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 14.0 mL of 6.30×10-3 M Pb(NO3)2 upon addition of 13.1 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

Answer:

All the 8.82*10^-5 moles of lead present is removed.

Explanation:

The total amount of lead II ion present is less than the amount of sulphide ion present and they react in 1:1 ratio so all the lead II ions are removed from the solution. See attached image.