}

Explanation

Solution

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 Ksp{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 AgBrO3=0.0072gAgBr{O_3} = 0.0072g
Molecular mass of AgBrO3=236 gmol−1AgBr{O_3} = 236\,gmo{l^{ - 1}}
Therefore, number of moles of AgBrO3=given massmolecular massAgBr{O_3} = \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}
Substituting values:
⇒n=0.0072236\Rightarrow n = \dfrac{{0.0072}}{{236}}
⇒n=3.05×10−5  moles\Rightarrow n = 3.05 \times {10^{ - 5}}\;{\text{moles}}
Given volume of solution=1000  mL = 1000\;mL
⇒V=1L\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 AgBrO3AgBr{O_3} in aqueous solution will be as follows:
Concentration C=nVC = \dfrac{n}{V}
Substituting values:
⇒C=3.05×10−51\Rightarrow C = \dfrac{{3.05 \times {{10}^{ - 5}}}}{1}
⇒C=3.05×10−5molL−1\Rightarrow C = 3.05 \times {10^{ - 5}}mol{L^{ - 1}}
Now, the dissociation of AgBrO3AgBr{O_3} in aqueous solution takes place as follows:
AgBrO3⇌Ag++BrO3−AgBr{O_3} \rightleftharpoons A{g^ + } + BrO_3^ -
ICE table for the concentration of molecules in the given reaction is as follows:

| [AgBrO3]\left[ {AgBr{O_3}} \right]| [Ag+]\left[ {A{g^ + }} \right]| [BrO3−]\left[ {BrO_3^ - } \right]
---|---|---|---
Initial concentration| 3.05×10−5molL−13.05 \times {10^{ - 5}}mol{L^{ - 1}}| 00| 00
Change| −3.05×10−5 - 3.05 \times {10^{ - 5}}| +3.05×10−5 + 3.05 \times {10^{ - 5}}| +3.05×10−5 + 3.05 \times {10^{ - 5}}
Concentration at equilibrium| 00| 3.05×10−5molL−13.05 \times {10^{ - 5}}mol{L^{ - 1}}| 3.05×10−5molL−13.05 \times {10^{ - 5}}mol{L^{ - 1}}

Therefore, the solubility product of AgBrO3AgBr{O_3} can be expressed as follows:
Ksp=[Ag+][BrO3−]{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {BrO_3^ - } \right]
Substituting the values of final concentration from ICE table:
⇒Ksp=3.05×10−5×3.05×10−5\Rightarrow {K_{sp}} = 3.05 \times {10^{ - 5}} \times 3.05 \times {10^{ - 5}}
⇒Ksp=9.3×10−10\Rightarrow {K_{sp}} = 9.3 \times {10^{ - 10}}
Hence, the solubility product of AgBrO3AgBr{O_3} under given conditions =9.3×10−10 = 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.