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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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Consider a solution of CH3COONH4 which is a salt of weak acid weak base. Theequilibrium involved in the solutions are :If we add these three reactions, then the net reaction isBoth CH3C00- and NH4 get hydrolysed independently and their hydrolysis depends on(i) their initial concentration(ii) the value of Kh which is Since both of the ions were produced from the same salt, their initial concentrations are same . Therefore unless untial the value of Kw/Kaand Kb is same, the degree of hydrolysis of ion cant be same.To explain why we assume that degree of hydrolysis of cation and anion is same, we need to now look at the third reaction i.e., combination of H+ and OH ions. It is obvious that this reaction happens only because one reaction produced H+ ion and the other produced OH ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this thing in mind that the equilibrium which has smaller value of equilibrium conxtant is affected more by the common ion effect. For the same reason if for any reason a reaction is made to occur to a greater extent by the consumption of any one of the product ion, the reaction with the smaller value of equilibrium constant tends to get affected more.Therefore we conclude that firstly the hydrolysis of both the ions ocurs more in the presence of each other (due to consumption of the product ions) than in each other is absence. Secondly the hydrolysis of the ion which occurs to a lesser extent (due to smaller value of Kh) is affected more than the one whose Kh is greater. Hence we can see that the degree of hydrolysis of both the ions would be close to each other when they are getting hydrolysed in the presence of each other.Q.In a solution of NaHCO3 , the amphiprotic anion can under ionization to form H+ ion and hydrolysis to from OH ion.To calculat PH, suitable approximation is

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