the solubility of MN(OH)2 is 3.04 x 10^-4 gram per 100. milliliters of solution at 25 degrees C Write the balance me chemical equation for the disassociation of MN(OH)2 (s) in a aqueous solution.

Question

Calculate the molar solubility 

Calculate the value of solubility product constant Ksp for MN(OH)2 at 25 degrees celcius

Answer 1

Mn(OH)₂(s) <==> Mn²⁺(aq) + 2OH^-(aq) ... balanced equation for dissociation

molar solubility = moles solute/liter solution. To find this value, we obtain moles of Mn(OH)₂ as follows:

3.04x10^-4 g Mn(OH)₂ x 1 mole / 88.95 g = 3.412x10^-6 moles

This is in 100 ml = 0.100 liters

Molar solubility = 3.412x10^-6 moles / 0.1 liters = 3.4x10^-5 moles/liter = 3.4x10^-5 M

Ksp = [Mn²⁺][OH-]²

From molar solubility and the balanced dissociation we know that [Mn²⁺] = 3.4x10^-5 M and that the

[OH-] = 2 x 3.4x10^-5 M = 6.8x10^-5 M. Plugging these values into the Ksp expression we obtain...

Ksp = (3.4x10^-5)(6.8x10^-5)2 = 1.6x10^-13