Equilibrium Question 741
Question: If $ pK_{b} $ A for fluoride ion at $ 25{}^\circ C $ is 10.83, (he ionization constant of hydrofluoric acid in water at this temperature is
Options:
A) $ 1.74\times {10^{-5}} $
B) $ 3.52\times {10^{-3}} $
C) $ 6.75\times {10^{-4}} $
D) $ 5.38\times {10^{-2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ HF+H_2O\rightarrow {F^{-}}+H_3{O^{+}} $
$ K_{a}\times K_{b}=K_{w} $
$ pK_{a}+pK_{b}=pK_{w} $
$ pK_{a}=14-10.83\Rightarrow pK_{a}=3.17 $
$ K_{a}=6.75\times {10^{-4}} $
