0.10 M NH3 50.00 mL + 0.10 M NH4Cl 25.00 mL pH

0.10 M NH3 50.00 mL + 0.10 M NH4Cl 25.00 mL pH

NH3의 Kb = 1.8×10^(-5)

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH3 with 25.00 mL of 0.10 M NH4Cl? Assume that the volume of the solutions are additive and that Kb = 1.8×10^(-5) for NH3.

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> 0.10 M × 50.00 mL = 5 mmol NH3

> 0.10 M × 25.00 mL = 2.5 mmol NH4Cl

( 참고: MV=mol https://ywpop.tistory.com/7787 )

Henderson-Hasselbalch 식

pOH = pKb + log([짝산]/[약염기])

( 참고 https://ywpop.tistory.com/1926 )

= (–log(1.8×10^(-5))) + log(2.5 / 5)

= 4.44

pH = 14 – pOH

= 14 – 4.44

= 9.56

답: pH = 9.56

[키워드] NH3 완충용액의 pH 기준