A 25.0 mL sample of drinking water suspected to contain lead(II) ions is treated with excess 0.800 M sodium chloride. A precipitate forms, which is filtered, dried, and weighed. 1.36 g of precipitate are collected. What was the concentration of lead in the drinking water sample? (Assume that the lead(II) ions were in the form of lead(II) nitrate in the water.)

Convert 1.36g PbCl2 to Pb.

1.36 x (atomic mass Pb/molar mass PbCl2) = ? g Pb in 25 mL. Convert that to what ever concentration units you want. The problem doesn't specify. Note: I don't think it makes any difference what form the Pb is as long as it's soluble.

To determine the concentration of lead in the drinking water sample, we can use the concept of stoichiometry and the amount of precipitate formed.

Here's how you can approach this problem:

1. Write the balanced chemical equation for the reaction between lead(II) nitrate and sodium chloride. The balanced equation is:
Pb(NO3)2 + 2NaCl -> PbCl2 + 2NaNO3

2. Determine the molar mass of lead(II) chloride (PbCl2). The molar mass of Pb is 207.2 g/mol, and the molar mass of Cl is 35.45 g/mol. Adding these together gives a molar mass of 207.2 + 2(35.45) = 278.1 g/mol for PbCl2.

3. Calculate the number of moles of lead(II) chloride (PbCl2) formed. To do this, divide the mass of the precipitate (1.36 g) by the molar mass of PbCl2 (278.1 g/mol). So, the number of moles of PbCl2 is 1.36 g / 278.1 g/mol = 0.00489 mol.

4. Based on the balanced equation, we know that 1 mole of PbCl2 is formed from 1 mole of Pb(NO3)2. Therefore, the number of moles of lead(II) nitrate (Pb(NO3)2) in the water sample is also 0.00489 mol.

5. Now, we need to calculate the volume of the drinking water sample that contains 0.00489 mol of Pb(NO3)2. From the given information, we know that the volume of the sample is 25.0 mL.

6. Convert the volume of the sample from milliliters to liters by dividing by 1000. So, the volume of the sample is 25.0 mL / 1000 = 0.025 L.

7. Finally, calculate the concentration of lead(II) ions in the drinking water sample by dividing the number of moles of Pb(NO3)2 (0.00489 mol) by the volume of the sample (0.025 L). So, the concentration of lead in the drinking water is 0.00489 mol / 0.025 L = 0.1956 M.

Therefore, the concentration of lead in the drinking water sample is 0.1956 M.