A solution contains 1.50 g of dissolved Pb
2+ ions. How many grams of NaI must be added to the solution to completely precipitate all of the dissolved Pb
2+ as PbI2?
Pb^2+ + 2I^- ==> PbI2.
mols Pb^2+ = 1.50g/atomic mass Pb
mols I^- = twice that.
g NaI = mols I^- x molar mass NaI
To find out how many grams of NaI must be added to completely precipitate all of the dissolved Pb²⁺ as PbI₂, we need to use stoichiometry and the concept of mole ratios.
First, let's write down the balanced chemical equation for the reaction between Pb²⁺ ions and NaI to form PbI₂:
Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
From the equation, we can see that one mole of Pb²⁺ reacts with two moles of I⁻ to form one mole of PbI₂.
Now, let's calculate the number of moles of Pb²⁺ in the solution. We're given that the solution contains 1.50 g of dissolved Pb²⁺ ions, but we need to convert grams to moles. To do this, we divide the mass by the molar mass of Pb²⁺, which is 207.2 g/mol:
Number of moles of Pb²⁺ = 1.50 g / 207.2 g/mol = 0.00724 mol
Since the stoichiometric ratio between Pb²⁺ and NaI is 1:2, we need twice the moles of I⁻ ions compared to the moles of Pb²⁺ ions.
Number of moles of I⁻ = 2 * Number of moles of Pb²⁺ = 2 * 0.00724 mol = 0.0145 mol
Now, we need to convert the moles of I⁻ to grams of NaI. To do this, we multiply the moles of I⁻ by the molar mass of NaI, which is 149.9 g/mol:
Mass of NaI = 0.0145 mol * 149.9 g/mol = 2.17 g
Therefore, you need to add approximately 2.17 grams of NaI to the solution to completely precipitate all of the dissolved Pb²⁺ as PbI₂.