Why is - K - a2- - - nbsp- - K - a1- nbsp- for nbsp- - H-2SO-4-

H_2SO_4 ionizes in two steps: H_2SO_4→ H^++HSO^ _4,K_a1 HSO^ _4→ H^++SO^2 _4,K_a2 It is seen that #160; Ka1 gt; gt; #160; Ka2This is bec ...

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Standard XII

Chemistry

Hydrides of Nitrogen

Why is K...

Question

Why is $K_{a 2}$ $≪$ $K_{a 1}$ for $H_{2} S O_{4}$

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Solution

$H_{2} S O_{4}$ ionizes in two steps:
$H_{2} S O_{4} \to H^{+} + H S O_{4}^{-}, K_{a 1}$
$H S O_{4}^{-} \to H^{+} + S O_{4}^{2 -}, K_{a 2}$

It is seen that Ka1 >> Ka2

This is because a neutral $H_{2} S O_{4}$ has a much higher tendency to lose a proton than the negatively charged $H S O_{4}^{-}$ . Thus, the former is a much stronger acid than the latter.

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Similar questions

Why is for H₂SO₄ in water?

Q. Why is

K a_{2} << K a_{1}

for

H_{2} S O_{4}

in water?

Q. For a polybasic acid, the dissociation constants have different value for each steps, e.g.,

H_{3} A ⇌ H^{+} + H_{2} A^{-}; K = K_{a_{1}}

H_{2} A^{-} ⇌ H^{+} + H A^{2 -}; K = K_{a_{2}}

H A^{2 -} ⇌ H^{+} + A^{3 -}; K = K_{a_{3}}

What is the observed trend of dissociation constants in succesive stage?

100 mL

0.1 M NaOH

solution is titrated with

100 mL

0.05 M H_{2} S O_{4}

solution. The

p H

of the resulting solution is :

(for H_{2} S O_{4}, K a_{1} = \infty, K a_{2} = 10^{- 2})

Q. If

K_{a_{1}}

and

K_{a_{2}}

H_{2} S O_{4}

are

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and

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Why is $K_{a 2}$ $≪$ $K_{a 1}$ for $H_{2} S O_{4}$

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Why is Ka2 ≪ Ka1 for H2SO4

H2SO4 ionizes in two steps:H2SO4→H++HSO−4,Ka1HSO−4→H++SO2−4,Ka2It is seen that Ka1 >> Ka2This is because a neutral H2SO4 has a much higher tendency to lose a proton than the negatively charged HSO−4 . Thus, the former is a much stronger acid than the latter.

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Why is $K_{a 2}$ $≪$ $K_{a 1}$ for $H_{2} S O_{4}$