0.25 M NH4Cl 용액의 pH. Kb=1.75e-5

0.25 M NH4Cl 용액의 pH. Kb=1.75e-5

 

 

Calculate the pH and pOH for 0.25 M ammonium chloride (NH4Cl).

NH3의 Kb = 1.75×10^(-5)

 

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▶ 참고: 0.1 M NH4Cl 용액의 pH. Kb=1.0e-5

[ https://ywpop.tistory.com/8402 ]

 

 

 

NH4^+(aq) + H2O(l) ⇌ NH3(aq) + H3O^+(aq)

Kh = Ka = [NH3][H3O^+] / [NH4^+]

= (x) (x) / (C–x)

( 참고 https://ywpop.tistory.com/4294 )

 

 

 

Ka = Kw / Kb

( 참고 https://ywpop.tistory.com/2937 )

 

= (1.0×10^(–14)) / (1.75×10^(-5))

= 5.71×10^(-10)

 

 

 

C–x ≒ C 라 근사처리하면,

x = [Ka × C]^(1/2)

= [(5.71×10^(-10)) × 0.25]^(1/2)

= 1.1948×10^(-5) M = [H^+]

 

 

 

pH = –log[H^+]

= –log(1.1948×10^(-5))

= 4.92

 

 

 

pOH = 14.00 – 4.92

= 9.08

 

 

 

 

[키워드] NH4Cl 용액의 pH 기준문서