Question

If 12.8 g lead(II) sulfate (303.3 g/mol) precipitates when excess potassium chloride is added to 1.65 L of a water sample, what is molar concentration of Pb2+ in the sample?

Answer 1

The molar concentration of Pb2+ in the water sample is 0.0256 M.

Step-by-step explanation:

To find the molar concentration of Pb2+ in the water sample, we first need to determine the amount of lead(II) sulfate that precipitates when excess potassium chloride is added. Given that the molar mass of lead(II) sulfate is 303.3 g/mol and 12.8 g of it precipitates, we can calculate the number of moles using the formula:

moles = mass / molar mass

moles = 12.8 g / 303.3 g/mol

moles = 0.0422 mol

Since the lead(II) sulfate dissolves into Pb2+ and SO4²-, and all of the lead(II) sulfate precipitates, the molar concentration of Pb2+ in the sample is equal to the calculated moles divided by the volume of the sample, which is 1.65 L:

Molar concentration of Pb2+ = moles / volume

Molar concentration of Pb2+ = 0.0422 mol / 1.65 L

Molar concentration of Pb2+ = 0.0256 M

Answer 2

`M=0.0256M`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`PbSO_4(aq)+2KCl(aq)\rightarrow PbCl_2(s)+K_2SO_4(aq)`

In such a way, since all the lead (II) is converted due to the excess of potassium chloride, the moles of lead (II) in the sample are computed from the mass of lead (II) sulfate:

`n_(Pb^(2+))=12.8gPbSO_4*(1molPbSO_4)/(303.3gPbSO_4) *(1molPb^(2+))/(1molPbSO_4) \\\\n_(Pb^(2+))=0.0422molPb^(2+)`

Thus, since volume of the solution is 1.65 due to the fact that the addition of the reactants is not enough to significantly modify the reaction's volume, the resulting molar concentration of the lead (II) ions is:

`M=(n_(Pb^(2+)))/(V)=(0.0422molPb^(2+))/(1.65L)\\ \\M=0.0256M`

Regards.