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Catalytic Hydrogenation

The hydrolysi...

Question

The hydrolysis constant $K_{H}$ for the reaction $H_{2} P O_{4}^{-} + H_{2} O \to H_{3} P O_{4} + O H^{-}$ is $1.4 \times 10^{- 12}$ . The ionisation constant for $H_{3} P O_{4} + H_{2} O \to H_{2} P O_{4}^{-} + H_{3} O^{+}$ is:

A

7.14 \times 10^{- 3}

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B

1.4 \times 10^{- 12}

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C

7.14 \times 10^{- 12}

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D

1.4 \times 10^{- 3}

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Solution

The correct option is A $7.14 \times 10^{- 3}$
The ion $H_{2} P O_{4}^{-}$ is present in the salt $H_{2} P O_{4}^{-}$ . It is a salt obtained from mixture $N a O H + H_{3} P O_{4}$
The expression for the hydrolysis constant is as given below,
$K_{H} = \frac{K_{w}}{K_{a (H_{3} P O_{4}})}$
Substitute values in the above expression, we get
$K_{H} = \frac{10^{- 14}}{1.4 \times 10^{- 12}} = 7.14 \times 10^{- 3}$

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Similar questions

Q. The hydrolytic constant

k_{h}

for the hydrolytic equilibrium

H_{2} P O_{4}^{-} + H_{2} O \to H_{3} P O_{4} + O H^{-}

1.4 \times 10^{- 12}

, the value of ionization constant for the reaction

H_{3} P O_{4} + H_{2} O \to H_{2} P O_{4}^{-} H_{3} O^{+}

Q. Three reactions involving

H_{2} P O_{4}^{-}

H_{2} P O_{4} + H_{2} O \to H_{3} O^{+} + H_{2} P O_{4}^{-}

H_{2} P O_{4}^{-} + H_{2} O \to H P O_{4}^{2 -} + H_{3} O^{+}

H_{2} P O_{4}^{-} + O H^{-} \to H_{3} P O_{4} + O^{2 -}

In which of the given does

H_{2} P O_{4}^{-}

Q. Three reactions involving

H_{2} P O_{4}^{-}

are given below.
I.

H_{3} P O_{4} + H_{2} O \to H_{3} O^{+} + H_{2} P O_{4}^{-}

H_{2} P O_{4}^{-} + H_{2} O \to H P O_{4}^{2 -} + H_{3} O^{+}

H_{2} P O_{4}^{-} + O H^{-} \to H_{3} P O_{4} + O^{2 -}

In which of the above does

H_{2} P O_{4}^{-}

act as an acid?

Q. Calculate the pH of

0.1

M solution of (i)

N a H C O_{3}

N a_{2} H P O_{4}

N a H_{2} P O_{4}

C O_{2} + H_{2} O ⇌ H^{+} + C O_{3}^{2 -}

K_{1} = 4.2 \times 10^{- 7}

M

H C O_{3}^{-} ⇌ H^{+} + C O_{3}^{2 -}

K_{2} = 4.8 \times 10^{- 11}

M

H_{2} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-}

K_{1} = 7.5 \times 10^{- 3}

M

H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{2 -}

K_{2} = 6.2 \times 10^{- 8}

M

H P O_{4}^{2 -} ⇌ H^{+} + P O_{4}^{3 -}

K_{3} = 1.0 \times 10^{- 12}

Q. Consider the equilibrium reactions,

H_{3} P O_{4} ⇌ K_{1} H^{+} + H_{2} P O_{4}^{-}

H_{2} P O_{4} ⇌ K_{1} H^{+} + H P O_{4}^{- 2}

H_{- 2} P O_{4} ⇌ K_{1} H^{+} + H P O_{4}^{- 3}

The equilbrium constant,

K

for the following dissociation

H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{- 3}

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