Essay About Mass Of Pbcl2 And Pbcl2

Solubility Product of Lead Chloride
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Title : Solubility product of Lead Chloride
Aim : to determine the solubility and the solubility product of lead chloride in pure water
Method/ material : refer to the handout
Result
Copy the table below and supply the necessary experimental data with proper calculations
Mass of weighing tray + PbCl2
4.47g
4.66g
Mass of weighing tray after transfer of PbCl2
1.472g
1.66g
Mass of PbCl2
Mass of sintered crucible
30.45g
31.11g
Mass of sintered crucible +residue
31.07g
Mass of residue
0.62g
0.89g
Mass of PbCl2 dissolved
2.38g
2.11g
Amount of PbCl2 dissolved
8.56 x (10^-3) mol
7.59 x (10^-3) mol
Concentration of PbCl2 dissolved
0.0428M
0.038M
Mass of PbCl2 : (Mass of weighing tray+PbCl2) – (Mass of weighing tray after transfer of PbCl2)
A : 4.47g – 1.472g = 3 g
B : 4.66g – 1.66g = 3 g
Mass of residue : (Mass of sintered crucible +residue) – (Mass of sintered crucible)
A : 31.07g – 30.45g = 0.62g
B : 32g – 31.11g = 0.89g
Mass of PbCl2 dissolved : (Mass of PbCl2) – (Mass of residue)
A : 3g – 0.62g = 2.38 g
B : 3g – 0.89g = 2.11g
Molar weight of PbCl2
Pb : 207.2 g/mol
Cl : 35.45 g/mol
207.2 g/mol + (2 x 35.45 g/mol) = 278.1 g/mol
Amount of PbCl2 dissolved : Mass of PbCl2 dissolved / Molar weight of PbCl2
A : 2.38g / (278.1g/mol) = 8.56 x (10^-3) mol
B : 2.11g / (278.1g/mol) = 7.59 x (10^-3) mol
Concentration of PbCl2 dissolved ( in 200ml of distilled water)
A : 8.56 x (10^-3) mol / 200ml x 1L/1000ml = 0.0428 mol/L = 0.0428 M
B : 7.59 x (10^-3) mol / 200ml x 1L/1000ml = 0.03795 mol/L = 0.038M
Calculate the Ksp for PbCl2
PbCl2 <-> Pb(2+) + 2 Cl(-)
1 : 1 : 2
Ksp = [Pb(2+)] x [Cl(-)]^2
Ksp of A : [Pb(2+)] = 0.0428
[Cl(-)] = 0.0428 x 2 = 0.0856 M
= [0.0428] x [0.0856]^2
Ksp of A = 3.14 x (10^-4) mol(3) /dm(9)
Ksp of B : [Pb(2+)] = 0.038M
[Cl(-)] = 0.038M x 2 = 0.076 M
= [0.038] x [0.076]^2
Ksp of B = 2.2 x (10^-4) mol(3) / dm(9)
Any change (procedure 6)
When we allow the lead chloride to cool to room temperature,