Question

Heavy metal ions like lead(II) can be precipitated from laboratory wastewater by adding sodium sulfide, Na2S. Will all the lead be removed from 13.6 mL of 7.00×10-3 M Pb(NO3)2 upon addition of 13.3 mL of 0.0123 M Na2S? If all the lead is removed, how many moles of lead is this? If not, how many moles of Pb remain?

Answer 1

Given Volume of Pb(NO₃)₂ = 13.6 mL

Molarity of  Pb(NO₃)₂ = 7.00* 10^-3 M = 7.00* 10^-3 mol/L

Moles of Pb(NO₃)₂ = Molarity * VOlume = 7.00* 10^-3 mol/L * 13.6 mL = 9.52 * 10^-5 moles

Moles of Na₂S = 13.3 mL *  0.0123 M = 1.64 * 10^-4 moles

Reaction is :

Pb(NO₃)₂ (aq) + Na₂S (aq) 2 NaNO₃ (aq) + PbS (s)

From reaction,

1 mol of Na₂S will react with 1 mol of Pb(NO₃)₂

9.52 * 10^-5 moles of Na₂S will react with 9.52 * 10^-5 mol of Pb(NO₃)₂

But we have  1.64 * 10^-4 moles of  Na₂S  

So, all the Pb(NO₃)₂ will precipitate.

Thus, moles of Pb removed = 9.52 * 10^-5 mol