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Fluorite into a 0.352 M aqueous soi or this probletm (a) Set up the equilibrium

ID: 637009 • Letter: F

Question

Fluorite into a 0.352 M aqueous soi or this probletm (a) Set up the equilibrium variable to be solved by the geologist's finding, decided to try dissolving a lump of (b) Insert the variable used in the eqilbrium table into the Ksp expression (e) Calcu solution. (Use the Newton-Raphson procedure if necessary). late the equilibrium concentration of Ca? and F ions within the chemist's (d) Did the geologist's pond or the chemist's solution have a higher concentration of calcium ion? (e) Name and briefly describe the principle that explains this difference in calcium ion concentration.

Explanation / Answer

a) Fluorite contains calcium fluoride, CaF2. CaF2 is sparingly soluble in water and the solubility if given as

CaF2 (s) <======> Ca2+ (aq) + 2 F- (aq); Ksp = [Ca2+][F-]2 = 3.45*10-11

Sodium fluoride, NaF ionizes completely in aqueous solution as below.

NaF (aq) -------> Na+ (aq) + F- (aq)

Since 1 mole NaF = 1 mole F-, therefore,

[F-] = 0.352 M.

Note that [F-] from ionization of NaF is much higher than [F-] obtained from CaF2 because CaF2 is sparingly soluble in water. Therefore, we can approximate the total [F-] in the aqueous solution of NaF containing fluorite to be 0.352 M.

Thus, we set up the equilibrium table as below.

CaF2 (s) <=======> Ca2+ (aq) + 2 F- (aq)

initial                                      -                                      

equilibrium                                                              x                0.352

Let the concentration of Ca2+ in solution = solubility of CaF2 be x M. We are required to determine the numerical value of x.

b) Write down the Ksp expression for the dissolution equation as

Ksp = [Ca2+][F-]2 = (x).(0.352)2

c) Now solve the above expression.

Ksp = (x).(0.123904)

=====> 3.45*10-11 = (x).(0.123904)

=====> x = (3.45*10-11)/(0.123904) = 2.7844*10-10 ? 2.78*10-10

The equilibrium concentrations of Ca2+ and F- ions in the chemist’s solution are 2.78*10-10 M and 0.352 M respectively. Note that we have started by saying that the concentration of F- obtained from the ionization of NaF is much higher than that obtained from CaF2 (ans).

d – e) Need to know the value obtained by the geologist.