First slide

Solubility product(KSP)

Question

Which of the following salt is least soluble ?

Difficult

A

Bi₂S₃ (K_sp = 10^-17)
B

CuS (K_sp= 10^-36)
C

MnS (K_sp = 10^-15)
D

Ag₂S(K_sp = 10^-50)

Solution

$\large \left( 1 \right)B{i_2}{S_3}\left( s \right) \rightleftharpoons \mathop {B{i_2}{S_3}\left( {aq} \right)}\limits_{{S_l}} \xrightarrow{{}}\mathop {2B{i^{ + 3}}\left( {aq} \right)}\limits_{2{S_l}} + \mathop {3{S^{ - 2}}\left( {aq} \right)}\limits_{3{S_1}}$
           $\large {K_{sp}} = {\left( {2{S_1}} \right)^2} \times {\left( {3{S_1}} \right)^3} = 108S_1^5$
          $\large \Rightarrow {10^{ - 17}} = 108S_1^5 \to \left( i \right)$
$\large \left( 2 \right)Cus\left( s \right) \rightleftharpoons \mathop {CuS\left( {aq} \right)}\limits_{{S_2}} \xrightarrow{{}}\mathop {C{u^{ + 2}}\left( {aq} \right)}\limits_{{S_2}} + \mathop {{S^{ - 2}}\left( {aq} \right)}\limits_{{S_2}}$
           $\large {K_{sp}} = {{S_1}} \times {S_2}=\left ({S_2} \right )^2$
           $\large \Rightarrow {10^{ - 36}} = {\left( {{S_2}} \right)^2} \to \left( {ii} \right)$
$\large \left( 3 \right)MnS\left( s \right) \rightleftharpoons \mathop {MnS\left( {aq} \right)}\limits_{{S_3}} \xrightarrow{{}}\mathop {M{n^{ + 2}}\left( {aq} \right)}\limits_{{S_3}} + \mathop {{S^{ - 2}}\left( {aq} \right)}\limits_{{S_3}}$
           $\large {K_{sp}} = {{S_3}} \times {S_3}$
           $\large \Rightarrow {10^{ - 15}} = S_{3}^{2} \to \left( {iii} \right)$
$\large \left( 4 \right)A{g_2}S\left( s \right) \rightleftharpoons \mathop {A{g_2}S\left( {aq} \right)}\limits_{{S_4}} \xrightarrow{{}}\mathop {2A{g^ + }\left( {aq} \right)}\limits_{2{S_4}} + \mathop {{S^{ - 2}}\left( {aq} \right)}\limits_{{S_4}}$
        $\large {10^{ - 50}} = {K_{SP}} = {\left( {2{S_4}} \right)^2} \times {S_4} = 4S_4^3 \to \left( {iv} \right)$
By observing (i), (ii), (iii) and (iv), the solubility of CuS(s) will be the least.

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