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Preparation of Chlorine

Let hydrolysi...

Question

Let hydrolysis constant (Kh) of NH4+ be y. Then equilibrium constant (Ka) for conjugate acid of NH3 at 25°C is : (1) 1 x 10-14 y ( 1 by y into 10 to the power - 14) (2) 10–14 (3) y (4) y × 10–14

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Solution

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(K_{h})

K_{a} =

1.8 \times

10^{- 5}

K_{b} =

2 \times

10^{- 5}

K_{w} = 10^{- 14}

Calculate the hydrolysis constant $(K_{h})$ of the salt containing $N O_{2}$ . Given the

$K_{a}$ for $H N O_{2}$ = $4.5 \times 10^{- 10}$

y -

5

y -

- 7,

25^{\circ} C

3.83 \times 10^{- 10}

1.75 \times 10^{- 5}

1 \times 10^{- 14}

\oplus N H_{4}

5.6 \times 10^{- 10}

25^{\circ} C

\oplus N H_{4}

⊖ O H

N H_{3}

H_{2} O

25^{\circ} C

3.4 \times 10^{10} L m o l^{- 1} s^{- 1}

N H_{3}

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