Values of dissociation constant, Ka are given as follows: Acid Ka HCN 6.2 × 10-10 HF 7.2 × 10-4 HNO2 4.0 × 10-4 Correct order of increasing base strength of the base CN- F- and NO2- will be :

Question

Values of dissociation constant, Ka are given as follows: Acid Ka HCN 6.2 × 10-10 HF 7.2 × 10-4 HNO2 4.0 × 10-4 Correct order of increasing base strength of the base CN- F- and NO2- will be : (A) F- < CN- < NO2- (B)  NO2- < CN- < F-  (C) F- < NO2- < CN- (D) NO2-< F- < CN-

Tardigrade Admin · Accepted Answer

Higher the value of Ka lower will be the value of pKa i.e. higher will be the acidic nature. Further since CN- , F- and NO2- are conjugate base of the acids HCN, HF and HNO2 respectively hence the correct order of base strength will be F- < NO2-< CN- (∴ stronger the acid weaker will be its conjugate base)

Acid	$K_{a}$
HCN	$6.2 \times 1 0^{- 10}$
HF	$7.2 \times 1 0^{- 4}$
$H N O_{2}$	$4.0 \times 1 0^{- 4}$

Q. Values of dissociation constant, $K_{a}$ are given as follows:

Acid $K_{a}$

HCN $6.2 \times 1 0^{- 10}$

HF $7.2 \times 1 0^{- 4}$

$H N O_{2}$ $4.0 \times 1 0^{- 4}$

Correct order of increasing base strength of the base $C N^{-} F^{-}$ and $N O_{2}^{-}$ will be :

$F^{-} < C N^{-} < N O_{2}^{-}$

$N O_{2}^{-} < C N^{-} < F^{-}$

$F^{-} < N O_{2}^{-} < C N^{-}$

$N O_{2}^{-} < F^{-} < C N^{-}$

Q. Values of dissociation constant, Ka​ are given as follows: Acid Ka​ HCN 6.2×10−10 HF 7.2×10−4 HNO2​ 4.0×10−4 Correct order of increasing base strength of the base CN−F− and NO2−​ will be :

F−<CN−<NO2−​

NO2−​<CN−<F−

F−<NO2−​<CN−

NO2−​<F−<CN−

Q. Values of dissociation constant, $K_{a}$ are given as follows:

Acid $K_{a}$

HCN $6.2 \times 1 0^{- 10}$

HF $7.2 \times 1 0^{- 4}$

$H N O_{2}$ $4.0 \times 1 0^{- 4}$

Correct order of increasing base strength of the base $C N^{-} F^{-}$ and $N O_{2}^{-}$ will be :

$F^{-} < C N^{-} < N O_{2}^{-}$

$N O_{2}^{-} < C N^{-} < F^{-}$

$F^{-} < N O_{2}^{-} < C N^{-}$

$N O_{2}^{-} < F^{-} < C N^{-}$