Equilibrium Question 851

Question: If $ pK_{b} $ for fluoride ion at $ 25{}^\circ C $ is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is

Options:

A) $ 3.52\times {10^{-3}} $

B) $ 6.75\times {10^{-4}} $

C) $ 5.38\times {10^{-2}} $

D) $ 1.74\times {10^{-5}} $

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Answer:

Correct Answer: B

Solution:

$ K_{w}=K_{a}\times K_{b} $

$ K_{b}={10^{-10.83}}=1.48\times {10^{-11}} $

$ \therefore K_{a}=\frac{K_{w}}{K_{b}}=\frac{{10^{-14}}}{1.48\times {10^{-11}}}=6.75\times {10^{-4}} $



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