Ionisation constant of NH 4 - in water is 5-6- 10 -10 at 25 o C- The rate constant for reaction of NH 4 - and OH - to form NH 3 and H 2 O at 25 o C is 3-4- 10 10 litre- mol -1 - sec -1- Calculate rate constant for proton transfer from water to NH 3-

NH _ 3 + H _ 2 O k _ f k _ b ⇌ #160; #160; NH _ 4 ^ + + OH ^ ; K _ b =3.4× 10 ^ 10 NH _ 4 ^ + + H _ 2 O⇌ NH _ 4 OH+ H ^ + ; K _ a =5.6× 10 ^ 1 ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

Ionisation co...

Question

Ionisation constant of ${N H}_{4}^{+}$ in water is $5.6 \times 10^{- 10}$ at $25^{o} C$ . The rate constant for reaction of ${N H}_{4}^{+}$ and ${O H}^{-}$ to form ${N H}_{3}$ and $H_{2} O$ at $25^{o} C$ is $3.4 \times 10^{10} l i t r e . {m o l}^{- 1} . {s e c}^{- 1}$ . Calculate rate constant for proton transfer from water to ${N H}_{3}$ .

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Solution

${N H}_{3} + H_{2} O_{f} {N H}_{4}^{+} + {O H}^{-}; K_{b} = 3.4 \times 10^{10}$
${N H}_{4}^{+} + H_{2} O ⇌ {N H}_{4} O H + H^{+}; K_{a} = 5.6 \times 10^{- 10}$
$K_{B a s e} = \frac{k_{f}}{k_{b}} = \frac{K_{w}}{K_{A c i d}} . . . (i)$
$K_{A c i d} \times K_{B a s e} = K_{w} \frac{k_{f}}{3.4 \times 10^{10}} = \frac{10^{- 14}}{5.6 \times 10^{- 10}}$
$k_{f} = 6.07 \times 10^{5}$

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Similar questions

Q. The ionization constant of

\oplus N H_{4}

ion in water is

5.6 \times 10^{- 10}

25^{\circ} C

. The rate constant the reaction of

\oplus N H_{4}

and

⊖ O H

ion to form

N H_{3}

and

H_{2} O

25^{\circ} C

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

. Calculate the rate constant for proton transfer from water to

N H_{3}

The ionization constant of $N H_{4}^{+}$ in water is $2.8 \times 10^{- 10}$ at $25^{\circ} C$ .The rate constant for the reaction of $N H_{4}^{+}$ and $O H^{-}$ to form $N H_{3}$ and $H_{2} O$ at $25^{\circ} C$ is $1.4 L m o l^{- 1} s^{- 1}$ ,the rate constant per proton transfer from water to $N H_{3}$ will be