Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it.

Question

Suppose 

6.29g

 of nickel(II) chloride is dissolved in 50.mL

 of a 0.70 M

 aqueous solution of potassium carbonate.Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it.

Be sure your answer has the correct number of significant digits. 

Answer 1

NiCl₂(aq) + K₂CO₃(aq) ==> NiCO₃(s) + 2KCl(aq) ... balanced equation

moles NiCl₂ present = 6.29 g x 1 mol / 129.6 g = 0.0485 mols NiCl₂

moles K₂CO₃ present = 50. ml x 1 L / 1000 ml x 0.70 mol/L = 0.0350 moles K₂CO₃

Limiting reactant = K₂CO₃ since the mol ratio is 1:1 and there are fewer mols K₂CO₃

This means NiCl₂ is in excess.

Moles NiCl₂ used = 0.0350 mols K₂CO₃ x 1 mol NiCl₂ / mol K₂CO₃ = 0.0350 mols NiCl₂ used

Moles NiCl₂ left over = 0.0485 - 0,0350 = 0.0135 mols NiCl₂ left over

Moles Cl^- left over = 0.0135 mol NiCl₂ x 2 mol Cl^- / mol NiCl₂ = 0.0270 moles Cl^-

Final molarity of Cl^- = 0.0270 mol Cl^- / 50. ml x 1000 ml / L = 0.54 mol Cl^-/L = 0.54 M Cl^- (2 sig. figs.)