Question

The ionization constant of $HF,HCOOH$ and $HCN$ at $298K$ are $6.8\times {10}^{-4},1.8\times {10}^{-4}\text{and}4.8\times {10}^{-9}$ respectively. The ionization constants of the corresponding conjugate bases, respectively are:

• A. $2.08\times {10}^{-6},1.47\times {10}^{-11},5.6\times {10}^{-11}$
• B. $1.47\times {10}^{-11},5.6\times {10}^{-11},2.08\times {10}^{-6}$
• C. $1.47\times {10}^{-11},2.08\times {10}^{-6},5.6\times {10}^{-11}$
• D. $5.6\times {10}^{-11},1.47\times {10}^{-11},2.08\times {10}^{-6}$

Answer 1

The correct option is B $1.47\times {10}^{-11},5.6\times {10}^{-11},2.08\times {10}^{-6}$

Conjugate base for $HF$ is $F^{-}$
For $F^{-}$,
$k_b=\frac{k_w}{k_a}=\frac{k_w}{k_a}=\frac{{10}^{-14}}{6.8\times {10}^{-4}}=1.47\times {10}^{-11}$
Conjugate base for $HCOOH$ is $HCO{O}^{-}$
$k_b=\frac{k_w}{k_a}=\frac{k_w}{k_a}=\frac{{10}^{-14}}{1.8\times {10}^{-4}}=5.6\times {10}^{-11}$
Conjugate base for $HCN$ is $C{N}^{-}$
For $C{N}^{-}$,
$k_b=\frac{k_w}{k_a}=\frac{k_w}{k_a}=\frac{{10}^{-14}}{4.8\times {10}^{-9}}=2.08\times {10}^{-6}$