
At 25$^{0}C$ , the molar conductance of 0.007M hydrofluoric acid is 150 mho $c{{m}^{2}}mo{{l}^{-1}}$ and its ${{\wedge }_{m}}^{0}$ =500 mho $c{{m}^{2}}mo{{l}^{-1}}$. The value of the dissociation constant of the acid at the given concentration at 25 $^{0}C$ is
(A) $7X{{10}^{-4}}M$
(B) $7X{{10}^{-5}}M$
(C) $9X{{10}^{-3}}M$
(D) $9X{{10}^{-4}}M$
Answer
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Hint: Molar conductance is the conductivity of an electrolyte divided by the molar concentration. The molar conductance of a solution is represented in lambda and the molarity of a solution is represented in M.
Complete step by step solution:
Given in the question: The molar conductance of 0.07M hydrofluoric acid is = 150 mho $c{{m}^{2}}mo{{l}^{-1}}$
The ${{\wedge }_{m}}^{0}$ of 0.07M hydrofluoric acid is =500 mho $c{{m}^{2}}mo{{l}^{-1}}$
We have to find the value of the dissociation constant of the acid at the given concentration at 25 $^{0}C$
The degree of dissociation which is also represented by alpha is calculated as:
Degree of dissociation or $\alpha =\dfrac{{{\lambda }_{c}}}{{{\lambda }_{m}}}=\dfrac{150}{500}=0.3$
Given in the question the concentration i.e. C = 0.007M
And the reaction involved in the dissociation of hydrofluoric acid is mentioned below:
\[HF\to {{H}^{+}}+{{F}^{-}}\]
We can write that $C-C\alpha =C(1-\alpha )$
The value of the dissociation constant for the reaction will be :
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{F}^{-}}]}{[HF]}=\dfrac{C{{\alpha }^{2}}}{(1-\alpha )}\]
After substituting the value of alpha i.e. degree of dissociation in the above equation we get
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{F}^{-}}]}{[HF]}=\dfrac{0.07{{(0.3)}^{2}}}{(1-0.3)}=9{{(10)}^{-4}}M\]
Hence the correct answer for the above question is option (D).
Note: The electrolytes are called strong electrolytes if the ions in the electrolytes are strongly dissociated in the solution when it is present in its aqueous form. The ions of the strong electrolytes are very good conductors of electricity and the value of degree of dissociation for the strong electrolytes is equal to 1. Similarly the weak electrolytes dissociates partly or does not dissociates in the solution in its aqueous form and the value of degree of dissociation for the weak electrolytes is always less than 1.
Complete step by step solution:
Given in the question: The molar conductance of 0.07M hydrofluoric acid is = 150 mho $c{{m}^{2}}mo{{l}^{-1}}$
The ${{\wedge }_{m}}^{0}$ of 0.07M hydrofluoric acid is =500 mho $c{{m}^{2}}mo{{l}^{-1}}$
We have to find the value of the dissociation constant of the acid at the given concentration at 25 $^{0}C$
The degree of dissociation which is also represented by alpha is calculated as:
Degree of dissociation or $\alpha =\dfrac{{{\lambda }_{c}}}{{{\lambda }_{m}}}=\dfrac{150}{500}=0.3$
Given in the question the concentration i.e. C = 0.007M
And the reaction involved in the dissociation of hydrofluoric acid is mentioned below:
\[HF\to {{H}^{+}}+{{F}^{-}}\]
| C | 0 | 0 | Initial concentration |
| $C-C\alpha $ | $C\alpha $ | $C\alpha $ | Concentration at time t |
We can write that $C-C\alpha =C(1-\alpha )$
The value of the dissociation constant for the reaction will be :
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{F}^{-}}]}{[HF]}=\dfrac{C{{\alpha }^{2}}}{(1-\alpha )}\]
After substituting the value of alpha i.e. degree of dissociation in the above equation we get
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{F}^{-}}]}{[HF]}=\dfrac{0.07{{(0.3)}^{2}}}{(1-0.3)}=9{{(10)}^{-4}}M\]
Hence the correct answer for the above question is option (D).
Note: The electrolytes are called strong electrolytes if the ions in the electrolytes are strongly dissociated in the solution when it is present in its aqueous form. The ions of the strong electrolytes are very good conductors of electricity and the value of degree of dissociation for the strong electrolytes is equal to 1. Similarly the weak electrolytes dissociates partly or does not dissociates in the solution in its aqueous form and the value of degree of dissociation for the weak electrolytes is always less than 1.
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