Question

For the reactions shown below, we added 2.25 mL of 0.0450 M HCl to a test tube containing one of the two cations (Ag⁺ or Pb²⁺) and recovered 0.0141 g of precipitate.

AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)
Pb(NO₃)₂(aq) + 2 HCl(aq) → PbCl₂(s) + 2 HNO₃(aq)

How much precipitate in moles would be recovered theoretically if the ion was Ag⁺? (Enter an unrounded value. Use at least one more digit than given.)
mol

How much precipitate in moles would be recovered theoretically if the ion was Pb²⁺? (Enter an unrounded value. Use at least one more digit than given.)
mol

How much precipitate in grams would be recovered theoretically if the ion was Ag⁺?
g

How much precipitate in grams would be recovered theoretically if the ion was Pb²⁺?

g

Based on the precipitate amount recovered, which of the two ions was in the unknown?

Ag⁺

Pb²⁺  

Answer 1

Assume that Ag⁺ and Cl^- react completely,  Ag⁺ + Cl^- → AgCl,

recovered precipitate = 0.0141gms , If this is divided with the molar mass of Agcl (143.32,units of molar mass is gm/mol) you would obtain experimental grams of precipitate recovered when Ag+ ion is used,
0.0141/143.32 =9.83812* 10^-5 moles.

To find experimental gms of precipitate when Ag⁺ ion is used , you can perform calculations
Similarly for PbCl₂, Pb²⁺+ Cl ₂^2-→ PbCl₂ (molar mass of  PbCl₂ is 278 gm/mol)
Therefore, experimental grams of precipitate recovered when Pb²⁺ is the ion will be,
0.0141/278 = 5.071942*10^-5 moles

---Actual Solution starts from here--
Assuming that there is 1 lit of AgNO₃ , from the given info it is clear that moles of HCl will be,= 2.25*10^-3 (lit)*0.0450(mol/lit) = 1.0125*10^-4 moles, the first reaction given in the equation is balanced so,

1 mole of   AgNO₃ ----------------- 1mole of Agcl
1.0125*10^-4 moles of AgNO₃ ------------1.0125*10^-4 moles mole of AgCl (theoretical moles when Ag+ ion is used)
multiply this with molar mass of Agcl, which will give you 1.0125*10^-4 * 143.32= 0.01451115 gms. ( theoretical gms when Ag+ ion is used)

Similarly do this for Pb2+, ( reaction with 2 moles of Hcl, please observe)
1 mole of Pb(NO3)₂    --------------- 1 mole of Pbcl₂
5.0625*10^-5 moles of Pb(NO3)₂ ---------------- 5.0625*10^-5 moles  Pbcl₂ (theoretical moles when Pb²⁺ ion is used)
Therfore, 5.0625*10^-5 * 278.1= 0.0140788125 gms (theoretical moles when Pb²⁺ ion is used)

Observe the difference between theoretical and experimental gms of precipitate recovered, for Ag+ ion it is 4.11150*10^-5 gms, for Pb2+ ion it is 2.11875*10^-5 gms (theoretical is more than actual). Therefore Pb⁺² ion will be unknown