Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

In a saturated solution of the sparingly soluble strong electrolyte \[AgI{O_3}\](molecular mass =283) the equilibrium which sets in is\[AgI{O_3} \rightleftharpoons A{g^ + }(aq) + I{O_3}^ - (aq)\]. 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. \[1.0 \times {10^{ - 4}}g\]

seo-qna
Last updated date: 13th Jun 2024
Total views: 51.6k
Views today: 0.51k
Answer
VerifiedVerified
51.6k+ views
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.