The high dipole moment of water and its ease in forming hydrogen bonds make it an excellent analysis. A molecule is soluble in water if it can interact with its molecules through hydrogen bonds or ion-dipole interactions.

With anions that have oxygen they can form hydrogen bonds, since oxygen acts as their acceptor. The attraction of the anion on the water dipole must be taken into account. The same goes for Cl-F, which have solitary electron pairs and can act as hydrogen bridge acceptors. On the other hand, cations such as Na+, K+, Ca++ or Mg++ are surrounded by water molecules to which they are joined by dipole ion interactions while oxygen atoms are oriented towards the catión.

Explanation:

5 0

3 years ago

Part A - Calculate the molar solubility in water

Andrei [34K]

Part 1)

when the balanced equation for this reaction is:

and by using ICE table

Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)

initial 0 0

change +X +2X

Equ X 2X

When KSp expression = [Mg2+][OH-]

when we have KSp = 5.61 x 10^-11

and when we assumed [Mg2+] = X

and [ OH-] = (2X)^2

when we assume X is the value of molar solubility of Mg(OH)2

so, by substitution:

5.61 x 10^-11 = 4X^3

∴ X = 2.4 x 10^-4 M

∴ molar solubility of Mg(OH)2 = X = 2.4 x 10^-4 M

Part 2)

the molar solubility of Mg(OH)2 in 0.16 m NaOH we assumed it = X

by using the ICE table:

Mg(OH)2(s) → Mg2+(aq) + 2OH-

initial 0 0.16m

change +X +2X

equ X (0.16+2x)

when Ksp = [mg2+][OH-]^2

5.61 x 10^-11 = X * (0.16+2X)^2 by solving for X

∴ X = 1.3 x 10^-5 M

∴ the molar solubility = X = 1.3 x 10^-5 M

3 0

4 years ago