Autoprotolysis constant of a weak acid HA is 4×10−8. Hence, [A−]is

(a) HA+HA[H2A−]K[H2A−]2∴[A−]⇌H2A−+A−=[A−]=[H2A−][A−]=4×10−8=[A−]2=4×10−6=2×10−4M

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Autoprotolysis constant of a weak acid {H} A is 4 10^{-8}. He

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Solving time: 3 mins

Autoprotolysis constant of a weak acid $H A$ is $4 \times 1 0^{- 8}$ . Hence, $[A^{-}]$ is

2 \times 1 0^{- 4} M

0.25 \times 1 0^{- 6} M

16 \times 1 0^{- 16} M

4 \times 1 0^{- 8} M

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Autoprotolysis constant of a weak acid

H A

4 \times 1 0^{- 8}

. Hence,

[A^{-}]

12 mins ago

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12 mins ago

Text solutionVerified

(a)

HA + HA [H_{2} A^{-}] K [H_{2} A^{-}]^{2} ∴ [A^{-}] ⇌ H_{2} A^{-} + A^{-} = [A^{-}] = [H_{2} A^{-}] [A^{-}] = 4 \times 1 0^{- 8} = [A^{-}]^{2} = 4 \times 1 0^{- 6} = 2 \times 1 0^{- 4} M

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Question Text	Autoprotolysis constant of a weak acid $H A$ is $4 \times 1 0^{- 8}$ . Hence, $[A^{-}]$ is
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1
Upvotes	80

Autoprotolysis constant of a weak acid $H A$ is $4 \times 1 0^{- 8}$ . Hence, $[A^{-}]$ is

Text solutionVerified

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Autoprotolysis constant of a weak acid HA is 4×10−8. Hence, [A−]is

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Autoprotolysis constant of a weak acid $H A$ is $4 \times 1 0^{- 8}$ . Hence, $[A^{-}]$ is