Question

Consider these two equilibrium processes.    

(i)      CaF₂ (s)    ↔   Ca²⁺   +   2 F−                  K^o_sp = 3.2 x 10⁻¹¹

            (ii)     HF     ↔    H+    +     F⁻                             K^o_a    = 6.8 x 10⁻⁴   

(a) (0.6 pt) Use equations (i) and (ii), and their K’s to find K^o_overall for this process:   

(iii)       CaF₂ (s)       + 2 H+      ↔     Ca²⁺   +   2 HF           K_eq = ??

(b) (0.6 pt) Find K_{sp, conditional} for reaction (iii) when the pH is such that the fraction of C_HF that is F⁻ is 0.25. (i.e., its alpha-function, a₁, is 0.25)

(c) (0.6 pt) Calculate the pH at which a₁ (fraction of F⁻) is 0.25.

Please explain!!

Answer 1

Solution: Given data CaF,(s)-Ca2+2F Kosp- 3.2x10-11 Koa-6.8x104 i) a.) From the above given reactions F ion is in common, then overall reaction constant can be calculated as overall K 3.2*10 *6.8*104 K 2.176*10 i) Given reaction CaF2(s)+2H-Ca2+2HF Kequillibrium can be calculated for any reaction as verall [AT [B] For the above reaction Keq is calculated as From the above orders Keq obtained is Keq-6.7 * 10-4