2012 hsc q20MC (1 Viewer)

[aanthnnyyy]

All the lead ions present in a 50ml solution were precipitated by reaction with excess chloride ions. Mass of dried precipitate was 0.595g. What was concentration of lead in original solution : answ- 8.87g/L

I think I'm overthinking a basic question any ideas !!!!!!

 

[Librah]

Answer is correct. 0.595(207.2/207.2+2(35.45)) to get mass of lead ions. Then divide by 0.05.

 

[photastic]

Pb2+ + 2Cl- ⟶ PbCl2(s)
n(PbCl2) = m/M = 0.595/278.1 = 0.00214 mol
n(Pb2+) = 0.00214 mol
m(Pb2+) = 0.00214 x 207.2 = 0.4434 g
c(Pb2+) = 0.4434/0.0500 = 8.87 g/L

 

[aanthnnyyy]

Zzzz thanks guys I knew it was pb2+ idk why I kept trying 4+