An aqueous solution of ammonium acetate is

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Ammonium acetate $(CH_3COONH_4)$ is a salt formed from a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.For a salt of a weak acid and a weak ba

Vedclass · Accepted Answer

Almost neutral

Vedclass · Answer

(D) Ammonium acetate $(CH_3COONH_4)$ is a salt formed from a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.For a salt of a weak acid and a weak base,the $pH$ of the aqueous solution is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.Since the dissociation constants for acetic acid $(K_a = 1.8 	imes 10^{-5})$ and ammonium hydroxide $(K_b = 1.8 	imes 10^{-5})$ are approximately equal,$pK_a \approx pK_b$.Therefore,$pH \approx 7$,making the solution almost neutral.

Similar Questions

Why is the following anionic hydrolysis correct?
$X^{-} + H_2O \rightleftharpoons HX + OH^{-}$

$A$ white salt is readily soluble in water and gives a colourless solution with a $pH$ of about $9$. The salt would be

The degree of hydrolysis of a $0.01 \ M$ ammonium acetate solution is = ........ [ $K_h = 3.175 \times 10^{-5}$ ]

When a salt of a weak acid and a weak base is dissolved in water at $25\,^{\circ}C$,the $pH$ of the resulting solution will always :-

The dissociation constant of $NH_4OH$ is $1.8 \times 10^{-5}$. Calculate the hydrolysis constant of $NH_4Cl$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module