In a basic buffer 0.0025 mole of NH 4 Cl and 0.15 mole of NH 4 OH are present. The pH of the solution will be: ( p Kb=4.74)

Question

In a basic buffer 0.0025 mole of NH 4 Cl and 0.15 mole of NH 4 OH are present. The pH of the solution will be: ( p Kb=4.74) (A) 11.04 (B) 10.24 (C) 6.62 (D) 5.48

Tardigrade Admin · Accepted Answer

From Handerson's equation pOH =p Kb+ log ([ textsalt ]/[ textbase]) =4.74+ log (0.0025 / V/0.15 / V) =4.74+ log (1/60) =4.74-1.78=2.96 ∴ pH =14-2.96=11.04

Q. In a basic buffer $0.0025$ mole of $N H_{4} Cl$ and $0.15$ mole of $N H_{4} O H$ are present. The $p H$ of the solution will be : $(p K_{b} = 4.74)$

11.04

10.24

6.62

5.48

From Handerson's equation
$pO H = p K_{b} + lo g \frac{[ salt ]}{[ base ]}$
$= 4.74 + lo g \frac{0.0025/ V}{0.15/ V}$
$= 4.74 + lo g \frac{1}{60}$
$= 4.74 - 1.78 = 2.96$
$∴ p H = 14 - 2.96 = 11.04$

Q. In a basic buffer 0.0025 mole of NH4​Cl and 0.15 mole of NH4​OH are present. The pH of the solution will be : (pKb​=4.74)