Lead Nitrate VS

Pb(NO₃)₂ = 331.2

For a 0.05m solution Dissolve 16.5 g of lead(ii) nitrate in sufficient water to produce 1000 mL.

Ascertain its exact concentration in the following manner. To 50 mL of the solution add 300 mL of water and carry out the method for the complexometric titration of lead, Appendix VIII D. Each mL of 0.1m disodium edetate VS is equivalent to 33.12 mg of Pb(NO₃)₂.

For a 0.1m solution Dissolve 33.0 g of lead(ii) nitrate in sufficient water to produce 1000 mL.

Ascertain its exact concentration in the following manner. To 20 mL of the solution add 300 mL of water and carry out the method for the complexometric titration of lead, Appendix VIII D. Each mL of disodium edetate VS is equivalent to 33.12 mg of Pb(NO₃)₂.