0.090 M KBr 0.010 M HBr pH

0.090 M KBr 0.010 M HBr pH

 

 

Calculate the pH of 0.010 M HBr in 0.090 M KBr.

 

 

0.090 M KBr에 녹인 0.010 M HBr의 pH를 계산하시오.

 

 

 

Activity, 활동도의 정의

a_C = [C] × γ_C

 

> a_C: C의 활동도

> [C]: C의 몰농도

> γ_C: C의 활동도 계수 (activity coefficient)

( 참고 https://ywpop.tistory.com/2684 )

 

 

 

HBr(aq) → H^+(aq) + Br^-(aq)

[HBr] = [H^+] = [Br^-] = 0.010 M

 

KBr(aq) → K^+(aq) + Br^-(aq)

[KBr] = [K^+] = [Br^-] = 0.090 M

 

 

 

ionic strength, μ

μ = (1/2) ( [H^+]×(+1)^2 + [Br^-]×(–1)^2

+ [K^+]×(+1)^2 + [Br^-]×(–1)^2 )

= (1/2) ( (0.010)×(+1)^2 + (0.010)×(–1)^2

+ (0.090)×(+1)^2 + (0.090)×(–1)^2 )

= 0.10

( 참고 https://ywpop.tistory.com/19070 )

 

 

 

[도표] Activity coefficients for aqueous solutions at 25℃

 

 

---> The activity coefficient of H^+ is 0.83.

 

 

 

pH = –log(a_H^+)

= –log([H^+] × γ_H^+)

= –log(0.010 × 0.83)

= 2.0809

= 2.08

 

 

 

답: pH = 2.08

 

 

 

 

[참고] Calculate the pH of 0.10 M HBr,

using activity coefficients.

 

 

μ = (1/2) ( [H^+]×(+1)^2 + [Br^-]×(–1)^2 )

= (1/2) ( (0.10)×(+1)^2 + (0.10)×(–1)^2 )

= 0.10

---> The activity coefficient of H^+ is 0.83.

 

 

Remember that pH is –log(a_H^+), not –log[H^+]

 

 

pH = –log([H^+] × γ_H^+)

= –log(0.10 × 0.83)

= 1.0809

= 1.08

 

 

 

 

[키워드] 활동도 기준, 활동도 계수 기준, 이온 강도 기준, 이온 세기 기준

 

[FAQ] [①24/06/20]