Question

1. A solution of magnesium hydroxide is prepared by dissolving 361.824 mg of Mg(OH)2 in 273.5 mL of water. What is the equilibrium concentration of OH- in this solution?
For magnesium hydroxide, Ksp = 7.1 x 10-12.
Assume no volume change.

2. The above solution is filtered to remove all undissolved magnesium hydroxide solid. What mass of solid Mg(OH)2 should be recovered from the solution?
Express your answer in units of mg.

3. After the above solution was filtered and all undissolved magnesium hydroxide solid removed, 564.1 mg of tin chloride is added.
SnCl2 completely dissociates in aqueous solution. Hydroxide reacts with tin (II) to form a Sn(OH)2complex.
What is the equilibrium concentration of Sn2+ in this solution?
The formation constants for the tin-hydroxide complexes are: Kf1 = 2.5 x 1010, Kf2 = 3.2 x 1013.

4. What is the equilibrium concentration of the SnOH+ complex in this same solution?

5. What is the equilibrium concentration of Sn(OH)2 in this same solution?

6. What is the ionic strength of the above solution?

Answer 1

1.
Mg (OH)2 <-----> Mg2+ + 2OH-
                                        s               2s
Ksp = [Mg2+] [OH-]^2
7.1*10^-12 = s * (2s)^2
7.1*10^-12 = 4*s^3
s = 1.21*10^-4 M
[OH-] = 2s
              = 2* 1.21*10^-4
              = 2.42*10^-4 M
Answer: 2.42*10^-4 M

2.
Mg(OH)2 dissolved = 1.21*10^-4 mL
V = 273.5 mL = 0.2735 L
number of moles of Mg(OH)2 = M*V
                           = 1.21*10^-4 * 0.2735
                           = 3.31*10^-5 mol
So,
amount of Mg(OH)2 dissolved = number of moles * molar mass
                                                               = (3.31*10^-5) * 58.32
                                                               = 1.931*10^-3 g
                                                                = 1.931 mg
Amount of Mg(OH)2 to be recovered = 361.824 mg - 1.931 mg = 359.893 mg
Answer: 359.893 mg

only 1 question at a time please