1) The reaction will be
Here, PbS will precipitate out.
2)
Total Volume = 6.80 mL + 3.20 mL = 10.00 mL
milli moles of Pb(NO3)2 = Volume * Molarity = 6.80 mL * 1.30 M = 8.84 mmmol
milli moles of (NH4)2S = Volume * Molarity = 3.20 mL * 5.0 M = 16.0 mmmol
Concentration of NO3-
1 mmol of Pb(NO3)2 will give 2 mmol of NO3- which will remain soluble.
Thus, mmol of NO3- = 2 * 8.84 mmol = 17.68 mmol
Concentration of NO3- = Moles/Volume = 17.68 mmol / 10.00 mL = 1.768 M
Concentration of NH4+
1 mmol of (NH4)2Swill give 2 mmol of NH4+ which will remain soluble.
Thus, mmol of NH4+ = 2 * 16.00 mmol = 32.00 mmol
Concentration of NH4+ = Moles/Volume = 32.00 mmol/10.00 mL = 3.200 M
Concentration of Pb2+
1 mmol of Pb(NO3)2 will give 1 mmol of Pb2+
Thus, mmol of Pb2+= 8.84 mmol
But all of it will react with 8.84 mmol S2- to form precipitate and none of it will remain.
Concentration of Pb2+ =0 M
Concentration of S2-
1 mmol of (NH4)2Swill give 1 mmol of S2- which will remain soluble.
Thus, mmol of S2- = 16.00 mmol
But 8.84 mmol will react with Pb2+.
Moles remaining = 16.00 mmol - 8.84 mmol = 7.16 mmol
Concentration of S2- = Moles/Volume = 7.16 mmol/10.00 mL = 0.716 M
Total concentration of all soluble ions = 1.768 M + 3.200 M + 0.716 M = 5.684 M = 5.68 M (3 sig figure)
c)
Theoretical moles of PbS produced = 8.84 mmol = 8.84 * 10-3 mol
Molar Mass of PbS = 239.3 g/mol
Theoretical Mass = (8.84 * 10-3 mol) * (239.3 g/mol) = 2.115 g
Actual Mass = 2.01 g
% Yield = Actual Mass / Theoretical Mass * 100 = 2.01 g / 2.115 g * 100= 95.03%
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