Question

At ${25^o}C$ what will the solubility of silver carbonate be in $0.1M\,N{a_2}C{O_3}$ solutions. At this temperature ${K_{sp}}$ of silver carbonate is $4\, \times \,{10^{ - 13}}$?
A.$2\, \times \,{10^{ - 7}}$
B.$2\, \times \,{10^{ - 6}}$
C.${10^{ - 6}}$
D.${10^{ - 7}}$

Answer 1

Here, we can use the formula of solubility product ${K_{sp}}$. It is a kind of equilibrium constant for the dissolution of a solid substance into an aqueous solution. The equilibrium is formed between the solid solute and its constituent ions that are dissociated in the solutions.

Complete answer:
Solubility is the property of a solute particle to get dissolved in a solvent to form a solution. When the ionic compound is dissolved in water, it dissociates into ions i.e cations and anions.
Basically, the solubility product is the type of equilibrium constant for the dissolution of solid substance into an aqueous solution. It is denoted as ${K_{sp}}$. It depends upon the lattice enthalpy of salts and solvation enthalpy of ions in the solutions. It also depends on the temperature. When the temperature increases, the solubility increases and hence, the solubility product increases. Solubility product ${K_{sp}}$indicates the extent of solid that can be dissolved in liquids.
The general dissolution reaction in aqueous solution is:
$aA(s)\,\,\,\,\,\overset {} \leftrightarrows \,\,\,\,cC\,(aq) + \,\,dD\,(aq)$
${K_{sp}}\, = \,{[C]^c}{[D]^d}$
The concentration of solids are included as changes in their concentration are significant.
According to the question, the following reaction occurs:
Let, solubility= $x\,\,moles/litres$
$A{g_2}C{O_3}\,\,\,\xrightarrow{{}}\,\,2A{g^{ + \,}}\, + \,\,C{O_3}^{2 - }$
$\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x$
$N{a_2}C{O_3}\,\,\,\xrightarrow{{}}\,\,2N{a^{ + \,}}\, + \,\,C{O_3}^{2 - }$
$0.1M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.2M\,\,\,\,\,\,\,\,\,0.1M$
${K_{sp}} = \,[A{g^ + }][C{O_3}^{2 - }]$
$[A{g^ + }] = 2x$
$[C{O_3}^{2 - }] = \,0.1\, + \,x$
${K_{sp}}$for silver carbonate = $4\, \times \,{10^{ - 13}}$
${K_{sp}}\, = \,(2x)\,{(0.1 + x)^2}$
$4\, \times \,{10^{ - 13}}\, = \,{(2x)^2}\,(0.1 + x)$
$x \ll \,0.1$, neglect it.
$4\, \times \,{10^{ - 13}}\, = \,0.4{x^2}$
By solving this equation we get,
$x\, = {10^{ - 6}}$
The solubility of silver carbonate in $0.1M\,N{a_2}C{O_3}$ solutions = $x\, = {10^{ - 6}}$
Hence, the correct answer is option C.

Note:
It must be remembered that the solubility and solubility product are temperature dependent..
There are three types of categories on the basis of their solubility
-If Solubility$> 0.1M$, then the salts are completely soluble.
-If $0.01M <$Solubility$< 0.1M$, then the salts are slightly soluble.
-If Solubility $< 0.1M$ , then the salts are sparingly soluble.