Question

The $p{K}_b$ for fluorides ion at $25{}^{\circ }C$ is 10.83, if K is the ionization constant of hydrofluoric acid at this temperature then the value of log K is:

• A. $+3.17$
• B. $-24.83$
• C. $-3.17$
• D. $+24.83$

Answer 1

The correct option is C $-3.17$

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

$K_b=\frac{\left[HF\right]\left[O{H}^{-}\right]}{\left[F^{-}\right]}\dots \dots \left(1\right)$

$K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\dots \dots \left(2\right)$

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

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

$K=\frac{\left[H_3{O}^{+}\right]\left[F^{-}\right]}{\left[HF\right]}$

Multiplying of eqn (1) and (2)

$K_b.K=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]=K_w$

$\frac{K_w}{K_b}=K$

Taking log on both the sides

$logK=log{K}_w-log{K}_b$

$=-p{K}_w+p{K}_b=-14+10.83$

$=-3.17$