A buffer solution is prepared in which the concentration of NH 3 is 0.30 M and the concentration of NH 4+is 0.20 M. If the equilibrium constant, K b for NH 3 equals 1.8 × 10-5, what is the p H of the solution?

Question

A buffer solution is prepared in which the concentration of NH 3 is 0.30 M and the concentration of NH 4+is 0.20 M. If the equilibrium constant, K b for NH 3 equals 1.8 × 10-5, what is the p H of the solution? (A) 8.73 (B) 9.08 (C) 9.44 (D) 11.72

Tardigrade Admin · Accepted Answer

Given, Kb=1.8 × 10-5 p Kb=- log Kb =- log 1.8 × 10-5=4.74 p O H=p Kb+ log ([ text salt ]/[ text base ]) =4.74+ log (0.20/0.30) =4.74-0.176=4.56 p H+p O H=14 p H=14-4.56=9.44

Q. A buffer solution is prepared in which the concentration of $N H_{3}$ is $0.30 M$ and the concentration of $N H_{4}^{+}$ is $0.20 M$ . If the equilibrium constant, $K_{b}$ for $N H_{3}$ equals $1.8 \times 1 0^{- 5}$ , what is the $p H$ of the solution?

$8.73$

$9.08$

$9.44$

$11.72$

Given, $K_{b} = 1.8 \times 1 0^{- 5}$
$p K_{b} = - lo g K_{b}$
$= - lo g 1.8 \times 1 0^{- 5} = 4.74$
$pO H = p K_{b} + lo g \frac{[ salt ]}{[ base ]}$
$= 4.74 + lo g \frac{0.20}{0.30}$
$= 4.74 - 0.176 = 4.56 p H + pO H = 14$
$p H = 14 - 4.56 = 9.44$

Q. A buffer solution is prepared in which the concentration of NH3​ is 0.30M and the concentration of NH4+​is 0.20M. If the equilibrium constant, Kb​ for NH3​ equals 1.8×10−5, what is the pH of the solution?

8.73

9.08

9.44

11.72