Calculate the concentration (in M) of ammonium ions when 10.3 mL of 0.803 M ammonium carbonate and 12.7 mL of 0.984 M ammonium chloride are combined.

Answer 1

Ammonium carbonate = (NH₄)₂CO₃

Ammonium chloride = NH₄Cl

We will find the moles of NH₄⁺ ions from each source, and then add them together to get the TOTAL moles of NH₄⁺ ions. We will then divide that by the total volume (in liters) to get the final molar concentration of NH₄⁺.

Moles NH₄⁺ from (NH₄)₂CO₃:

Note: each mole of (NH₄)₂CO3 produces 2 moles NH₄⁺ since (NH₄)₂CO₃ ==> 2NH₄⁺ + CO₃^2-

10.3 ml (NH₄)₂CO₃ x 1 L/1000 ml x 0.803 mol/L x 2 mol NH₄⁺/mol (NH₄)₂CO₃ = 0.01654 moles NH₄⁺

Moles NH₄⁺ from NH₄Cl:

Note: each mole of NH₄Cl produces 1 mole NH₄⁺ since NH₄Cl ==> NH₄⁺ + Cl^-

12.7 ml NH₄Cl x 1 L/1000 ml x 0.984 mol/L x 1 mol NH₄⁺/mol NH₄Cl = 0.01250 moles NH₄⁺

Total moles of NH₄⁺ = 0.01654 + 0.01250 = 0.02904 moles

Total volume = 10.3 ml + 12.7 ml = 23 ml = 0.023 L

Concentration of NH₄⁺ ions = 0.02904 moles NH₄⁺ / 0.023 L = 1.26 M NH₄⁺