45.5 mL of a 1.25 M solution of ammonium chloride and 23.8 mL of a 1.55 M solution of lead(II) nitrate are added together. Does a reaction occur?

Question

If a reaction occurred, what is the molarity of each free ion when the reaction is over? Summarize values on the table below. Clearly show all work used to determine each amount.

Ion

Initial # of moles

Moles used up in the reaction

Moles left unused

Molarity

Answer 1

Write a balanced equation:

2NH₄Cl(aq) + Pb(NO₃)₂(aq) ==> 2NH₄NO₃(aq) + PbCl₂(s)

A reaction does occur since a precipitate of PbCl₂ is formed.

moles NH₄Cl initially present = 45.5 ml x 1 L/1000 ml x 1.25 mol/L = 0.0569 mols NH₄Cl

moles Pb(NO₃)₂ initially present = 23.8 ml x 1 L/1000 ml x 1.55 mol/L = 0.0369 mols Pb(NO₃)₂

LIMITING REACTANT = NH₄Cl since it takes twice as many moles as it does Pb(NO₃)₂ but don't have that.

Net ionic equation: Pb²⁺(aq) + 2Cl^-(aq) ==> PbCl₂(s)

Since NH₄Cl is limiting ALL OF THE Cl^- will precipitate as PbCl₂ so there will be NO FREE Cl^-

Spectator ions (NH₄⁺ and NO₃^- will not take part in the reaction so they will be present as follows:

NH₄⁺ = 0.0569 mols NH₄⁺

NO₃^- = 2 x 0.0369 mols = 0.0738 mols NO₃^-

To find mols Pb²⁺ remaining in solution, we find mols used up as follows:

0.0569 mols NH₄Cl x 1 mol Pb(NO₃)₂ / 2 mols NH₄Cl = 0.0285 mols Pb(NO₃)₂ used up in the reaction

Mols Pb²⁺ left = 0.0369 mols - 0.0285 mols = 0.0084 mols Pb²⁺ left

SUMMARY TABLE

ION...............Initial # mols.................moles used up....................moles left unused..............Molarity*

NH₄⁺.............0.0569...............................0.........................................0.0569.........................0.8211 M

NO₃^-..............0.0738...............................0.........................................0.0738.........................1.065 M

Pb²⁺..............0.0369...............................0.0285.................................0.0084.........................0.1212 M

Cl^-..................0.0569..............................0.0569......................................0...............................0 M

*Molarity is calculated as moles of ion / liter of solution. Since there are 45.5 ml + 23.8 ml = 69.3 ml, the moles of ions were divided by 0.0693 liters to obtain moles/liter (M)