Question

The $p{K}_b$ for fluoride ion at $25^{\circ }C$ is $10.83,$ the ionizatic constant of hydrofluoric acid at this temperature is

• A. $1.74\times 10^{-5}$
• B. $3.52\times 10^{-3}$
• C. $6.76\times 10^{-4}$
• D. $5.38\times 10^{-2}$

Answer 1

Answer: (C)

$F^{-}+H_{2}O\rightleftharpoons HF+O{H}^{-}$

$K_b=\frac{[HF]\left[O{H}^{-}\right]}{\left[F^{-}\right]}$ ____(i)

$K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]=10^{-14}$ ____(ii)

Dissociation of $HF$ in water is represented by the equation,

$HF+H_{2}O\rightleftharpoons H_3{O}^{+}+F^{-}$

$K_a=\frac{\left[H_3{O}^{+}\right]\left[F^{-}\right]}{[HF]}$ ____(iii)

Multiplying eqn (i) and (iii),

$K_b\cdot K_a=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]=K_w\frac{K_w}{K_b}=K_a$

Taking log on both sides

$\log K_a=\log K_w-\log K_b=-p{K}_w+p{K}_b=-14+10.83=-3.17$

or, $K_a=6.76\times 10^{-4}$