A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10-5 , what is the pH of this solution? (log 2.7 = 0.43 )

Question

A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10-5 , what is the pH of this solution? (log 2.7 = 0.43 )  (A) 9.43 (B) 11.72 (C) 8.73 (D) 9.08

Tardigrade Admin · Accepted Answer

pOH = pK b+ log ([ text salt] / text [base] ) =- log Kb+ log ([ text salt] / text [base] ) =- log 1.8 × 10-5+ log (0.20/0.30) =5-0.25+(-0.176) =4.75-0.176=4.57 ∴ pH =14-4.57=9.43

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 this solution? $(l o g 2.7 = 0.43)$

9.43

11.72

8.73

9.08

$pO H = p K_{b} + lo g \frac{[ salt]}{[base]}$
$= - lo g K_{b} + lo g \frac{[ salt]}{[base]}$
$= - lo g 1.8 \times 1 0^{- 5} + lo g \frac{0.20}{0.30}$
$= 5 - 0.25 + (- 0.176)$
$= 4.75 - 0.176 = 4.57$
$∴ p H = 14 - 4.57 = 9.43$

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 this solution? (log2.7=0.43)