Question

Given,\[{{\text{K}}_{{\text{sp}}}}\] of \[{\text{AgCl}}\] is \[1 \times {10^{ - 10}}\]. Its solubility in \[{\text{0}}{\text{.1M}}\]\[{\text{KN}}{{\text{O}}_{\text{3}}}\]? Will be
A.${10^{ - 5}}moles/lit$
B.$ > {10^{ - 5}}moles/lit$
C.$ < {10^{ - 5}}moles/lit$
D.None of these

Answer 1

Hint: We need to know that the chemical formula of silver sulphite is $A{g_2}S{O_3}(s)$. The \[{{\text{K}}_{{\text{sp}}}}\] is known as the solubility product of the compound. The solubility product of the compound is defined as the product of the molar concentration of the product in the form of constituent ions formed in the decomposition reaction. In these each constituent ion is raised to the power of its stoichiometric coefficient in the equilibrium balanced reaction. For the decomposition of the reaction form cation and anion. The positive charge of the ion is called cation. The negative charge of the ion is called an anion. The sum of the formed ions is equal to the zero that means neutral.
Formula used:
The stability product of a compound is equal to the product of the molar concentration of each ion is multiplied by the number of formed ions in the decomposition reaction.
For example, the equilibrium reaction
\[{X_m}{Y_n} \to m{X^{n + }} + {\text{ }}n{Y^{m - }}\]
Here, two ions are formed in the decomposition reaction of \[{X_m}{Y_n}\] are \[\;{X^{n + }}\] and \[{Y^{ - m}}\]. The n number \[{Y^{ - m}}\] and m number of \[\;{X^{n + }}\] are formed.
Hence, the solubility product of the above decomposition reaction is,
\[{K_{sp}} = {\left[ {{X^{n + }}} \right]^m}{\left[ {{Y^{m - }}} \right]^n}\]

Complete answer:
Solubility product of the \[AgCl\] is
The \[{{\text{K}}_{{\text{sp}}}}\] of \[{\text{AgCl}}\] is \[1 \times {10^{ - 10}}\]
The concentration of \[{\text{KN}}{{\text{O}}_{\text{3}}}\] is \[{\text{0}}{\text{.1M}}\].
We changed the formula for find the concentration is
The concentration is equal to the ratio between \[{{\text{K}}_{{\text{sp}}}}\] and concentration of \[{\text{KN}}{{\text{O}}_{\text{3}}}\].
The concentration of \[{\text{AgCl}}\] is calculate below,
 \[ = \dfrac{{1 \times {{10}^{ - 10}}}}{{0.1}} = 1 \times {10^{ - 9}}M\]
According to the above discussion, we conclude the concentration will be \[1 \times {10^{ - 9}}{\text{M}}\].

Hence, option D is the correct answer.

Note:
We need to know that the solubility product of the compound is used to find the molar concentration of ions. If we find the one molar concentration of ion, it is easy to calculate the molar concentration of other ions in the equilibrium reaction. The solubility product of the compound is divided by the molar concentration of ions to find the molar concentration of other ions. The solubility product of the compound is used to find the nature of the reaction and precipitation condition. If the solubility product of the compound is greater than ionic product means, precipitation will occur and the solution in the form of supersaturation. If the solubility product of the compound is lesser than ionic product means, no precipitation and solution in the form of unsaturation. If the solubility product of the compound is equal to the ionic product, the reaction in the state of equilibrium and solution in form of saturation.