Homework question

Question

50.0 ml of 10.0 M sodium hydroxide is added to a 1.00 L solution containing 0.100M nickel (III) nitrate, 0.100M tin (II) nitrate, and 0.100M zinc nitrate. How many grams of tin (II) hydroxide precipitates from solution

Answer 1

3NaOH(aq) + Ni(NO₃)₃(aq) => Ni(OH)₃(s) + 3NaNO₃(aq)

2NaOH(aq) + Sn(NO₃)₂(aq) => Sn(OH)₂(s) + 2NaNO₃(aq)

2NaOH(aq) + Zn(NO₃)₂(aq) => Zn(OH)₂(s) + 2NaNO₃(aq)

0.050 L x 10.0 mol/L = 0.50 moles NaOH added to the solution.

moles of Ni³⁺, Sn²⁺ and Zn²⁺ = 1.0 L x 0.100 mol/L = 0.100 moles of each cation.

moles OH- needed to precipitate the cations:

3 mol OH-/mol Ni x 0.1 mol Ni = 0.3 moles OH- needed

2 mol OH-/mol Sn x 0.1 mol Sn = 0.2 moles OH- needed

2 mol OH-/mol Zn x 0.1 mol Zn = 0.2 moles OH- needed

Total moles needed to precipitate all cations = 0.7 moles. But only have 0.5 moles OH- present.

The mass of each precipitate will depend on the Ksp values for the individual insoluble salts and it isn't possible to find the mass of tin(II) hydroxide precipitated without knowing the Ksp values.