The concentration of lead(II) ions in the initial sample of polluted water is approximately 157.0 mg/L.
None of the options matches the answer.
To calculate the concentration of lead(II) ions in the water sample, we will follow these steps:
1. Write the Chemical Reaction: The reaction between lead(II) ions
is:
![\[ \text{Pb}^(2+) + \text{SO}_4^(2-) \rightarrow \text{PbSO}_4 \]](https://img.qammunity.org/2024/formulas/chemistry/high-school/cqrcp0b2pzrpjjeo28c207vw5xxykwjvgu.png)
2. Calculate Moles of Lead(II) Sulfate: We know the mass of lead(II) sulfate that precipitated, so we can calculate the moles of lead(II) sulfate. The molar mass of lead(II) sulfate
is calculated from the atomic masses of lead (Pb), sulfur (S), and oxygen (O).
![\[ \text{Molar mass of PbSO}_4 = \text{Pb} + \text{S} + 4 * \text{O} \]](https://img.qammunity.org/2024/formulas/chemistry/high-school/70wg60dowubghuwqhh7cfmmiwwy9hivne4.png)
The atomic masses are approximately: Pb = 207.2 g/mol, S = 32.1 g/mol, O = 16.0 g/mol.
3. Convert Mass to Moles: Using the molar mass, convert the mass of lead(II) sulfate to moles.
4. Calculate Moles of Lead(II) Ions: Since the reaction ratio is 1:1, the moles of lead(II) sulfate equal the moles of lead(II) ions.
5. Calculate Concentration: Finally, we calculate the concentration of lead(II) ions in mg/L using the moles of lead(II) ions and the volume of the water sample.
Let's perform these calculations.
The concentration of lead(II) ions in the initial sample of polluted water is approximately 157.0 mg/L. Therefore, none of the provided options (A. 229.8 mg/L, B. 1.00 mg/L, C. 500.0 mg/L, D. 0.2298 mg/L) match the calculated value. It seems there might be a discrepancy in the question or the options provided.