Question

(a) If the molar solubility of ScF3 at 25 oC is 6.81e-07 mol/L, what is the Ksp at this temperature? Ksp = (b) It is found that 0.00354 g of Ag2CO3 dissolves per 100 mL of aqueous solution at 25 oC. Calculate the solubility-product constant for Ag2CO3. Ksp = (c) The Ksp of ZnSe at 25 oC is 3.60e-26. What is the molar solubility of ZnSe? solubility =

Answer 1

Consider a reaction , ScF ₃ (s)   Sc ³⁺ (aq) + 3 F ^- (aq)

Equilibrium constant for above reaction is K sp = [Sc ³⁺ ] [F ^- ] ³

If S is the solubility of ScF ₃  in mol / L , then [Sc ³⁺ ] = S mol / L and [F ^- ] = 3 S mol / L.

K sp = [Sc ³⁺ ] [F ^- ] ³ = S (3 S ) ³

K sp = S 27 S ³

K sp = 27 S ⁴

We have, S = 6.81 10 ^-07 M

K sp = 27 (6.81 10 ^-07) ⁴

K sp =5.81 10 ^-24

b) Solubility of Ag₂CO₃ = 0.00354 g / 100 ml

We have relation , 1 L = 10 ( 100 ml )

Solubility of Ag₂CO₃ = 0.00354 g / 100 ml   ( 10 ( 100 ml ) / 1 L ) = 0.0354 g / L

Molar Mass of Ag₂CO₃ = ( 2 107.87) + 12.01 + ( 3 16.00) = 275.75 g / mol

Molar solubility of Ag₂CO₃ = ( 0.0354 g / L) / ( 275.75 g / mol) = 1.284 10 ^-04 mol / L

Consider a reaction , Ag₂CO ₃ (s)   2 Ag ⁺ (aq) + CO ₃^2- (aq)

Equilibrium constant for above reaction is K sp = [Ag ⁺ ] ² [CO ₃^2- ]

If S is the solubility of Ag₂CO ₃ in mol / L , then [Ag ⁺ ] = 2 S mol / L and [CO ₃^2- ] = S mol / L.

K sp =[Ag ⁺ ] ² [CO ₃^2- ]= (2 S ) ² S = 4 S ² S = 4 S ³

We have calculated S = 1.284 10 ^-04 mol / L.

K sp = 4 (1.284 10 ^-04 )  ³ = 8.467 10 ^-12

ANSWER : K sp of Ag₂CO ₃ = 8.467 10 ^-12

C)

Consider a reaction , ZnSe (s)   Zn ²⁺ (aq) + Se ^2- (aq)

Equilibrium constant for above reaction is K sp = [Zn ²⁺ ] [Se ^2- ] = 3.60 10 ^-26

If S is the solubility of ZnSe in mol / L , then [Zn ²⁺ ] = S mol / L and [Se ^2- ] = S mol / L.

K sp = [Zn ²⁺ ] [Se ^2- ] = S S = S ²

S =

K sp

S =

V3.60 x 10e - 26

S = 1.897 10 ^-13 M

ANSWER : Molar solubility of ZnSe is 1.897 10 ^-13 M