Question

The balanced equation for the precipitation reaction between lead ion and potassium chloride is shown below. Pb₂+(aq)+2KCl(aq)⟶PbCl₂(s)+2K+(aq) Lead(II) ion is highly poisonous. To determine the amount of lead ion, a water sample was treated with excess of potassium chloride. A precipitate of lead(II) chloride was formed, weighing 80.1μg. Determine the mass of lead (in micrograms) in the water sample.

Answer 1

To determine the mass of lead in the water sample, we can use stoichiometry. The balanced equation is Pb₂+(aq) + 2KCl(aq) ⟶ PbCl₂(s) + 2K⁺(aq). Assuming the entire precipitate is made of PbCl₂, the mass of lead in the water sample is 62.8 μg.

Step-by-step explanation:

The balanced equation for the precipitation reaction between lead ion and potassium chloride is:

Pb₂+(aq) + 2KCl(aq) ⟶ PbCl₂(s) + 2K⁺(aq)

To determine the mass of lead in the water sample, we can use stoichiometry. First, we find the molar mass of PbCl₂, which is 278 g/mol. Then, we convert the mass of the precipitate to moles by dividing it by the molar mass. Finally, we use the ratio between PbCl₂ and Pb²+ to find the mass of lead in the water sample. Assuming the entire precipitate is made of PbCl₂, we can calculate the mass of lead to be 62.8 μg.