Question

Calculate the solubility of CaF2 in a solution that is buffered at H 0.0045 M with HF] = 0.96 M. Kp (CaF2) 3.9 x 10-11 K, (HF) =7.2 x 10 Solubility g/L

Answer 1

HF HtF

$K_a=\frac{[H^+] \times [ F^- ]}{[HF]}=7.2 \times 10^{-4}$... ...(1)

Substitute values in equation (1)

$\\ \ \\ \ \frac{[H^+] \times [ F^- ]}{[HF]}=7.2 \times 10^{-4} \\ \ \\ \ \frac{0.0045 \times [ F^- ]}{0.96}=7.2 \times 10^{-4} \\ \ \\ \ [F^-]=0.154 \ M$

Let S mol/L be the solubility of calcium fluoride

$CaF_2 \rightleftharpoons Ca^{2+}+2F^-$

$K_{sp }= [Ca^{2+}] \times [F^-]^2=3.9 \times 10^{-11}$

Substitute values in above expression

$S \times (0.154)^2=3.9 \times 10^{-11}$

$S =1.65 \times 10^{-9} \ M$

The solubility of calcium fluoride is $1.65 \times 10^{-9} \ mol/L$

The molar mass of calcium fluoride is 78.1 g/mol

Solubility $=1.65 \times 10^{-9} \ mol/L \times 78.1 \ g/mol=1.29 \times 10^{-7} \ g/L$