Dissociation constants of C H3COOH and N H4OH are 1.8× 10- 5 each at 25° C . The equilibrium constant for the reaction of C H3COOH and N H4OH will be -

Question

Dissociation constants of C H3COOH and N H4OH are 1.8× 10- 5 each at 25° C . The equilibrium constant for the reaction of C H3COOH and N H4OH will be - (A) (1.8 × 1.8/104) (B) (1.8/1 0- 9) (C) 1.8 × 1.8 × 104 (D) 3.24 × 10- 10

Tardigrade Admin · Accepted Answer

CH 3 COOH + NH 4 OH leftharpoons CH 3 COO -+ NH 4++ H 2 O K eq =( CH 3 COO - NH 4 +/ CH 3 COOHNH 4 OH )=( K a K b / K W )=1.82 × 104

Q. Dissociation constants of $C H_{3} COO H$ and $N H_{4} O H$ are $1.8 \times 1 0^{- 5}$ each at $2 5^{\circ} C$ . The equilibrium constant for the reaction of $C H_{3} COO H$ and $N H_{4} O H$ will be -

$\frac{1.8 \times 1.8}{1 0 ^{4}}$

$\frac{1.8}{1 0 ^{- 9}}$

$1.8 \times 1.8 \times 1 0^{4}$

$3.24 \times 1 0^{- 10}$

$C H_{3} COO H + N H_{4} O H ⇌ C H_{3} CO O^{-} + N H_{4}^{+} + H_{2} O$
$K_{e q} = \frac{C H _{3} CO O ^{-} N H _{4} ^{+}}{C H _{3} COO H N H _{4} O H} = \frac{K _{a} K _{b}}{K _{W}} = 1. 8^{2} \times 1 0^{4}$

Q. Dissociation constants of CH3​COOH and NH4​OH are 1.8×10−5 each at 25∘C . The equilibrium constant for the reaction of CH3​COOH and NH4​OH will be -

1041.8×1.8​

10−91.8​

1.8×1.8×104

3.24×10−10