Consider of the following two acids and their ionization constants. Which conjugate base ( $H C O O^{-}$ $HCOO^-$ or $C N^{-}$ $CN^-$ ) is stronger?

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Consider of the following two acids and their ionization constants. Which conjugate base ( $H C O O^{-}$ $HCOO^-$ or $C N^{-}$ $CN^-$ ) is stronger?

$H C O O H$ $HCOOH$
$K_{a} = 1.7 x 10^{- 4}$ $K_a = 1.7 x 10^-4$

$H C N$ $HCN$
$K_{a} = 4.9 x 10^{- 10}$ $K_a = 4.9 x 10^-10$

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Answer 1 · 2017-01-01T17:32:57

The stronger the acid, the weaker the conjugate base..........

Explanation:

Formic acid is clearly a stronger acid than prussic acid. That is the given equilibrium lies farther to the right:

$H X (a q) + H_{2} O (l) ⇌ H_{3} O^{+} + X^{-}$ $HX(aq) + H_2O(l)rightleftharpoonsH_3O^+ + X^-$

And $K_{a} = \frac{[H_{3} O^{+}] [X^{-}]}{[H X (a q)]}$ $K_a=([H_3O^+][X^-])/([HX(aq)])$

SInce $K_{a}$ $K_a$ for formic acid is five orders of magnitude greater than $H C \equiv N$ $HC-=N$ , it follows that cyanide ion is a MUCH stronger base than formate.

Given the figures, we could even calculate $K_{b}$ $K_b$ for the following reaction:

$A^{-} + H_{2} O (l) ⇌ H A (a q) + H O^{-}$ $A^(-) + H_2O(l) rightleftharpoons HA(aq) + HO^-$

We know that in water $p K_{a} + p K_{b} = 14$ $pK_a+pK_b=14$ . And $p K_{b}$ $pK_b$ for $N \equiv C^{-}$ $N-=C^-$ will THUS BE SMALLER than that for $formate ion$ $"formate ion"$ .

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Consider of the following two acids and their ionization constants. Which conjugate base ( $H C O O^{-}$ $HCOO^-$ or $C N^{-}$ $CN^-$ ) is stronger?

$H C O O H$ $HCOOH$
$K_{a} = 1.7 x 10^{- 4}$ $K_a = 1.7 x 10^-4$

$H C N$ $HCN$
$K_{a} = 4.9 x 10^{- 10}$ $K_a = 4.9 x 10^-10$

1 Answer

Explanation:

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Consider of the following two acids and their ionization constants. Which conjugate base (HCOO−HCOO^- or CN−CN^-) is stronger?

HCOOHHCOOH Ka=1.7x10−4K_a = 1.7 x 10^-4 HCNHCN Ka=4.9x10−10K_a = 4.9 x 10^-10

1 Answer

Explanation:

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Consider of the following two acids and their ionization constants. Which conjugate base ( $H C O O^{-}$ $HCOO^-$ or $C N^{-}$ $CN^-$ ) is stronger?

$H C O O H$ $HCOOH$
$K_{a} = 1.7 x 10^{- 4}$ $K_a = 1.7 x 10^-4$

$H C N$ $HCN$
$K_{a} = 4.9 x 10^{- 10}$ $K_a = 4.9 x 10^-10$