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 13.4 mL of 7.50×10-3 M Pb(NO3)2 upon addition of 13.9 mL of 0.0121 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

Ans: - Given that The lead be removed boom 134 ml de 7.5 0x10 31 Pb(NO3)2 upon adding of 13.9mL of 0.0121 M Nags The balanced equation is Nags + Pb(NO3), - pbs+ 2Nanoz I mole cts of with mole to give of PbCNO3)2 сть I mole of pbs moles a Pb(NO3)2 requires I mole at Nazs moles at pb (NO)2 = 0.0075 x 13.4 1000 0.00019 05 - moles do Nags = 0.0121x 13.9 1000 0.00016819 Excess is was and hence all the PbCNO3)2 is removed in the bosm of pbs moles at pbs bormed The moles of pbs formed = 0.00016819