Question

The values of \[{K_{SP}}\] of following sparingly soluble salts \[Ni{(OH)_2}\], $Ce{(OH)_4}$,AgCN, $A{l_2}{(S{O_4})_2}$ are respectively \[2 \times {10^{ - 15}},4 \times {10^{ - 35}},6 \times {10^{ - 17}}\] and \[3.2 \times {10^{ - 34}}\] respectively. Which salt is more soluble?
A. \[Ni{(OH)_2}\]
B. $Ce{(OH)_4}$
C. $A{l_2}{(S{O_4})_2}$
D.AgCN

Answer 1

Hint: A salt may give on dissociation two or more than two anions and cations carrying different charges. A solid salt of the general formula \[{M_x}^{a + } + {X_y}^{b - }\] − with molar solubility S in equilibrium with its saturated solution. And its solubility constant \[{K_{SP}} = \left[ {{M^{a + }}} \right] + \left[ {{X^{b - }}} \right]\]

Complete step by step answer:
 \[Ni{(OH)_2} \rightleftharpoons \left[ {N{i^{2 + }}} \right] + \left[ {2O{H^ - }} \right]\]
Let the number of moles of \[Ni{(OH)_2}\] be ‘s’, \[\left[ {N{i^{2 + }}} \right]\] be ‘s’ and \[\left[ {2O{H^ - }} \right]\] be 2s
Given is, \[{K_{SP}} = 2 \times {10^{ - 15}}\]
 \[ \Rightarrow s \times {\left( {2s} \right)^2} = {K_{SP}}\]
 \[ \Rightarrow 4{s^3} = {K_{SP}}\]
$\Rightarrow$ \[s = {\left( {\dfrac{{{K_{SP}}}}{4}} \right)^{\dfrac{1}{3}}}\]
$\Rightarrow$ \[s = {\left( {2 \times \dfrac{{{{10}^{ - 15}}}}{4}} \right)^{\dfrac{1}{3}}} = 5.8 \times {10^{ - 5}}mol\,{L^{ - 1}}\]

 $Ce{(OH)_4} \rightleftharpoons \left[ {C{e^{4 + }}} \right] + \left[ {4O{H^ - }} \right]$
Let the number of moles of $Ce{(OH)_4}$ be , \[\left[ {C{e^{4 + }}} \right]\] be , and \[\left[ {4O{H^ - }} \right]\] be ,
Given is, \[{K_{SP}} = 4 \times {10^{ - 35}}\]
 \[ \Rightarrow {s_2} \times {\left( {4{s_2}} \right)^2} = {K_{SP}}\]
 \[ \Rightarrow {\left( {8{s_3}} \right)^3} = {K_{SP}}\]
$\Rightarrow$ \[s = {\left( {\dfrac{{{K_{SP}}}}{8}} \right)^{\dfrac{1}{3}}}\]
$\Rightarrow$ \[s = {\left( {4 \times \dfrac{{{{10}^{ - 35}}}}{8}} \right)^{\dfrac{1}{3}}} = 1.7099 \times {10^{ - 12}}mol\,{L^{ - 1}}\]

 \[AgCN \rightleftharpoons \left[ {A{g^ + }} \right] + \left[ {C{N^ - }} \right]\]
Let the number of moles of AgCN be , \[\left[ {A{g^ + }} \right]\] be , and \[\left[ {C{N^ - }} \right]\] be ,
Given is, \[{K_{SP}} = 6 \times {10^{ - 17}}\]
 \[ \Rightarrow {\left( {{s_3}} \right)^2} = {K_{SP}}\]
 \[ \Rightarrow {s_3} = \surd {K_{SP}}\]
$\Rightarrow$ \[{S_3} = \surd 6 \times {10^{ - 17}}\]
$\Rightarrow$ \[{S_3} = 7.8 \times {10^{ - 9}}mol\,{L^{ - 1}}\]

 $A{l_2}{(S{O_4})_2} \rightleftharpoons \left[ {2A{l^{2 + }}} \right] + [2S{O_4}^{2 - }]$
Let the number of moles of $A{l_2}{(S{O_4})_2}$ be , \[\left[ {2A{l^{2 + }}} \right]\] be , and $[2S{O_4}^{2 - }]$ be,
Given is, \[{K_{SP}} = 3.2 \times {10^{ - 34}}\]
\[ \Rightarrow {\left( {2{s_4}} \right)^2} \times {\left( {2{s_4}} \right)^2} = {K_{SP}}\]
\[ \Rightarrow 16{s_4}^4 = {K_{SP}}\]
$\Rightarrow$ \[{S_3} = {\left( {3.2 \times \dfrac{{{{10}^{ - 34}}}}{{16}}} \right)^{\dfrac{1}{4}}}\]
$\Rightarrow$ \[{S_3} = 11.89 \times {10^{ - 9}}mol\,{L^{ - 1}}\]
Hence, the salt which is most soluble is \[Ni{(OH)_2}\].

Therefore, the correct answer is option (A).

Note: The term \[{K_{SP}}\] in the equation can also be represented by \[{Q_{SP}}\] when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions \[{K_{SP}} = {Q_{SP}}\] but otherwise it gives the direction of the processes of precipitation or dissolution.