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Question: Hydrofluoric acid is weak acid. At \({25^ \circ }C\) , the molar conductivity of \(0.002M\) HF is \[...

Hydrofluoric acid is weak acid. At 25∘C{25^ \circ }C , the molar conductivity of 0.002M0.002M HF is 176.2Ω−1cm2mol−1176.2{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} . If it's ∧∘m=405Ω−1cm2mol−1{ \wedge ^ \circ }_m = 405{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} . Equilibrium constant at given concentration is:
A. 6.7×10−4M6.7 \times {10^{ - 4}}M
B. 3.2×10−4M3.2 \times {10^{ - 4}}M
C. 6.7×10−5M6.7 \times {10^{ - 5}}M
D. 3.2×10−5M3.2 \times {10^{ - 5}}M

Explanation

Solution

There are various quantities in this question but the question can be solved by the formula for degree of dissociation. Equilibrium constant of chemical reaction is the value of reaction quotient at chemical equilibrium. It is a state that is reached after required time and the composition has no more tendency for further change.

Complete step by step solution: Given values for calculating equilibrium constant at given concentration are:
Concentration= 0.002M0.002M
Molar conductivity= 176.2Ω−1cm2mol−1176.2{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}}
Limiting molar conductivity= 405Ω−1cm2mol−1405{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}}
Temperature= 25∘C{25^ \circ }C
Now, Equilibrium constant is given as,
Degree of dissociation states that how many ions have been dissociated from the entire solution and is given by the following formula,
α=MolarConductivityLimitingMolarConductivity\alpha = \dfrac{{MolarConductivity}}{{LimitingMolarConductivity}}
α=∧c∧∘m=176.2405.2=0.435\alpha = \dfrac{{{ \wedge _c}}}{{{ \wedge ^ \circ }_m}} = \dfrac{{176.2}}{{405.2}} = 0.435
The chemical reaction is given as follows:
HF→H++F−HF \to {H^ + } + {F^ - }
Now, we will find the equilibrium constant for the reaction. For that we need to know the initial and final concentrations of the equation. The initial concentrations of HFHF , H+{H^ + } and F−{F^ - } are cc , 00 and 00 . Concentrations at equilibrium are c−cαc - c\alpha , cαc\alpha and cαc\alpha .
K=[H+][F−1][HF]=Cα×CαC−Cα=Cα21−α=0.002M×0.43521−0.435=6.7×10−4MK = \dfrac{{[{H^ + }][{F^{ - 1}}]}}{{[HF]}} = \dfrac{{C\alpha \times C\alpha }}{{C - C\alpha }} = \dfrac{{C{\alpha ^2}}}{{1 - \alpha }} = 0.002M \times \dfrac{{{{0.435}^2}}}{{1 - 0.435}} = 6.7 \times {10^{ - 4}}M

Therefore, the correct answer is option A.

Note: Units are important and have to be kept in mind. Here, the units used are MM and Ω−1cm2mol−1{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} which are of concentration and molar conductivity. Hydrofluoric acid is a solution of hydrogen fluoride in water. It is colourless, acidic and highly corrosive. It is mainly used in pharmaceutical industries and in production of PTFE i.e. Teflon.