What is the molar solubility of MgF2 in a 0.36 M Mg(NO3)2 solution? For MgF2, Ksp = 8.4 × 10^–8

2.4 × 10⁻⁴ M

Explanation:

Step 1: Calculate the concentration of Mg²⁺ coming from Mg(NO₃)₂

Mg(NO₃)₂ is a strong electrolyte and the molar ratio of Mg(NO₃)₂ to Mg²⁺ is 1:1. The initial molar concentration of Mg²⁺ is 1/1 × 0.36 M = 0.36 M.

Step 2: Make an ICE chart for the solution of MgF₂

        MgF₂(s) ⇄ Mg²⁺(aq) + 2 F⁻(aq)

I                           0.36             0

C                           +S             +2S

E                         0.36+S         2S

The solubility product constant is:

Ksp = [Mg²⁺] × [F⁻]² = (0.36+S) × (2S)²

Since S <<< 0.36, 0.36+S ≈ 0.36.

Ksp = 0.36 × 4S² = 8.4 × 10⁻⁸

S = 2.4 × 10⁻⁴ M

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.