NH3 gas 275 mL water pH 9.56 buffer 15.0 g NH4Cl 25℃ 0.981 atm

Ammonia gas is bubbled into 275 mL of water

to make an aqueous solution of ammonia.

To prepare a buffer with a pH of 9.56,

15.0 g of NH4Cl are added.

How many liters of NH3 at 25℃ and 0.981 atm

should be used to prepare the buffer?

Assume no volume changes and

ignore the vapor pressure of water.

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> NH3의 Kb = 1.75×10^(-5)

NH4Cl의 몰질량 = 53.49 g/mol 이므로,

15.0 g / (53.49 g/mol) = 0.2804 mol NH4Cl

pOH = 14 – 9.56

= 4.44

Henderson-Hasselbalch 식

pOH = pKb + log([짝산]/[약염기])

( 참고 https://ywpop.tistory.com/1926 )

4.44 = (–log(1.75×10^(-5))) + log(0.2804 / ?)

log(0.2804 / ?) = 4.44 – (–log(1.75×10^(-5)))

0.2804 / ? = 10^[4.44 – (–log(1.75×10^(-5)))]

? = 0.2804 / 10^[4.44 – (–log(1.75×10^(-5)))]

= 0.582 mol NH3

PV = nRT 로부터,

V = nRT / P

= (0.582) (0.08206) (273.15 + 25) / (0.981)

= 14.5 L NH3

답: 14.5 L

[키워드] NH3 완충용액 기준, NH3 buffer 기준