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 + OH ^{-}$

$K_{b}=\frac{[ HF ]\left[ OH ^{-}\right]}{\left[ F ^{-}\right]}$ ____(i)

$K_{w}=\left[ H _{3} O ^{+}\right]\left[ OH ^{-}\right]=10^{-14}$ ____(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),

$K_{b} \cdot K_{a}=\left[ H _{3} O ^{+}\right]\left[ OH ^{-}\right]=K_{w} \frac{K_{w}}{K_{b}}=K_{a}$

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}$