Find the ions in this soulution

Question

50mL of 0.20mol L^-1 sodium hydroxide was mixed with 50ml 0.30mol L_1 magnesium chloride to form a precipitate of magnesium hydroxide.

calculate the concentration of

sodium ions

magnesium ions

and chlorine ions

Answer 1

Compounds to be considered:

sodium hydroxide = NaOH

magnesium chloride = MgCl₂

magnesium hydroxide = Mg(OH)₂

Reaction taking place:

2NaOH(aq) + MgCl₂(aq) ==> Mg(OH)₂(s) + 2NaCl(aq)

mols NaOH = 50 ml x 1 L/1000 ml x 0.20 mol/L = 0.01 mols NaOH

mols MgCl2 = 50 ml x 1 L/1000 ml x 0.30 mol/L = 0.015 mols MgCl2

Limiting reactant = MgCl₂ since the mol ratio in balanced equation is 2 : 1 for NaOH : MgCl₂

This means all of the Mg²⁺ ion will end up in the precipitate of Mg(OH)₂(s) and all of the Cl^- ion will end up as NaCl(aq). All of the Na⁺ ion will be free in solution and all of the Cl^- ion will also be free in solution. This can be seen by looking at the net ionic equation for this reaction:

Mg²⁺(aq) + 2OH^-(aq) ==> Mg(OH)₂(s) ... note that the Na⁺ and Cl^- ions are spectators.

Final volume of solution after mixing = 100 mls = 0.1 L

Final [Na+] = 0.01 mols / 0.1 L = 0.1 M

Final [Cl-] = 0.01 mols / 0.1 L = 0.1 M

[Mg²⁺] = 0 (ignoring the solubility of Mg(OH)₂ since the Ksp was not provided). If you want to find this, look up the Ksp and solve for [Mg²⁺] from Ksp = [Mg²⁺][OH^-]²