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Nataly [62]
3 years ago
7

The Ksp of CaF2 is 5.3 mc003-1.jpg 10-9. What is the molar solubility of CaF2 in 0.050 M Ca(NO3)2?

Chemistry
1 answer:
faltersainse [42]3 years ago
4 0
<span>Calcium-fluoride and calcium-nitrite dissolve and dissociate as follows:
CaF₂(s) ⇄ Ca²⁺(aq) + 2 F⁻(aq)  Ksp = [Ca</span>²⁺] x ([F⁻])^2 = 5.3 x 10^-9
Ca(NO₃)₂(s) ⇄ Ca²⁺(aq) + 2 NO₃⁻(aq)  [Ca²⁺]₀ = [<span>Ca(NO₃)₂] = 0.050 M

If s is the molar solubility of</span><span> calcium-fluoride, then, at equilibrium, we can write:
[Ca²⁺] = s +</span><span>[Ca²⁺]₀ = s + 0.050 M
[F⁻]= 2s

5.3 x 10^-9 = (s + 0.050 M) x (2s)^2

S is very small compared to 0.050 M so the s inside the first brackets can be ignored:
</span>
5.3 x 10^-9 = (0.050 M) x (2s)^2 <span>
<span>5.3 x 10^-9 / 0.050 M = 4 x s^2
</span>1.06 x 10^-7 = 4 x s^2</span>
<span>1.06 x 10^-7 / 4 = s^2
2.65 x 10^-8 = s^2
s = √</span><span><span>2.65 x 10^-8
s = 1.63 x 10^-4</span>

The molar solubility of calcium-fluoride in 0.050 M of calcium-nitrite is 1.63 x 10 ^-4 moles / liter.</span>














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