Question

B. Solubility Equilibrium: Finding a Value for Ksp
1. Volume 0.30 M Pb(NO3)2 = 5.0 mL.  
Number of moles Pb2+ = [M x V(stock sln)] = 0.0015 moles
2.
a. Volume 0.30 M HCl used: 5.0 mL
b. Number of moles Cl- added: 0.0015 moles
3. Observations:
a. in hot water: the solid settled on the bottom then completely dissolved
b. in cold water: No change, it stayed dissolved
4. Volume H2O added to dissolve PbCl2: 4.0 mL
a. Total volume of solution: 12.0 mL
b. Explain why PbCl2 did not precipitate immediately on addition of HCl.
c. Explain your observations in step 3 (hot versus cold water).
d. Explain why the PbCl2 dissolved when water was added in Step 5.
e. Given the number of moles of Pb2+ and Cl- in the final solution in Step 4, and the volume of that solution, calculate [Pb2+] and [Cl-] in that solution.

Can anyone help with 4b-e please. My answers from lab are in bold for the 1st part(1-4a).

Answer 1

4.

(a). Total volume of solution = 5.0 + 5.0 + 4.0

= 14.0 mL

(b). PbCl2 does not precipitate immediately because initially the Cl- ion concentration was low. For a precipitate to occur, Qsp = [Pb2+][Cl-]² should be greater than Ksp value of PbCl2.

(d). PbCl2(s)    Pb2+(aq)     +    2Cl-(aq)

On adding water, the concentration of Pb2+ and Cl- ions on the right hand side will decrease, so, according to Le Chatelier's Principle, more PbCl2 will dissolve in order to reduce the effect and maintain the equilibrium.

(e). Moles of Pb2+ = 0.0015

Moles of Cl- = 0.0015

Volume of solution = 14 mL = 0.014 L

[Pb2+] = 0.0015 / 0.014

= 0.107 M

[Cl-] = 0.0015 / 0.014

= 0.107 M