Question

A 80.0 mL aliquot of a 0.250 M Ca(NO_3)_2 solution was mixed with 60.0 mL of a 0.500 M NaF solution. K_sp(CaF_2) = 3.90 times 10^-11. Calculate the mass of CaF_2 that was formed. Determine the equilibrium concentrations of Ca^2+ and F^- ions remaining in the resultant solution.
Please help!!!!!

Answer 1

a.

Balanced equation:
Ca(NO3)2 + 2NaF → CaF2 + 2NaNO3

Number of Ca(NO3)2= molarity * volume in L

= 0.250* 0.080

= 0.02 Moles Ca(NO3)2

Number of NaF = molarity * volume in L

= 0.500* 0.060

= 0.03 Moles NaF

Here NaF is limiting agent.

Moles of CaF2:

0.03 Moles NaF * 1 mole CaF2 / 2 Mole NaF

= 0.015 mole CaF2

Amount of CaF2 = Number of moles * molar mass

= 0.015 moel CaF2 *78.07 g/mol

= 1.17 g CaF2

Equilibrium concentrations of Ca2+ and F- ion

CaF2(s)⇔ Ca2+ (aq)+ 2F- (aq)

Given that Ksp = 3.90*10^-11

Ksp = [Ca2+] * [F-]

How does Ca(F)2 dissociate:
Ca(F)2 ↔ Ca2+ + 2F-
Let X = [Ca2+]
Because you dissociate to for 2F-, the equation become

Ksp = [X] * [2X]²
3.90*10^-11 = 4X³
X³ = 9.75*10^-12
X = 2.14*10^-4

X = [Ca2+]= 2.14*10^-4

2X= [F-] =2*2.14*10^-4

= 4.28*10^-4