Question

At ${30^0}C$, in which of the one litre solution, the solubility of \[A{g_2}C{O_3}\] ​ (solubility product \[ = 8 \times {10^{ - 12}}\]) will be maximum
A.0.05 M \[N{a_2}C{O_3}\]
B.Pure water
C.0.1 M NaBr
D.0.05 M \[N{H_3}\]

Answer 1

Hint:First of all, we should know the definition of solubility, it is the weight of solute in grams present in \[100g\] of solvent. It is denoted by s and is expressed in mole per litre or gram per litre. Solubility is inversely proportional to the concentrations of common ions or number of common ions. And we also know \[AgN{O_3}\]​, \[N{a_2}C{O_3}\]​ causes a common ion effect which decreases the solubility.

Complete answer:
Given in the question is,
\[{K_{SP}} = 8 \times {10^{ - 12}}\]
Temperature $ = {30^0}C$
In pure water, the reaction will be:
\[A{g_2}C{O_3} \Leftrightarrow 2A{g^ + } + C{O_3}^{2 - }\]
In \[N{H_3}\] ​, the reaction will be: (\[A{g^ + }\]from complex with\[N{H_3}\])
\[A{g_2}C{O_3} \Leftrightarrow 2A{g^ + } + C{O_3}^{2 - }\]
\[2A{g^ + } + 4N{H_3} \Leftrightarrow 2Ag{(N{H_3})^{2 + }}\]
Now, overall reaction: \[A{g_2}C{O_3} + 4N{H_3} \Leftrightarrow 2Ag{(N{H_3})^{2 + }} + C{O_3}^{2 - }\]
Here a complex formation is taking place and due to the formation of the complex reaction between silver ions and ammonia shifts the solubility equilibrium in the forward direction which in turns increases the solubility.
Due to the presence of common ion in these reactions, the solubility is suppressed or it decreases. So, the solubility of \[A{g_2}C{O_3}\]​ would be greatest in 1 L of \[0.05\,\,M\;N{H_3}\]due to the absence of common ion either \[A{g^ + }\]or\[C{O_3}^{2 - }\].
Hence, \[A{g_2}C{O_3}\] will be most soluble in \[N{H_3}\]

Therefore, the correct answer is option (D).
Note:
The solubility product constant which is also denoted by\[{K_{SP}}\], is the equilibrium constant for the equilibrium which is established between a slightly soluble ionic compound and its ions in a saturated aqueous solution.