Question

Will a precipitate form when 50.0 mL of 1.2 x 10^-3 M Pb(NO3)2 are added to 50.0ml of 2.0x 10^-4 M Na2S

if so identify the precipitate

Answer 1

The precipitate that will be formed is PbS

Step-by-step explanation:

Step 1: Data given

Volume of a 1.2 * 10^-3 M Pb(NO3)2 = 50.0 mL = 0.05 L

Volume of a 2.0 * 10^-4 M Na2S = 50.0 mL = 0.05 L

Step 2: Pb(NO3)2 + Na2S → 2NaNO3 + PbS

NaNO3 will dissociate in 2Na+ and 2NO3-

PbS might precipitate

Step 3: Calculate moles Pb^2+

Moles = volume * molarity

Moles Pb^2+ = 0.050 * 1.2 * 10^-3 = 6 * 10^-5 moles

Step 4: Calculate moles S^2-

Moles S2- = 0.050 L * 2.0 * 10^-4 = 1 * 10^-5

Step 5: Calculate total volume

Total volume = 50 mL + 50 mL = 100 mL = 0.100 L

Step 6: Calculate molarity

Molarity = number of moles / volume

[Pb2+] = 6*10^-5 moles / 0.100 L =6 * 10^-4 M

[S2-] = 1 * 10^-5 moles / 0.100L = 1 * 10^-4 M

Qsp = [Pb2+] [S2-]

Qsp = ( 6*10^-4 ) ( 1 * 10^-4 ) = 6*10^-8

Ksp of PbS = 3* 10^-28

Qsp >>>> Ksp this means a precipitate (PbS) will be formed