The concentration of Pb2+ in a sample of wastewater is to be determined by using
gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of
sodium carbonate, forming the insoluble lead (II) carbonate (267.2092 g/mol) according to
the balanced equation given below. The solid lead (II) carbonate is dried, and its mass is
measured to be 0.4078 g. What was the concentration of Pb2+ in the original wastewater
sample?
See your other posts above.
0.01526 M
To determine the concentration of Pb2+ in the wastewater sample using gravimetric analysis, we need to follow these steps:
1. Write the balanced chemical equation:
Pb2+ (aq) + CO32- (aq) -> PbCO3 (s)
2. Calculate the number of moles of PbCO3 using its molar mass:
mol PbCO3 = mass PbCO3 / molar mass PbCO3
mol PbCO3 = 0.4078 g / 267.2092 g/mol
mol PbCO3 = 0.0015274 mol
3. Determine the stoichiometry ratio between Pb2+ and PbCO3 in the balanced equation.
From the balanced equation, we can see that one mole of Pb2+ corresponds to one mole of PbCO3. Therefore, the number of moles of Pb2+ is also 0.0015274 mol.
4. Calculate the volume of the original wastewater sample:
We know that the volume of the sample is 100.0 mL or 0.1000 L.
5. Calculate the concentration of Pb2+ in the original wastewater sample:
Concentration (C) = moles / volume
C = 0.0015274 mol / 0.1000 L
C = 0.01527 mol/L
The concentration of Pb2+ in the original wastewater sample is 0.01527 mol/L.