solubility of CaF2​ at pH 3.0 is 3.86×10^(-x) M

solubility of CaF2 at pH 3.0 is 3.86×10^(-x) M

 

 

The solubility of CaF2 in a solution

buffered at pH = 3.0 is 3.86×10^(-x) M.

Calculate the value of x.

 

Ka for HF = 6.3×10^(-4) and

Ksp of CaF2 = 3.45×10^(-11)

 

---------------------------------------------------

▶ 참고: Ksp와 몰용해도

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

---------------------------------------------------

 

CaF2(s) ⇌ Ca^2+(aq) + 2F^-(aq)

 

Ksp = [Ca^2+] [F^-]^2 = 3.45×10^(-11)

 

[F^-] = 2 × [Ca^2+]

 

 

 

 

HF(aq) ⇌ H^+(aq) + F^-(aq)

 

Ka = [H^+] [F^-] / [HF] = 6.3×10^(-4)

 

 

 

 

pH 3.0 ---> [H^+] = 10^(-3) M

 

CaF2에서 이온화된 F^-와 H^+가 반응해서, HF 생성.

 

F^-(aq) + H^+(aq) ⇌ HF(aq)

 

K = [HF] / [F^-] [H^+] = (1 / (6.3×10^(-4)))

 

[HF] / [F^-] = (1 / (6.3×10^(-4))) × (10^(-3)) = 1.587

 

[HF] = 1.587 × [F^-]

 

 

 

 

[HF] + [F^-] = 2 × [Ca^2+]

 

 

 

 

1.587 × [F^-] + [F^-] = 2 × [Ca^2+]

 

2.587 × [F^-] = 2 × [Ca^2+]

 

[F^-] = (2 / 2.587) × [Ca^2+]

 

[F^-] = 0.7731 × [Ca^2+]

 

 

 

 

Ksp = [Ca^2+] [F^-]^2 = 3.45×10^(-11)

 

Let, s = [Ca^2+]

 

(s) (0.7731s)^2 = 3.45×10^(-11)

 

s = [ 3.45×10^(-11) / 0.7731^2 ]^(1/3)

= 0.000386 M

= 3.86×10^(-4) M

 

 

 

답: x = 4

 

 

 

 

[키워드] CaF2의 몰용해도 기준문서, CaF2의 Ksp 기준문서, why 2 × [Ca^2+] = [HF] + [F^-], why 2[Ca^2+] = [HF] + [F^-], why [HF] + [F^-] = 2 × [Ca^2+], why [HF] + [F^-] = 2[Ca^2+], how 2 × [Ca^2+] = [HF] + [F^-], how 2[Ca^2+] = [HF] + [F^-], how [HF] + [F^-] = 2 × [Ca^2+], how [HF] + [F^-] = 2[Ca^2+], MBE = mass balance equation, MBE dic