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Degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionization constant would be?
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Degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionization ...


Introduction:
The degree of dissociation is a measure of the extent to which a solute dissociates into ions in a solution. It is defined as the ratio of the number of moles of dissociated solute to the initial number of moles of the solute. In this case, we are given that the degree of dissociation of a 0.1 M HCN (hydrogen cyanide) solution is 0.01%.

Calculating the Degree of Dissociation:
To calculate the degree of dissociation, we can use the formula:
Degree of Dissociation (α) = (Number of moles of dissociated solute) / (Initial number of moles of solute)
Given that the concentration of the HCN solution is 0.1 M, we can assume that the initial number of moles of HCN is also 0.1 mol.

Using the given degree of dissociation of 0.01%, we can calculate the number of moles of dissociated solute by multiplying the initial number of moles by the degree of dissociation:
Number of moles of dissociated solute = 0.1 mol x (0.01 / 100) = 0.0001 mol

Therefore, the degree of dissociation of the 0.1 M HCN solution is 0.0001 mol.

Calculating the Ionization Constant:
The ionization constant, or the dissociation constant (Ka), is a measure of the strength of an acid or base in solution. It is defined as the ratio of the concentration of the products to the concentration of the reactants in a dissociation reaction.

In the case of HCN, the dissociation can be represented by the equation:
HCN ⇌ H+ + CN-

The ionization constant (Ka) is given by the equation:
Ka = [H+][CN-] / [HCN]

Given that the concentration of HCN is 0.1 M and the degree of dissociation is 0.01%, we can substitute these values into the equation to calculate the ionization constant.

The concentration of HCN that is dissociated is:
[HCN] dissociated = 0.0001 mol / 1 L = 0.0001 M

The concentration of H+ and CN- ions is equal to the concentration of dissociated HCN:
[H+] = [CN-] = 0.0001 M

Substituting these values into the equation for Ka:
Ka = (0.0001 M)(0.0001 M) / (0.1 M) = 0.00000001 / 0.1 = 0.0000001

Therefore, the ionization constant (Ka) for the 0.1 M HCN solution with a degree of dissociation of 0.01% is 0.0000001.

Conclusion:
In this case, we calculated the degree of dissociation of a 0.1 M HCN solution to be 0.0001 mol. Using this value, we calculated the ionization constant (Ka) to be 0.0000001. The ionization constant provides information about the strength of the acid in solution and can be used to determine the extent of dissociation.
Community Answer
Degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionization ...
Use K=c (alpha^2) /1- alpha
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Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociationwhere n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.Where n is number of reactant particles associated to give 1 mole product.A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.Q.The equilibrium constant for the process can be expressed as

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociationwhere n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.Where n is number of reactant particles associated to give 1 mole product.A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.Q.The vant Hoff factor will be [ if we start with one mole of X ]

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociationwhere n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.Where n is number of reactant particles associated to give 1 mole product.A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.Q.The colligative properties observed will be

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Degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionization constant would be?
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