Question

The $p{K}_b$ for fluoride ion at ${25}^{\circ }C$ is $10.83$. The ionisation constant of hydrofluoric acid at this temperature is:
$antilog\overline{4}.83=6.76\times {10}^{-4}$

• 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

The correct option is C $6.76\times {10}^{-4}$

$p{K}_b=10.83$
$F^{-}+H_{2}O\rightleftharpoons HF+O{H}^{-}$
$K_b=\frac{\left[HF\right]\left[O{H}^{-}\right]}{\left[F^{-}\right]}...\left(1\right)$
We know,
$K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]={10}^{-14}...\left(2\right)$
Dissociation of HF in water is represented by the equation.
$HF+H_{2}O\rightleftharpoons H_3{O}^{+}+F^{-}$
$K_a=\frac{\left[H_3{O}^{+}\right]\left[F^{-}\right]}{\left[HF\right]}$
Dividing (2) by (1),
$\frac{K_w}{K_b}=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\left[F^{-}\right]}{\left[HF\right]\left[O{H}^{-}\right]}$
$\frac{K_w}{K_b}=K_a$
Taking log on both sides,
$\log K_a=\log K_w-\log K_b=-p{K}_w+p{K}_b$
$=-14+10.83=-3.17=-4+0.83=\overline{4}.83$
$K=antilog\overline{4}.83=6.76\times {10}^{-4}$