Question

Het Calculate the solubility of CaF2 in a solution that is buffered at [H+] = 0.0025 M with (HF) = 0.70 M. Ksp (CaF,) = 3.9 10-11 K. (HF) = 7.2 x 10-4 Solubility = g/L

Answer 1

First we have to find out concentration of F ^- ions in the buffer solution.

We have, Henderson's equation for acidic buffer pH = pKa + log [Salt] / [ Acid]

$\therefore$ pH = pKa + log [F ^- ] / [HF]

pH can be calculated from [H⁺]

pH = - log [H⁺] = - log 0.0025 = 2.60

pKa = - log Ka = - log 7.2 $\times$ 10 ^-04 = 3.14

$\therefore$ 2.60 = 3.14 + log [F ^- ] / 0.70

log [F ^- ] / 0.70 = 2.60 -3.14 = - 0.54

[F ^- ] / 0.70 = 10 ^-0.54 = 0.288

[F ^- ] = 0.288 ( 0.70 ) = 0.202 M

Consider a reaction , CaF ₂ (s)   Ca ²⁺ (aq) + 2 F ^- (aq)

Equilibrium constant for above reaction is K sp = [Ca ²⁺ ] [F ^- ] ² = 3.9 $\times$ 10 ^-11

If S is the solubility of CaF ₂ in mol / L , then [Ca ²⁺ ] = S mol / L and [F ^- ] = ( 2 S + 0.202 ) mol / L.

K sp = [Ca ²⁺ ] [F ^- ] ² = S $\times$ (2 S + 0.202 ) = 3.9 $\times$ 10 ^-11

Assume S is very small as compared to 0.202, hence we can write 2 S + 0.202 $\approx$ 0.202

S $\times$ ( 0.202 ) ² =   3.9 $\times$ 10 ^-11

S =   3.9 $\times$ 10 ^-11 / ( 0.202 ) ²

S =  9.56 $\times$ 10 ^-10 M

( Check assumption  ( 9.56 $\times$ 10 ^-10 / 0.202 ) 100 = 4.73 10 ^-07 %. This value is very very small compared 0.202. Hence, our assumption is correct)

Molar mass of CaF ₂ = 40.08 + ( 2 $\times$ 19.00 ) = 78.08 g/mol

Solubility in g / L =   9.56 $\times$ 10 ^-10 mol / L $\times$ 78.08 g / mol = 7.5 $\times$ 10 ^-08 g / L