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
A) To write the balanced chemical equation for Mn(OH)2 in aqueous solution, we need to consider the dissociation of the compound into its respective ions. Mn(OH)2 will dissociate into one Mn2+ ion and two OH- ions.
The balanced chemical equation is:
Mn(OH)2 (s) -> Mn2+ (aq) + 2 OH- (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 of solution to moles per liter.
Given: Solubility of Mn(OH)2 = 3.04 x 10^-4 g/100 ml
First, convert grams to moles:
Molar mass of Mn(OH)2 = 54.938 g/mol (Molar mass of Mn = 54.938 g/mol, Molar mass of OH- = 17.007 g/mol)
Moles of Mn(OH)2 = (3.04 x 10^-4 g / 100 ml) / (54.938 g/mol)
Next, convert ml to L:
1 L = 1000 ml
Volume of solution = 100 ml/1000 = 0.1 L
Molar solubility of Mn(OH)2 = (Moles of Mn(OH)2) / (Volume of solution)
C) To calculate the value of the solubility product constant (Ksp) for Mn(OH)2 at 25 degrees Celsius, we need to write the expression for Ksp using the balanced chemical equation.
From the balanced equation: Mn(OH)2 (s) -> Mn2+ (aq) + 2 OH- (aq)
The solubility product constant, Ksp, expression is:
Ksp = [Mn2+] * [OH-]^2
Substitute the molar solubility of Mn(OH)2 calculated in part B to find Ksp. Remember to square the concentration of OH- ions since its coefficient is 2 in the balanced equation.