AP chem question

Question

A solution of 100.0 mL of 0.200 M KOH mixed with a solution of 200.0 ml of 0.150M NiSo4.

a) Write the balanced chemical equation for the reaction that occurs

b) What precipitate forms?

c) What is the limiting reactant?

d) What is the concentration of each ion that remains in solution?

Answer 1

a) 2KOH + NiSO₄ ==>K₂SO₄ + Ni(OH)₂

b) The precipitate is Ni(OH)₂, nickel hydroxide

c) KOH is limiting. How do we know?

moles KOH = 100 ml x 1 L / 1000 ml x 0.2 mol/L = 0.02 mols

moles NiSO4 = 200 ml x 1 L / 1000 ml x 0.15 mol/L = 0.03 moles

From balanced equation (a) it takes 2 mols KOH for each 1 mol NiSO4. Not enough KOH

d) moles K₂SO₄ formed:

0.02 mol KOH x 1 mol K₂SO₄ / 2 mol KOH = 0.02 mols K₂SO₄ = 0.04 mols K⁺ + 0.02 mols SO₄^2-

moles Ni(OH)₂ formed:

0.02 mol KOH x 1 mol Ni(OH)₂ / 2 mol KOH = 0.01 mols Ni(OH)₂. No ions since it is a precipitate.

moles NiSO₄ remaining:

0.02 mol KOH x 1 mol NiSO₄ / 2 mol KOH = 0.01 mol NiSO₄ used

0.15 - 0.01 = 0.05 mols NiSO₄ remaining = 0.05 mols Ni²⁺ + 0.05 mols SO₄^2-

Summary of ions remaining in solution:

K⁺ = 0.04 mols

SO₄^2- = 0.02 mols + 0.05 mols = 0.07 mols

Ni²⁺ = 0.05 mols

Final volume = 100 ml + 200 ml = 300 mls = 0.3 L

Final concentration of ions remaining in solution:

K⁺ = 0.04 mols / 0.3 L = 0.133 M K⁺

SO₄^2- = 0.07 mols / 0.3 L = 0.233 M SO₄^2-

Ni²⁺ = 0.05 mols / 0.3 L = 0.167 M Ni²⁺

(Note: This does not include the [H⁺] and [OH^-] contributed by H2O.