Question

Mass of Al wire before reaction = 3.96 g Mass of Al wire after reaction = 3.65 g Mass of Al lost = 0.31 g Moles of Al lost = 0.011 moles Mass of Pb + filter paper = 4.26 g Mass of filter paper = 0.92 g Mass of Pb = 3.34 g Moles of Pb formed = 0.016 moles Recall that there were 100 mL of solution. 0.016 moles of Pb were removed from the solution. What was the [Pb+2] for the saturated solution of PbCl2? ___ Recall that there was 100 mL of solution. 0.016 moles of Pb were removed from the solution. Also recall the original equilibrium: PbCl2(s) Pb+2(aq) + 2(Cl-)(aq) Since there are 2 Cl ions formed for every Pb ion, what was the [Cl-] for the saturated solution of PbCl2? ___M

a. 0.011 M
b. 0.016 M
c. 0.032 M
d. 0.064 M

Answer 1

The concentration of [Pb+2] is 0.016 M and the concentration of [Cl-] is 0.032 M.

Step-by-step explanation:

The student is asking for the concentration of [Pb+2] and [Cl-] in the saturated solution of PbCl2. From the given information, we know that 0.016 moles of Pb were removed from the solution. Since there are 2 Cl- ions formed for every Pb ion, the [Cl-] concentration is twice the [Pb+2] concentration. The [Pb+2] for the saturated solution is therefore 0.016 M and the [Cl-] is 0.032 M.