Question

In a saturated solution of MgF₂ at 16 °C, the concentration of Mg²⁺ is 1.44x10 ^–3 M

The equilibrium is represented by the equation

MgF₂ -->Mg ²⁺ + 2 F^-.

A.Write the expression for the solubility product constant, Ksp.

B.Calculate the value of Ksp at 16°C for MgF₂

Note: Your answer is assumed to be reduced to the highest power possible.

C.Calculate the equilibrium concentration of Mg²⁺ in 1.000 L of saturated MgF2 solution  at 16°C to which 0.450 mole of solid KF has been added. The KF dissolves completely and causes no change in volume.

Note: Your answer is assumed to be reduced to the highest power possible.

D.Calculate Q to determine if a precipitate will form when 250.0 mL of a 0.00200 M solution of Mg(NO₃)₂ is mixed with 250.0 mL of a 0.00200 M NaF solution at 16°C.

Note: Your answer is assumed to be reduced to the highest power possible.

E.Does a precipitate form? yes or no

F.At 24°C the concentration of Mg²⁺ in a saturated solution of MgF₂ is 1.87 x 10 ^–3 M.

Is the dissolving of MgF₂ in water an endothermic or exothermic process. Give an  explanation to support your choice. choose one of the choice below to answer F.

	A) Endothermic An increase in temperature causes the equilibrium to shift to the reactants, which is shifting away from the energy of the reaction.
	B) Endothermic An increase in temperature causes the equilibrium to shift to the products, which is shifting towards the energy of the reaction.
	C) Exothermic An increase in temperature causes the equilibrium to shift to the products, which is shifting away from the energy of the reaction.
	D) Exothermic An increase in temperature causes the equilibrium to shift to the reactants, which is shifting away from the energy of the reaction.

Answer 1

0 A) MgF2(s) = Mg2+ (aq)+2F (aq) I 1 C -s +s +2s E 1-s S 2s Where s = molar solubility of MgF2 Solubility product, K., =[Mg2+],[F-1 B) K., =[Mg2+ ][F-7 =(s)(25) = 453 Given that [Mg2+] =1.44*10-ⓇM ::s=1.44x10-'M Hence solubility product K.= 4(1.44x10^M) = 1.19x10-8 Jeg

C)KF(aq) →K+ (aq)+F (aq) 0.450 mol 1.000L =0.450 M K=[Mg2+][F-7 [M9%)FFI Jeg _ 1.19x10-8 (0.450M) = 5.88×10 M D)Concentrations of Mg2+ and F-in mixture can be calculated as follows. (0.00200 M x 250.0 mL) Mg (250.0 mL + 250.0 mL) = 0.00100M [F-1-(0.00200 M Ⓡ 250.0mL) (250.0 mL +250.0 mL) = 0.00100M Reaction quotient, Q=[Mg2+[F-] =(0.00100 M )(0.00100 M)? = 1.00x109