Question

(a) If the molar solubility of Tl, at 25°C is 5.31e-08 mol/L, what is the Ken at this temperature? Ksp = (b) It is found that 4.68e-06 g of Nd2(CO3)3 dissolves per 100 mL of aqueous solution at 25 °C. Calculate the solubility-product constant for Nd2(CO3)3. Ksp = (c) The Ksp of BaF2 at 25 °C is 1.84e-07. What is the molar solubility of BaFZ? solubility = mol/L

Answer 1

a)

At equilibrium:

Tl2S <----> 2 Tl+ + S2-

   2s s

Ksp = [Tl+]^2[S2-]

Ksp = (2s)^2*(s)

Ksp = 4(s)^3

Ksp = 4(5.31*10^-8)^3

Ksp = 5.989*10^-22

Answer: 5.99*10^-22

b)

S = 4.68*10^-6 g / 100 mL

= 4.68*10^-6 g / 0.1 L

= 4.68*10^-5 g /L

Molar mass of Nd2(CO3)3,

MM = 2*MM(Nd) + 3*MM(C) + 9*MM(O)

= 2*144.2 + 3*12.01 + 9*16.0

= 468.43 g/mol

Molar mass of Nd2(CO3)3= 468.43 g/mol

s = 4.68*10^-5 g/L

To covert it to mol/L, divide it by molar mass

s = 4.68*10^-5 g/L / 468.43 g/mol

s = 9.991*10^-8 mol/L

At equilibrium:

Nd2(CO3)3 <----> 2 Nd3+ + 3 CO32-

   2s 3s

Ksp = [Nd3+]^2[CO32-]^3

Ksp = (2s)^2*(3s)^3

Ksp = 108(s)^5

Ksp = 108(9.991*10^-8)^5

Ksp = 1.075*10^-33

Answer: 1.07*10^-33

c)

At equilibrium:

BaF2 <----> Ba2+ + 2 F-

   s 2s

Ksp = [Ba2+][F-]^2

1.84*10^-7=(s)*(2s)^2

1.84*10^-7= 4(s)^3

s = 3.583*10^-3 M

Answer: 3.58*10^-3 M