Question

1) Write the four reactions for the dissociation of each metal hydroxide into ions, AgOH(s), Fe(OH)3(s), Cu(OH)2(s), Zn(OH)2(s). 2) Write the three other solubility product expressions for the reactions in question 1. Ksp = [Ag+][OH-] 3) Write the reactions of cobalt(II) ion with hydroxide ion and the solubility product expression for the reaction. 4) Calculate the mass of CuSO4.5H2O required to make 25.00 mL of a 0.1 M solution. 5) Calculate the hydroxide ion concentration for a solution with pH 10. I need some help in questions 2 to 5

Answer 1

2.)

Fe(OH)_3(s)     <=====> Fe³⁺_(aq) +     3OH^-_(aq)

Ksp = [Fe³⁺_(aq)][OH^-_(aq)]³

Cu(OH)_2(s)     <=======> Cu²⁺_(aq)   +    2OH^-_(aq)

Ksp = [Cu²⁺_(aq)][OH^-_(aq)]²

Zn(OH)_2(s)     <=======> Zn²⁺_(aq)   +    2OH^-_(aq)

Ksp = [Zn²⁺_(aq)][OH^-_(aq)]²

3.

Co²⁺_(aq)   +    2OH^-_(aq)     <=======> Co(OH)_2(s)   

Ksp = [Co(OH)_2(s) ] / [Zn²⁺_(aq)][OH^-_(aq)]²

4.

Molarity = number of moles of solute X 1000 / volume of solvent (mL)

Or number of moles of solute = Molarity X volume of solvent / 1000

number of moles of solute = 0.1 X 25 / 1000 = 2.5 / 1000

number of moles of solute = 0.0025

CuSO4.5H2O = 249.685 (Mw)

Weight of solute = Mw of solute X moles of solute

Weight of solute (CuSO4.5H2O) = 249.685 X 0.0025 = 0.6242 g

5. From simple realation between pH and pOH:

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 10

pOH = 4

Now, [OH-] = 10^-pOH

Hence, [OH-] = 10^-4