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Solubility Product

Correct relat...

Question

Correct relation between dissociation constant of a di-basic acid is:

A

K a_{1} = K a_{2}

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B

K a_{1} > K a_{2}

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C

K a_{1} < K a_{2}

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D

K a_{1} = \frac{1}{K a_{2}}

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Solution

The correct option is B $K a_{1} > K a_{2}$
$H_{2} A ⇌ H A^{-} + H^{+}$

$H A^{-} ⇌ A^{-} + H^{+}$

For a dibasic acid, first, we have to remove $H^{+}$ from $H_{2} A$ and then from $H A^{-}$ . It is easier to remove $H^{+}$ from $H_{2} A$ as it has a much higher tendency to lose proton in comparison to negatively charged $H A^{-}$ which will not want to lose its proton.

Therefore, $K a_{1} > K a_{2}$

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Similar questions

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?

K_{a_{1}}, K_{a_{2}}

K_{a_{3}}

be the first, second and third ionization constant of

H_{3} P O_{4}

K_{a_{1}} >> K_{a_{2}} >> K_{a_{3}}

which is/ are correct

K_{a 1}, K_{a 2}

K_{a 3}

are the respective ionisation constants for the following reactions;

H_{2} S ⇌ H^{+} H S^{-}

H S^{-} ⇌ H^{+} S^{2 -}

H_{2} S ⇌ 2 H^{+} S^{2 -}

The correct relationship between

K_{a 1}, K_{a 2}

K_{a 3}

K_{a_{1}}, K_{a_{2}}

K_{a_{3}}

be the first, second and third ionization constant of

H_{3} P O_{4}

K_{a_{1}} ≫ K_{a_{2}} ≫ K_{a_{3}}

, then which of the following is/are correct?

K_{a 2}

≪

K_{a 1}

H_{2} S O_{4}

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