1.2 M HC7H5O2 Ka 6.3×10^(-5) [H^+] [OH^-] pH pOH percent ionization

1.2 M HC7H5O2 Ka 6.3×10^(-5) [H^+] [OH^-] pH pOH percent ionization

 

 

For a solution of 1.2 M HC7H5O2 (Ka = 6.3×10^(-5)),

calculate the following:

[H^+], [OH^-], pH, pOH and the percent ionisation

 

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Ka = x^2 / (C–x)

( 참고 https://ywpop.tistory.com/4294 )

 

 

 

C–x ≒ C 라 근사처리하면,

x = [Ka × C]^(1/2)

= [6.3×10^(-5) × 1.2]^(1/2)

= 0.0087 M = [H^+]

 

 

 

[H^+] [OH^-] = 10^(-14) 이므로,

( 참고 https://ywpop.tistory.com/2704 )

 

[OH^-] = 10^(-14) / 0.0087

= 1.1×10^(-12) M

 

 

 

pH = –log[H^+]

= –log(0.0087)

= 2.1

 

 

 

pOH = –log[OH^-]

= –log(1.1×10^(-12))

= 12

 

또는

pOH = 14 – 2.1 = 11.9

 

 

 

이온화 백분율(%) = ([H^+]_평형 / [HA]_초기) × 100

( 참고 https://ywpop.tistory.com/7263 )

 

= (0.0087 / 1.2) × 100

= 0.73%

 

 

 

 

[키워드] [H^+] [OH^-] pH pOH 기준문서