Question

Use ksp=4.87×10−17 to calculate the solubility of iron (ii) hydroxide in pure water in grams per 100.0 ml of solution. g

Answer 1

The ksp = 4.87 × 10^-17
Therefore; 4.87 ×10^-17 = [Fe2+][OH-]^2 = (X)(2X)^2 = 4X^3
Hence; 
 4x^3 =4.87 × 10^-17 
 x = 2.30 × 10^-6 M
therefore the molarity is 2.3 ×10^-6 M
Thus;
The mass of Fe(OH)2 in 100 ml of water, will be given by:
  = 2.30 × 10^-6 mol/L = 2.3 ×× 10^-5 mole/100 mL
 =  89.86 g/mol × 2.3 × 10^-6 × 0.100 = 2.07 × 10^-5 g
= 2.07 × 10^-5 g 
 = 2.07 × 10^-5 g