Question

The $p{K}_b$ for fluoride ion at ${25}^{\circ }C$ is $10.83$. The ionization 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

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

$p{K}_b=10.83$

$K_b={10}^{-1083}=1.48\times {10}^{-11}$

$HF+H_{2}O⟶F^{-}+H_3{O}^{+}$........$K_a$

$K_a=$ ionization constant of $HF$

$\left(K_a\right)\left(K_b\right)=K_w$

$K_a=\frac{K_w}{K_b}$ $=\frac{{10}^{-14}}{1.48\times {10}^{-11}}=6.76\times {10}^{-4}$