Net Ionic Equation

Question

Predict the mass of precipitate (Ni(OH)_2(aq)) expected when 50.0 mL of 0.45 M nickel(II) sulfate solution (NiSO_4(aq)) is combined with 25.0 mL of 1.00 M sodium hydroxide solution (NaOH_(aq)).

Answer 1

Hi Kiyan S.,

There are three things you do when confronted with a precipitation reaction such as this one. 1) write the balanced equation for the precipitation, 2) check the moles of material available to see which reactant, if any, is in excess, and 3) check the solubility product of the precipitate, since the precipitation may not be quantitative for the limiting reagent.

Here, you have 0.0225 moles of Ni(2+), requiring 0.045 moles of hydroxide to exactly react, But, you only have 0.025 moles of hydroxide added.

So, you could take the lazy way out, and say just that you make 0.0125 moles of Ni(OH)₂ . (and calculate that up into grams). Bt what you should really do, since this is what nature does, is then "allow" that precipitate to dissolve as it may, to satisfy the solubility product that it has. Let's see.

Ksp = 2* 10^-15 , which might look like not much, but you should calculate through to make sure:

Ksp = [Ni²⁺]¹ [OH]² for the reaction Ni(OH)₂ = Ni²⁺ + 2OH^-

Now you apply "ICE": initial -- change -- equilibrium as a table:

material Ni²⁺ OH^-

initial (0.0225-0.0125)mol/0.075mL =0.133M 10^-7

change +x +x/2

equilibrium 0.133+x 10^-7 + x/2

2*10^-15 = (0.133+x)(10^-7 +x/2)

You solve that through as a quadratic equation, to obtain x: if positive, then a little nickel hydroxide dissolved; if negative, a little nickel hydroxide preciptated. And you adjust your total precipitate based on that. It looks to me as though you're right about at the solubility product for nickel hydroxide, and you certainly can't precipitate much more out -- the water is reluctant to supply it!

-- Cheers, -- Mr. d.