Question

The expression for the solubility product of $A{g_2}C{O_3}$ will be:
A. ${K_{sp}} = {s^2}$
B. ${K_{sp}} = 4{s^3}$
C. ${K_{sp}} = 27{s^4}$
D. ${K_{sp}} = s$

Answer 1

Hint:First we have to write the balanced equation clearly showing the ions involved.Since we are given the solubility of the salt, we can use this as the concentration unit for each term. From the reaction equation, we can find how many moles are dissociated at equilibrium and write this in terms of the solubility. From this, we can find the solubility product, which is nothing but the equilibrium constant which uses solubility as the concentration term.
Formulas used: ${K_{sp}} = {\left[ {{A^ + }} \right]^a}{\left[ {{B^ - }} \right]^b}$
Where ${A^ + }$ and ${B^ - }$ are the respective ions and $a,b$ denote their stoichiometric coefficients, and the brackets denote concentrations

Complete step by step answer:
The solubility product of a substance is defined as the product of concentration of its ions raised to their respective stoichiometric coefficients. Thus:
${K_{sp}} = {\left[ {{A^ + }} \right]^a}{\left[ {{B^ - }} \right]^b}$
Where ${A^ + }$ and ${B^ - }$ are the respective ions and $a,b$ denote their stoichiometric coefficients, and the brackets denote concentrations.
Let us now write the balanced chemical reaction equation involved when $A{g_2}C{O_3}$ is dissolved:
$A{g_2}C{O_3} \rightleftharpoons 2A{g^ + } + CO_3^{2 - }$
Hence, we can write ${K_{sp}} = {\left[ {A{g^ + }} \right]^2}{\left[ {CO_3^{2 - }} \right]^1}$
If we take the solubility of silver carbonate as $s$, we notice that at equilibrium, twice this amount (in moles) of silver ions will be soluble, and the same amount of carbonate ions will be soluble, since their respective stoichiometric coefficients are $1$ and $2$ respectively. As solubility is the same as concentration of the substance in the solution, we can use it in the expression for solubility product. As we have deduced, the solubility of $A{g^ + }$ will be twice that of $A{g_2}C{O_3}$, that is, $2s$ and the solubility of $CO_3^{2 - }$ will be the same as that of $A{g_2}C{O_3}$ ($s$). Substituting this in our equation, we have:
${K_{sp}} = {\left[ {2s} \right]^2}\left[ s \right]$
On solving this, we get:
${K_{sp}} = 4{s^2} \times s = 4{s^3}$
Hence, the solubility product is $4{s^3}$. Therefore, the correct option to be marked is B.

Note: Solubility product is dependent on temperature and usually increases as temperature is increased, since solubility of ions increase when temperature is raised. Note that solubility is dependent on a number of factors, the major ones being lattice enthalpy (the energy needed to break the intermolecular forces of attraction between the salt ions) and solvation enthalpy (energy released when a salt is dissolved in a solvent). Thus, for a salt to be soluble in a particular solvent, its solvation enthalpy in that solvent should be greater than lattice enthalpy.