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Question: The values of \[{K_{SP}}\] of following sparingly soluble salts \[Ni{(OH)_2}\], \(Ce{(OH)_4}\),AgCN,...

The values of KSP{K_{SP}} of following sparingly soluble salts Ni(OH)2Ni{(OH)_2}, Ce(OH)4Ce{(OH)_4},AgCN, Al2(SO4)2A{l_2}{(S{O_4})_2} are respectively 2×10−15,4×10−35,6×10−172 \times {10^{ - 15}},4 \times {10^{ - 35}},6 \times {10^{ - 17}} and 3.2×10−343.2 \times {10^{ - 34}} respectively. Which salt is more soluble?
A. Ni(OH)2Ni{(OH)_2}
B. Ce(OH)4Ce{(OH)_4}
C. Al2(SO4)2A{l_2}{(S{O_4})_2}
D.AgCN

Explanation

Solution

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

Complete step by step answer:
Ni(OH)2⇌[Ni2+]+[2OH−]Ni{(OH)_2} \rightleftharpoons \left[ {N{i^{2 + }}} \right] + \left[ {2O{H^ - }} \right]
Let the number of moles of Ni(OH)2Ni{(OH)_2} be ‘s’, [Ni2+]\left[ {N{i^{2 + }}} \right] be ‘s’ and [2OH−]\left[ {2O{H^ - }} \right] be 2s
Given is, KSP=2×10−15{K_{SP}} = 2 \times {10^{ - 15}}
⇒s×(2s)2=KSP\Rightarrow s \times {\left( {2s} \right)^2} = {K_{SP}}
⇒4s3=KSP\Rightarrow 4{s^3} = {K_{SP}}
⇒\Rightarrow s=(KSP4)13s = {\left( {\dfrac{{{K_{SP}}}}{4}} \right)^{\dfrac{1}{3}}}
⇒\Rightarrow s=(2×10−154)13=5.8×10−5mol L−1s = {\left( {2 \times \dfrac{{{{10}^{ - 15}}}}{4}} \right)^{\dfrac{1}{3}}} = 5.8 \times {10^{ - 5}}mol\,{L^{ - 1}}

Ce(OH)4⇌[Ce4+]+[4OH−]Ce{(OH)_4} \rightleftharpoons \left[ {C{e^{4 + }}} \right] + \left[ {4O{H^ - }} \right]
Let the number of moles of Ce(OH)4Ce{(OH)_4} be , [Ce4+]\left[ {C{e^{4 + }}} \right] be , and [4OH−]\left[ {4O{H^ - }} \right] be ,
Given is, KSP=4×10−35{K_{SP}} = 4 \times {10^{ - 35}}
⇒s2×(4s2)2=KSP\Rightarrow {s_2} \times {\left( {4{s_2}} \right)^2} = {K_{SP}}
⇒(8s3)3=KSP\Rightarrow {\left( {8{s_3}} \right)^3} = {K_{SP}}
⇒\Rightarrow s=(KSP8)13s = {\left( {\dfrac{{{K_{SP}}}}{8}} \right)^{\dfrac{1}{3}}}
⇒\Rightarrow s=(4×10−358)13=1.7099×10−12mol L−1s = {\left( {4 \times \dfrac{{{{10}^{ - 35}}}}{8}} \right)^{\dfrac{1}{3}}} = 1.7099 \times {10^{ - 12}}mol\,{L^{ - 1}}

AgCN⇌[Ag+]+[CN−]AgCN \rightleftharpoons \left[ {A{g^ + }} \right] + \left[ {C{N^ - }} \right]
Let the number of moles of AgCN be , [Ag+]\left[ {A{g^ + }} \right] be , and [CN−]\left[ {C{N^ - }} \right] be ,
Given is, KSP=6×10−17{K_{SP}} = 6 \times {10^{ - 17}}
⇒(s3)2=KSP\Rightarrow {\left( {{s_3}} \right)^2} = {K_{SP}}
⇒s3=√KSP\Rightarrow {s_3} = \surd {K_{SP}}
⇒\Rightarrow S3=√6×10−17{S_3} = \surd 6 \times {10^{ - 17}}
⇒\Rightarrow S3=7.8×10−9mol L−1{S_3} = 7.8 \times {10^{ - 9}}mol\,{L^{ - 1}}

Al2(SO4)2⇌[2Al2+]+[2SO42−]A{l_2}{(S{O_4})_2} \rightleftharpoons \left[ {2A{l^{2 + }}} \right] + [2S{O_4}^{2 - }]
Let the number of moles of Al2(SO4)2A{l_2}{(S{O_4})_2} be , [2Al2+]\left[ {2A{l^{2 + }}} \right] be , and [2SO42−][2S{O_4}^{2 - }] be,
Given is, KSP=3.2×10−34{K_{SP}} = 3.2 \times {10^{ - 34}}
⇒(2s4)2×(2s4)2=KSP\Rightarrow {\left( {2{s_4}} \right)^2} \times {\left( {2{s_4}} \right)^2} = {K_{SP}}
⇒16s44=KSP\Rightarrow 16{s_4}^4 = {K_{SP}}
⇒\Rightarrow S3=(3.2×10−3416)14{S_3} = {\left( {3.2 \times \dfrac{{{{10}^{ - 34}}}}{{16}}} \right)^{\dfrac{1}{4}}}
⇒\Rightarrow S3=11.89×10−9mol L−1{S_3} = 11.89 \times {10^{ - 9}}mol\,{L^{ - 1}}
Hence, the salt which is most soluble is Ni(OH)2Ni{(OH)_2}.

Therefore, the correct answer is option (A).

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