Question
If $$p{K_b}$$ for fluoride ion at $${25^ \circ }C$$ is $$10.83,$$ the ionisation constant of hydrofluoric acid in water at this temperature is
A.
$$3.52 \times {10^{ - 3}}$$
B.
$$6.75 \times {10^{ - 4}}$$
C.
$$5.38 \times {10^{ - 2}}$$
D.
$$1.74 \times {10^{ - 5}}$$
Answer :
$$6.75 \times {10^{ - 4}}$$
Solution :
$$\eqalign{
& {K_w} = {K_a} \times {K_b} \cr
& {K_b} = {10^{ - 10.83}} \cr
& = 1.48 \times {10^{ - 11}} \cr
& \therefore \,\,{K_a} = \frac{{{K_w}}}{{{K_b}}} \cr
& = \frac{{{{10}^{ - 14}}}}{{1.48 \times {{10}^{ - 11}}}} \cr
& = 6.75 \times {10^{ - 4}} \cr} $$