What volume of 0.100 m sodium chloride must be added to 75.0 ml of 0.200 m lead (ii) nitrate to precipitate all of the lead ions?

Question

Chemistry

Hamlet

18 June, 04:04

What volume of 0.100 m sodium chloride must be added to 75.0 ml of 0.200 m lead (ii) nitrate to precipitate all of the lead ions?

+1

Answers (1)

Know the Answer?

Zariah Murphy · Answer 1

Reaction for precipitation is

Pb (NO3) 2 (aq) + 2Nacl (aq) →PbCL2 (s) + 2NaNO3 (aq)

The given reaction is

The mol of Nacl required is = 2*mol of Pb (NO3) 2 which is present.

∴M (Nacl) * V (Nacl) = 2*M (Pb (NO3) 2 * V (Pb (NO3) 2

o. 100M*V (Nacl) = 2*0.200 MX75.0ML

V (Nacl = 300mL.