Question

A salt has the form M₂X₃ where M is a metal cation and X is a nonmetal anion. If ΔG_rxn is 80.40 kJ/mol, what is the concentration of the metal ion in a 100.0 mL aqueous solution? Assume the temperature is 25.0^oC.

Answer 1

Solubility equilibrium equati on: M2X (s) 3X- (aq) 2M3 (ag) |x2- Solubility equilibrium expression, K M3 4.2x 1024 Temperature, T= 25°C273= 298 K _ Gas constant, R=8.314 JKmol1 Stan dard free energy change of the reacti on, AG = - RTln K _ AG0 In Kg RT 80.40x 103 J/mol 32.45 8.314 JKmolx 298 K X -3245 Solubility product, Kpi 8.1x10-15 2M* (аq) + 3X2- (аq) Solubility equilibrium equati on: M2X (s [Mx2- Solubility equilibrium expression, K 8.1x10-15 = ICE table: 2M* (аg) M2X (s) + 3X- (аg) Initial concentrati on (M) 0 Change (M) +2s +3s Equilibrium concentration (M): 2s 3s Solubility product, -[M*T[x 8.1x10-15(2s)x(3s) = 108s _ _ 8.1x10-15 6.0x10 M 4 _ 108 (s) M 6.0x10- M According to the ICE table, molar solubility According to the ICE table, concentration of M3, M | = 2(s) M _ = 2(6.0x10 M) = 1.2x103 M

Answer 2

Solve
Use equations for the dissociation of the salt, Ksp, and Δ?∘rxn$\mathrm{\Delta }{G}_{\text{rxn}}^{\circ }$ to solve for the equilibrium concentration. 

We can write a balanced equation for the dissociation of the salt as

$$M2X3(s)+H2O(l)2M3+(aq)+3X−(aq)

Using the balanced equation, we can write the Ksp expression as

Ksp= [M3+]2[X–]3

From the Ksp, we can deduce an expression for the molar solubility using an equilibrium table.

$$M2X3(s)+H2O(l)2M3+(aq)+3X−(aq)
Initial                                   0                    0
Change                             +2s                +3s
Equilibrium                        +2s                +3s

Ksp= [M3+]2[X–]3 = (2s)2(3s)3= 108s5, where s is the molar solubility.

From Δ?∘$\mathrm{\Delta }{G}^{\circ }$, we can determine the value for Ksp using the equation

Δ?∘=−?????$\mathrm{\Delta }{G}^{\circ }=-RTlnK$

$$K=e−(ΔGoRT)=e−(45.40 kJmol×(1000 J1 kJ)(8.314 Jmol K)×(298.15 K))

Ksp = 1.106×10-8 

Using the molar solubility expression, we can calculate a numeric value for the molar solubility.

Ksp= 108s5 = 1.106×10-8

s = (1.106×10-8/108)(1/5)

From the stoichiometry, the concentration of M3+ is 2s or 0.02010 M.

During dissolution, there is an increase in disorder or entropy. Therefore, Δ$\mathrm{\Delta }$S is greater than zero. Because Δ$\mathrm{\Delta }$G is positive, the change in enthalpy, Δ$\mathrm{\Delta }$H, must be positive.