The solubility of Mn(OH)2 is 3.04 x 10^-4 gram per 100 ml of solution .

A) write the balanced chem equation for Mn(OH)2 in aqueous solution
B) calculate the molar solubility of Mn(OH)2 at 25 degrees celcius
C) calculate the value of the solubility product constant, Ksp, for Mn(OH)2 at 25 degrees celcius

mols Mn(OH)2 = grams/molar mass = approx 3E-6 but this is an estimate. You should recalculate this and all that follows.

Then M = mols/L = about 3E-6/0.1L = about 3E-5.

...........Mn(OH)2 ==> Mn^2 + 2OH^-
I..........solid........0.......0
C..........solid........x.......2x
E..........solid........x.......2x

Ksp = (Mn^2+)(OH^-)^2
(Mn^2+) = about 3E-5
(OH^-) = about 2*3E-5 = about 6E-5

Substitute this into Ksp expression and solve for Ksp.

A) To write the balanced chemical equation for Mn(OH)2 in aqueous solution, we need to consider the dissociation of Mn(OH)2 into its respective ions in water. The balanced equation is as follows:

Mn(OH)2 (s) ⇌ Mn2+ (aq) + 2OH- (aq)

B) To calculate the molar solubility of Mn(OH)2 at 25 degrees Celsius, we need to convert the given solubility in grams per 100 ml to moles per liter.

Given: Solubility of Mn(OH)2 = 3.04 x 10^-4 gram/100 ml = 3.04 x 10^-3 gram/1000 ml = 3.04 x 10^-3 gram/1 liter

The molar mass of Mn(OH)2 is:
Molar mass of Mn = 54.94 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of Mn(OH)2 = (54.94 g/mol) + 2(1.01 g/mol) + 2(16.00 g/mol) = 88.96 g/mol

To calculate the molar solubility, we need to divide the solubility in grams by the molar mass:
Molar solubility of Mn(OH)2 = (3.04 x 10^-3 g) / (88.96 g/mol) = 3.42 x 10^-5 mol/L

Therefore, the molar solubility of Mn(OH)2 at 25 degrees Celsius is 3.42 x 10^-5 mol/L.

C) To calculate the value of the solubility product constant (Ksp) for Mn(OH)2 at 25 degrees Celsius, we can use the molar solubility obtained in part B.

The balanced equation is:
Mn(OH)2 (s) ⇌ Mn2+ (aq) + 2OH- (aq)

Since Mn(OH)2 dissolves into one Mn2+ ion and two OH- ions, the expression for the solubility product constant is:

Ksp = [Mn2+] * [OH-]^2

Since the concentration of Mn2+ is equal to the molar solubility and the concentration of OH- is equal to twice the molar solubility:

Ksp = (3.42 x 10^-5 mol/L) * (2 * 3.42 x 10^-5 mol/L)^2

Ksp = 1.47 x 10^-14

Therefore, the value of the solubility product constant (Ksp) for Mn(OH)2 at 25 degrees Celsius is 1.47 x 10^-14.