Calculate the molar solubility of fe(oh)2 in pure water. (the value of ksp for fe(oh)2 is 4.87×10−17.) express the molar solubility in moles per liter to two significant figures. m

When Fe(OH)2 (S) ↔ Fe2+   +   2OH- is the reaction equation

let the molar solubility = X

when mol/L Fe(OH)2   gives X mol / L Fe2+ and 2X mol / L OH-

when the Ksp expression is:

Ksp = [Fe2+] [OH-]^2

when we assume [Fe2+] = X 

and [OH-] = 2X

and when we have Ksp = 4.87 x 10^-17

so by substitution:

∴ 4.87 x 10^-17 = X*(2X)^2

∴ 4X^3 = 4.87 x 10^-17

∴X = 2.3 x 10^-6

∴ the molar solubility = X = 2.3 x 10^-6  mol/L

2Cr(OH)3 + 3Br2 + 10OH- → 2CrO42- + 6Br- + 8H2O

Explanation: