A 0.25 M solution of a salt NaA has pH = 9.29

A 0.25 M solution of a salt NaA has pH = 9.29

 

 

A 0.25 M solution of a salt NaA has pH = 9.29.

What is the value of Ka for the parent acid HA?

 

 

 

NaA(aq) → Na^+(aq) + A^-(aq)

---> [NaA] = [A^-] = 0.25 M

 

 

 

pOH = 14 – pH

= 14 – 9.29

= 4.71

 

 

 

pOH = –log[OH^-] = 4.71 이므로,

[OH^-] = 10^(-4.71) M = x

 

 

 

약산의 짝염기(A^-)의 가수분해

A^-(aq) + H2O(l) ⇌ HA(aq) + OH^-(aq)

( 참고 https://ywpop.tistory.com/5502 )

 

Kh = Kb = (x) (x) / (C – x)

( 참고 https://ywpop.tistory.com/4294 )

 

 

 

C – x ≒ C 라 근사처리하면,

Kb = x^2 / C

= (10^(-4.71))^2 / 0.25

= 1.52×10^(-9)

 

 

 

Ka = Kw / Kb

= (1.0×10^(-14)) / (1.52×10^(-9))

= 6.58×10^(-6)

 

 

 

답: Ka = 6.6×10^(-6)

 

 

 

 

[키워드] 약산의 짝염기의 가수분해 기준