Question

If $p{K}_b$ for the fluoride ion at ${25}^{\circ }C$ is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is
Given : $antilog(-10.83)=1.479\times {10}^{-11}$

• 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 $6.76\times {10}^{-4}$

$p{K}_b=10.83,log{K}_b=-10.83\Rightarrow K_b=1.479\times {10}^{-11}{K}_a.K_b=K_{w}or{K}_a\times 1.479\times {10}^{-11}=1.0\times {10}^{-14}$
$\Rightarrow K_a=6.76\times {10}^{-4}$