First slide

Solubility product(KSP)

Question

Three sparingly soluble salts M₂X, MX and MX₃, have same value of solubility product. Their solubillities follow the order

Moderate

A

MX₃ > MX > M₂X
B

MX > MX₃ > M₂X
C

MX > M₂X > MX₃
D

MX₃ > M₂X > MX

Solution

$\large {M_2}X\left( s \right) \rightleftharpoons \mathop {{M_2}X\left( {aq} \right)}\limits_{{S_1}} \xrightarrow{{}}\mathop {2{M^ + }\left( {aq} \right)}\limits_{2{S_1}} + \mathop {{X^{ - 2}}\left( {aq} \right)}\limits_{{S_1}}$
$\large {K_{sp}} = {\left( {2{S_1}} \right)^2}{S_1} = 4S_1^3$
$\large MX\left( s \right) \rightleftharpoons \mathop {MX\left( {aq} \right)}\limits_{{S_2}} \xrightarrow{{}}\mathop {{M^ + }\left( {aq} \right)}\limits_{{S_2}} + \mathop {{X^ - }\left( {aq} \right)}\limits_{{S_2}}$
$\large {K_{sp}} = {\left( {{S_2}} \right)^2}$
$\large MX_3\left( s \right) \rightleftharpoons \mathop {{MX_3}\left( {aq} \right)}\limits_{{S_3}} \xrightarrow{{}}\mathop {{M^{+3} }\left( {aq} \right)}\limits_{{S_3}} + \mathop {3{X^ - }\left( {aq} \right)}\limits_{3{S_3}}$
$\large {K_{sp}} = S_3\times {\left( {3{S_3}} \right)^3}=27S_{3}^{4}$
$\large \\4S_1^3 = {\left( {{S_2}} \right)^2} = 27S_3^4 = {K_{sp}} = {10^{ - 12}}\\\\{S_3} > {S_1} > {S_2}$ Let Ksp =10^-12

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