Question

Will a precipitate occur when 12.0 mL of 0.00035M silver chlorate and 15.0mL of 0.00150M sodium carbonate are mixed?

Answer 1

The chemical reaction between the aqueous solutions of silver chlorate (AgClO3) and sodium carbonate (Na2CO3) is-

2AgClO3(aq) + Na2CO3(aq) ------------> 2NaClO3(aq) + Ag2CO3(s)

Here double displacement reaction takes place where the ions of both the salts get exchanged and two new salts are formed. Now among these 2 newly formed salts, NaClO3 is water soluble, where as Ag2CO3 is insoluble. So Ag2CO3 stays as precipitate in solution.

Now the chemical equation shows that reaction of 2 moles of AgClO3(aq) and 1 mole of Na2CO3(aq) results in formation of 1 mole Ag2CO3  

Now given

12.0 mL of 0.00035M silver chlorate taken

That means mols of silver chlorate taken = concentration * volume

= 0.00035M * 12.0 mL

= 0.00035 mol/1000 mL * 12.0 mL

= 0.0000042‬ mols

15.0mL of 0.00150M sodium carbonate taken

That means mols of Na2CO3 taken = concentration * volume

= 0.00150 M * 15.0 mL

= 0.00150 mol/1000 mL * 15.0 mL

= 0.0000225‬‬ mols

Now mols of AgClO3(aq) required for complete reaction = 0.0000225‬‬ mols * 2 = 0.000045‬ mols

We have mols of AgClO3(aq) present = 0.0000042‬ mols

That means here AgClO3(aq) is the limiting reagent. So the reaction will take place as per the mols of AgClO3(aq)

i.e

0.0000042‬ mols AgClO3(aq) + 0.0000042‬ mols/2 Na2CO3(aq) ------------> 0.0000042‬ mols NaClO3(aq) + 0.0000042‬ mols /2 Ag2CO3(s)

0.0000042‬ mols AgClO3(aq) + 0.0000021‬ mols Na2CO3(aq) ------------> 0.0000042‬ mols NaClO3(aq) + 0.0000021‬ mols Ag2CO3(s)

So finally 0.0000021‬ mols Ag2CO3(s) of precipitate will be produced in the solution