Question:

Solubility product constant ( $K_{sp}$ )of salts of types $ MX_1, MX_ 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 }$, respectively. Solubilities (mol dm$^{- 3}$ ) of the salts at temperature T" are in the order

Updated On: May 19, 2024
  • $MX > MX_2 > M_3 X $
  • $M_3X > MX_2 > MX $
  • $ MX_2 > M_2 X > MX_2 $
  • $MX > M_3 X > MX_2 $
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Approach Solution - 1

MX : $ K_{ s p } = S^2 = 4 \times 10^{ - 8 } $
$\Rightarrow S = 2 \times 10^{ - 4 } $
$ MX_2 : K_{ s p } = 4 \, S^3 = 3.2 \times 10^{ - 14 } \Rightarrow S = 2 \times 10^{ - 5 } $
$ M_3 \, X : K_{ sp } = 27 S^4 = 2.7 \times 10^{ - 15 } \Rightarrow S = 10^{ - 4 } $
Order of solubility is $MX > M_3 X > MX_2 $
Was this answer helpful?
0
1
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

To determine the solubilities of the salts in the order of \(MX\), \(MX_2\), and \(M_3X\), we can use the solubility product constants (\(K_{sp}\)) provided for each salt type. The solubility product constant, \(K_{sp}\), represents the equilibrium constant for the dissolution of a sparingly soluble salt. It is defined as the product of the concentrations of the ions raised to the power of their coefficients in the solubility equilibrium expression.The solubility equilibrium expression for the dissolution of a salt \(MX\) can be written as:\[ MX \rightleftharpoons M^+ + X^- \]And its solubility product constant \(K_{sp}\) is given as \(K_{sp} = [M^+][X^-]\). Similarly, for \(MX_2\), the equilibrium expression and \(K_{sp}\) are:\[ MX_2 \rightleftharpoons M^+ + 2X^- \]\[ K_{sp} = [M^+][X^-]^2 \]And for \(M_3X\), it is:\[ M_3X \rightleftharpoons 3M^+ + X^- \]\[ K_{sp} = [M^+]^3[X^-] \]Given the \(K_{sp}\) values: - \(K_{sp}(MX) = 4.0 \times 10^{-8}\)- \(K_{sp}(MX_2) = 3.2 \times 10^{-14}\)- \(K_{sp}(M_3X) = 2.7 \times 10^{-15}\)We can compare the values of \(K_{sp}\) to determine the order of solubilities.Since \(K_{sp}\) is directly related to the solubility of a salt, a higher value of \(K_{sp}\) corresponds to a higher solubility.So, in ascending order of solubility, So Correct Answer  would be  Option 4 i.e. \(MX\), \(M_3X\), and \(MX_2\).

Was this answer helpful?
0
0

Questions Asked in JEE Advanced exam

View More Questions

Concepts Used:

Equilibrium

An equilibrium represents a state in a process when the observable properties such as color, temperature, pressure, concentration etc do not show any change.

The word equilibrium means ‘balance’ which indicates that a chemical reaction represents a balance between the reactants and products taking part in the reaction. The equilibrium state is also noticed in certain physical processes such as the melting point of ice at 0℃, both ice and water are present at equilibrium.

In the case of physical processes such as the melting of solid, dissolution of salt in water etc., the equilibrium is called physical equilibrium while the equilibrium associated with chemical reaction is known as chemical equilibrium.

Equilibrium in Chemical changes

The chemical equilibrium in a reversible reaction is the state at which both forward and backward reactions occur at the same speed.

The stage of the reversible reaction at which the concentration of the reactants and products do not change with time is called the equilibrium state.

Read More: Calculating Equilibrium Concentration

Types of Chemical Equilibrium

There are two types of chemical equilibrium:

  • Homogeneous Equilibrium
  • Heterogeneous Equilibrium

Homogenous Chemical Equilibrium

In this type, the reactants and the products of chemical equilibrium are all in the same phase. Homogenous equilibrium can be further divided into two types: Reactions in which the number of molecules of the products is equal to the number of molecules of the reactants. For example,

  • H2 (g) + I2 (g) ⇌ 2HI (g)
  • N2 (g) + O2 (g) ⇌ 2NO (g)

Reactions in which the number of molecules of the products is not equal to the total number of reactant molecules. For example,

  • 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
  • COCl2 (g) ⇌ CO (g) + Cl2 (g)

Heterogeneous Chemical Equilibrium

In this type, the reactants and the products of chemical equilibrium are present in different phases. A few examples of heterogeneous equilibrium are listed below.

  • CO2 (g) + C (s) ⇌ 2CO (g) 
  • CaCO3 (s) ⇌ CaO (s) + CO2 (g)

Thus, the different types of chemical equilibrium are based on the phase of the reactants and products.

Check Out: Equilibrium Important Questions