Question

Dissociation constants of $C H_{3}COOH$ and $N H_{4}OH$ are $1.8\times 10^{- 5}$ each at $25^\circ C$ . The equilibrium constant for the reaction of $C H_{3}COOH$ and $N H_{4}OH$ will be -

• A. $\frac{1.8 \times 1.8}{10^{4}}$
• B. $\frac{1.8}{1 0^{- 9}}$
• C. $1.8 \times 1.8 \times 10^{4}$
• D. $3.24 \times 10^{- 10}$

Answer 1

Answer: (C)

$CH _3 COOH + NH _4 OH \rightleftharpoons CH _3 COO ^{-}+ NH _4^{+}+ H _2 O$
$K _{ eq }=\frac{ CH _3 COO ^{-} NH _4{ }^{+}}{ CH _3 COOHNH _4 OH }=\frac{ K _{ a } K _{ b }}{ K _{ W }}=1.8^2 \times 10^4$