A 100 mL 0.1 (M) solution of ammonium acetate is diluted by adding 100 ml of water. The pH of the resulting solution will be( pKa of acetic acid is nearly equal to pKb of NH4 OH)

Question

A 100 mL 0.1 (M) solution of ammonium acetate is diluted by adding 100 ml of water. The pH of the resulting solution will be( pKa of acetic acid is nearly equal to pKb of NH4 OH) (A) 4.9 (B) 5.0 (C) 7.0 (D) 10.0

Tardigrade Admin · Accepted Answer

Ammonium acetate is a salt of weak acid and weak base. When solutions of such salts diluted resulting

p H

of solution is calculated from following relationship

p H = 7 + \frac{1}{2} p K_{a} - \frac{1}{2} p K_{b}

∵ p K_{a} ≃ p K_{b}

(given)

∴ p H

of diluted solution

= 7

Q. A 100mL0.1(M) solution of ammonium acetate is diluted by adding 100ml of water. The pH of the resulting solution will be( pKa​ of acetic acid is nearly equal to pKb​ of NH4​OH)

4.9

5.0

7.0

10.0

Ammonium acetate is a salt of weak acid and weak base. When solutions of such salts diluted resulting pH of solution is calculated from following relationship pH=7+21​pKa​−21​pKb​ ∵pKa​≃pKb​ (given) ∴pH of diluted solution =7

Q. A $100 m L 0.1 (M)$ solution of ammonium acetate is diluted by adding $100 m l$ of water. The pH of the resulting solution will be( $p K_{a}$ of acetic acid is nearly equal to $p K_{b}$ of $N H_{4} O H$ )

Ammonium acetate is a salt of weak acid and weak base. When solutions of such salts diluted resulting $p H$ of solution is calculated from following relationship
$p H = 7 + \frac{1}{2} p K_{a} - \frac{1}{2} p K_{b}$
$∵ p K_{a} ≃ p K_{b}$ (given)
$∴ p H$ of diluted solution $= 7$