Solubility product constant (K_ sp ) of salts of types M X, M X_2 and M_3 X at temperature ' T ' are 4.0 × 10^ -8 , 3.2 × 10^ -14 and 2.7 × 10^ -15 , respectively. Solubilities (mol dm^ -3 ) of the salts at temperature

Correct answer: D. M X>M_3 X>M X_2 array lcl Salt & Solubility product & Solubility M X & S_1^2=4.0 × 10^ -8 & S_1=2 × 10^ -4 M X_2 & 4 S_2^3=3.2 × 10^ -14 & S_2=2 × 10^ -5 M_3 X & 27 S_3^4=2.7 × 10^ -15 & S_3=1 × 10^ -4 array Thus, solubility order =M X>M_3 X>M X_2 .

JEE Advanced 2008Paper 2

Exam: JEE Advanced
Year: 2008
Paper: Paper 2
Subject: Chemistry
Chapter: Ionic Equilibrium

JEE Advanced 2008 · Paper 2 · Chemistry · Ionic Equilibrium

Solubility product constant (K_ sp ) of salts of types M X, M X_2...

Solubility product constant

\left(K_{\mathrm{sp}}\right)

of salts of types

M X, M X_2 and M_3 X

at temperature '

T

' are

4.0 \times 10^{-8}, 3.2 \times 10^{-14}

and

2.7 \times 10^{-15}

, respectively. Solubilities

\left(\mathrm{mol} \mathrm{dm}^{-3}\right)

of the salts at temperature '

T

' are in the order

Options

A $M X>M X_2>M_3 X$
B $M_3 X>M X_2>M X$
C $M X_2>M_3 X < M X$
D $M X>M_3 X>M X_2$

Correct answer

D. $M X>M_3 X>M X_2$

Step-by-step solution

$\begin{array}{lcl}\text{ Salt } & \text{ Solubility product } & \text{ Solubility } \\ M X & S_1^2=4.0 \times 10^{-8} & S_1=2 \times 10^{-4} \\ M X_2 & 4 S_2^3=3.2 \times 10^{-14} & S_2=2 \times 10^{-5} \\ M_3 X & 27 S_3^4=2.7 \times 10^{-15} & S_3=1 \times 10^{-4}\end{array}

Thus, solubility order

=M X>M_3 X>M X_2

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