Question

Predict the products of the following reactions and then balance the equation. Make sure to put in the phases of the products. Pb(NO3)2 (aq) + Na3PO4 (aq) →

A. For the above equation, write the complete ionic and net ionic equations.

B. If you have a 2.5 M solution of lead (II) nitrate, and a 3.5 M solution of sodium phosphate, how much of the two solutions will you need to make 11.5 g of the product that precipitates?

C. To make the two solutions how many grams of each will you need?

D. What are the spectator ions and will be the final molarity of the compound made up of the spectator ions?

Answer 1

The balanced reaction becomes

3Pb(NO₃)₂ (aq)+2 Na₃PO₄ (aq)>> Pb₃(PO₄)₂ (s) + 6 NaNO₃ (aq)

The ionic equation becomes

3 Pb²⁺ + 6 NO^3- + 6 Na⁺ + 2 PO4^3- >> Pb3(PO4)2 + 6 Na+ + 6 NO^3-

Net ionic equation becomes

Pb2+ + 2 PO43- >> Pb3(PO4)2

3Pb(NO3)₂ (aq)+2 Na₃PO₄ (aq)>> Pb₃(PO₄)₂ (s) + 6 NaNO₃ (aq)

Product required = 11.5 gms

Product formed that gets precipitated =Pb₃(PO4)₂

Molecular weight of Pb₃(PO4)₂ = 811.5

Moles of Pb₃(PO4)₂= 11.5/811.5=0.014171

As per the balanced reaction

1 mole of Pb₃(PO4)₂ requires 3 moles of Pb(NO3)2 and 2 moles of Na3PO4.

0.014171moels of Pb3(PO4)2 requires 0.014171*3 =0.042514 moles of Pb(NO₃)₂ and 2*0.014171=0.028343 moles of Na₃PO₄.

Molecular weight of Pb(NO3)2 = 331 and that of Na3PO4= 259

Mass of Pb(NO3)2= 331*0.042514 =14.07 gm and mass of Na3PO4= 0.028343*259 gm=7.34 gm

Volume of 2.5 M solution of lead nitrate

Moles/ litres = 2.5

0.042514 moles/ lires = 2.5

Volume of lead nitrate= 0.042514/2.5 L=0.017006 L =17 ml

Volume of 3.5 M solution of sodium phosphate,

Moles/Litres=3.5

0.028343/Litres= 3.5

Volume = 0.028343/ 3.5 =0.008098 L =8,1 ml

d) NO^3- Na⁺ are spectator ions and the final molarity of the compound is not made of these ions.