Answer
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Hint: Try to recall that solubility product is defined as the product of molar concentrations of ions in a saturated solution raised to their stoichiometric coefficients. Now, by using this you can easily find the correct option from the given ones.
Complete step by step answer:
It is known to you that \[AgI{O_3}\] is a sparingly soluble salt and \[AgI{O_3}\] ionizes completely in the solution as: \[AgI{O_3} \to A{g^ + } + I{O_3}^ - \].
Calculation:
Given, \[{K_{sp}}\]= \[1.0 \times {10^{ - 8}}\]--------1
On complete ionization: \[AgI{O_3} \to A{g^ + } + I{O_3}^ - \].
So, let the solubility of \[\left[ {A{g^ + }} \right] = \left[ {I{O_3}^ - } \right] = s\].
\[\begin{gathered}
AgI{O_3} \to A{g^ + } + I{O_3}^ - \\
{\text{s 0 0}} \\
{\text{0 s s}} \\
\end{gathered} \].
Therefore, \[{K_{sp}} = \left( s \right)\left( s \right) = {s^2}\]---------2
Equating eq. 1 and 2 we get,
\[\begin{gathered}
{s^2} = 1.0 \times {10^{ - 8}} \\
or,s = {10^{ - 4}}mol/L \\
\end{gathered} \].
So, \[\left[ {AgI{O_3}} \right] = s = {10^{ - 4}}mol/L\]
Given, volume of solution =100 ml=0.1L
Let the number of moles of \[AgI{O_3}\] be n.
\[\begin{gathered}
\dfrac{n}{{0.1}} = {10^{ - 4}} \\
or,n = {10^{ - 5}} \\
\end{gathered} \]
Also, given molar mass of \[AgI{O_3}\]= 283
Let the mass of \[AgI{O_3}\] be x.
\[\begin{gathered}
\dfrac{x}{{283}} = {10^{ - 5}} \\
or,x = 283 \times {10^{ - 5}} \\
or,x = 2.83 \times {10^{ - 3}}g \\
\end{gathered} \]
Hence, from the above calculation we can easily conclude that option B is the correct option to the given question.
Note:It should be remembered to you that if to the solution of a weak electrolyte which ionizes to a small extent, a strong electrolyte having a common ion is added which ionizes almost completely, the ionization of weak electrolyte is further suppressed.Similarly, if the solution of a sparingly soluble salt if a soluble salt having a common ion is added, the solubility of the sparingly soluble salt further decreases.
Complete step by step answer:
It is known to you that \[AgI{O_3}\] is a sparingly soluble salt and \[AgI{O_3}\] ionizes completely in the solution as: \[AgI{O_3} \to A{g^ + } + I{O_3}^ - \].
Calculation:
Given, \[{K_{sp}}\]= \[1.0 \times {10^{ - 8}}\]--------1
On complete ionization: \[AgI{O_3} \to A{g^ + } + I{O_3}^ - \].
So, let the solubility of \[\left[ {A{g^ + }} \right] = \left[ {I{O_3}^ - } \right] = s\].
\[\begin{gathered}
AgI{O_3} \to A{g^ + } + I{O_3}^ - \\
{\text{s 0 0}} \\
{\text{0 s s}} \\
\end{gathered} \].
Therefore, \[{K_{sp}} = \left( s \right)\left( s \right) = {s^2}\]---------2
Equating eq. 1 and 2 we get,
\[\begin{gathered}
{s^2} = 1.0 \times {10^{ - 8}} \\
or,s = {10^{ - 4}}mol/L \\
\end{gathered} \].
So, \[\left[ {AgI{O_3}} \right] = s = {10^{ - 4}}mol/L\]
Given, volume of solution =100 ml=0.1L
Let the number of moles of \[AgI{O_3}\] be n.
\[\begin{gathered}
\dfrac{n}{{0.1}} = {10^{ - 4}} \\
or,n = {10^{ - 5}} \\
\end{gathered} \]
Also, given molar mass of \[AgI{O_3}\]= 283
Let the mass of \[AgI{O_3}\] be x.
\[\begin{gathered}
\dfrac{x}{{283}} = {10^{ - 5}} \\
or,x = 283 \times {10^{ - 5}} \\
or,x = 2.83 \times {10^{ - 3}}g \\
\end{gathered} \]
Hence, from the above calculation we can easily conclude that option B is the correct option to the given question.
Note:It should be remembered to you that if to the solution of a weak electrolyte which ionizes to a small extent, a strong electrolyte having a common ion is added which ionizes almost completely, the ionization of weak electrolyte is further suppressed.Similarly, if the solution of a sparingly soluble salt if a soluble salt having a common ion is added, the solubility of the sparingly soluble salt further decreases.
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