Question

What concentration of $S{O_3}^{2 - }$ is in equilibrium with $A{g_2}S{O_3}\left( s \right)$ and $1.80 \times {10^{ - 3}}MA{g^ + }$ ? The ${K_{sp}}$ of $A{g_2}S{O_3}$ is $1.50 \times {10^{ - 14}}$ .

Answer 1

We have to remember that the chemical formula of silver sulphide is $A{g_2}S{O_3}\left( s \right)$ . The ${K_{sp}}$ 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.

Formula used: As we know that the stability product of a compound is equal to the product of the molar concentration of respective ions is multiplied by the number of formed ions in the decomposition reaction.
For example, the equilibrium reaction
${X_m}{Y_n} \rightleftharpoons m{X^{n + }} + n{Y^{m - }}$
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 step by step answer:
We can write the decomposition of silver sulphide in water as,
$A{g_2}S{O_3} \rightleftharpoons A{g^ + } + S{O_3}^{2 - }$
Silver sulfate is soluble in water and then forms two moles of $A{g^ + }$ ion and one mole of $S{O_3}^{2 - }$ ion in the equilibrium reaction.
Solubility product of the reaction is
${K_{sp}} = {\left[ {A{g^ + }} \right]^2}{\left[ {S{O_3}} \right]^{2 - }}$
The ${K_{sp}}$ of $A{g_2}S{O_3}$ is $1.50 \times {10^{ - 14}}$
The concentration of $A{g^ + }$ is $1.80 \times {10^{ - 3}}M$ .
We changed the formula for find the concentration of $S{O_3}^{2 - }$ is
${\left[ {S{O_3}} \right]^{2 - }} = \dfrac{{{K_{sp}}}}{{{{\left[ {A{g^ + }} \right]}^2}}}$
Now we can substitute the given values we get,
$= \dfrac{{1.50 \times {{10}^{ - 14}}}}{{{{\left( {1.8 \times {{10}^{ - 3}}} \right)}^2}}}$
$\Rightarrow {\left[ {S{O_3}} \right]^{2 - }} = \dfrac{{1.5 \times {{10}^{ - 14}}}}{{3.24 \times {{10}^{ - 6}}}} = 0.463 \times {10^{ - 8}}$
On simplification we get,
$\Rightarrow {\left[ {S{O_3}} \right]^{2 - }} = 4.63 \times {10^{ - 9}}$
The concentration of $S{O_3}^{2 - }$ is formed when the decomposition reaction is $4.62 \times {10^{ - 9}}$ .

Hence, the concentration of $S{O_3}^{2 - }$ is $4.62 \times {10^{ - 9}}M$ .

Additional information:
- In the solubility product of the decomposition reactions each constituent ion raised to the power of its stoichiometric coefficient in the equilibrium balanced reaction. For the decomposition of the reaction, it forms cation and anion. The positive charge of the ion is called cation. The negative charge of the ion is called an anion. When the sum of the formed ions is equal to the zero, that means neutral.
- 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 the 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 form unsaturation. If the solubility product of the compound is equal to the ionic product means, the reaction in the state of equilibrium and solution in form of saturation.

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 ions it means 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.