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Question: \(\alpha \) of \(HA({{K}_{\alpha }}={{10}^{-5}})\) is \(0.01\) at one molar concentration, what is \...

α\alpha of HA(Kα=105)HA({{K}_{\alpha }}={{10}^{-5}}) is 0.010.01 at one molar concentration, what is α\alpha when concentration is 4M4M?
A.0.050.05
B.0.0050.005
C.0.040.04
D.0.00250.0025

Explanation

Solution

The molar dissociation constant for a weak or strong acid is denoted by Kα{{K}_{\alpha }} where, α\alpha is the degree of dissociation of that solution.
-The ionization of an acid, or the liberation of acid into its constituent ions, is directly proportional to the molar dissociation constant of that acid.

Complete answer:
We know that α\alpha is the value of the degree of dissociation of the given acid. We assume it is an acid because It has a proton, which it could dissociate when it is dissolved in water. The degree of dissociation measures the amount of the acid which dissociates into its constituent ions when dissolved in aqueous solution. The values which are being provided in the question are
α=0.01\alpha =0.01 when the concentration of the solution is 11 in terms of mole, as it is being mentioned that this value of degree of dissociation corresponds to one molar concentration. And the value of dissociation constant is also given as Kα=105{{K}_{\alpha }}={{10}^{-5}}. Now we have to calculate the degree of dissociation of the same solution when the concentration of the solution increases to 4M4M from one mole,
Therefore, the degree of dissociation for the solution with increased concentration can be calculated as:
Kα=Cα2{{K}_{\alpha }}=C{{\alpha }^{2}}
Here Kα{{K}_{\alpha }} is the molar dissociation constant, CC is the concentration of the acid and α\alpha is the degree of dissociation.
So now we will substitute the values of dissociation constant and the concentration which is being provided to us,
105=4α2{{10}^{-5}}=4{{\alpha }^{2}}
Now we will put the unknown value to the left side of the equation and the knowns on the other side in order to find the value of the degree of dissociation,
α2=0.25×105{{\alpha }^{2}}=0.25\times {{10}^{-5}}
=>α=0.05\alpha =0.05
Hence the value of dissociation constant came out to be 0.050.05 , which corresponds to the option A.
Thus, the correct answer is option (A).

Note:
-Higher the value of dissociation constant, greater the strength of the acid, and higher will be the value of its degree of dissociation.
-Lower the value of dissociation constant, weaker the acid. That means, the sooner the acid gets ionized in a solution, the lesser is the strength of the acid and vice versa.