BaF2의 몰용해도

BaF2의 몰용해도

 

 

What is the molar solubility of barium fluoride (BaF2) in water?

The solubility-product constant for BaF2 is 1.7×10^(-6) at 25℃.

 

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[참고] 용해도곱상수와 몰용해도. Ksp와 몰용해도

[ https://ywpop.tistory.com/8434 ]

 

 

 

BaF2(s) ⇌ Ba^2+(aq) + 2F^-(aq)

 

Ksp = [Ba^2+] [F^-]^2

 

1.7×10^(-6) = (x) (2x)^2 = 4x^3

 

x^3 = (1.7×10^(-6)) / 4

 

x = [(1.7×10^(-6)) / 4]^(1/3)

= 0.0075 M

 

 

 

답: 7.5×10^(-3) M

 

 

 

 

[ 관련 글 https://ywpop.tistory.com/471423 ]

What is the molar solubility of BaF2 in 0.30 M NaF?

Ksp of BaF2 = 1.0×10^(-6)

 

 

 

[키워드] BaF2의 몰용해도 기준