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}} $
Show Answer
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}} $
