Dissociation constants of C H 3 C O O H and N H 4 O H are 1.8 × 10 – 5 each at 25 ° C . The equilibrium constant for the reaction of C H 3 C O O H and N H 4 O H will be -

CH 3 COOH + NH 4 OH ⇌ CH 3 COO - + NH 4 + + H 2 O K h = CH 3 COO – NH 4 + CH 3 COOH NH 4 OH = K a K b K w = 1.8 2 × 10 4

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumHydrolysis of SaltsEasy2 minPYQ_2020

ChemistryEasysingle choice

Dissociation constants of ${C H}_{3} C O O H$ and ${N H}_{4} O H$ are $1.8 \times 10^{- 5}$ each at $25 ° C$ . The equilibrium constant for the reaction of ${C H}_{3} C O O H$ and ${N H}_{4} O H$ will be -

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Answer:

Solution:

${CH}_{3} COOH + {NH}_{4} OH ⇌ {CH}_{3} {COO}^{-} + {NH}_{4}^{+} + H_{2} O$

$K_{h} = \frac{[{CH}_{3} {COO}^{-}] [{NH}_{4}^{+}]}{[{CH}_{3} COOH] [{NH}_{4} OH]} = \frac{K_{a} K_{b}}{K_{w}} = {(1.8)}^{2} \times 10^{4}$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Hydrolysis of Salts

⏱ 2mℹ️ Source: PYQ_2020

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