If 32.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.545 g precipitate, what is the molarity of lead(II) ion in the original solution?
The precipate is Lead(II)iodide, hopefully, and you have .545grams lead iodide.
The easy way is to figure the number of moles that you have.
moles= massPbI2/molmassPbI2
Next, because the moles of Pb+2 ion is the same as the moles of PbI2 which was formed, then
Molarityoriginal= molesPb+2/.032
To find the molarity of the lead(II) ion in the original solution, we need to first calculate the number of moles of lead(II) ion present in the precipitate, and then use that information to determine the molarity.
Let's go step by step:
1. Convert the mass of the precipitate to moles:
- Determine the molar mass of lead(II) iodide (PbI2), which consists of one lead (Pb) atom with a molar mass of 207.2 g/mol and two iodine (I) atoms with a molar mass of 126.9 g/mol each.
- Add the molar masses of the individual atoms to find the molar mass of PbI2: 207.2 g/mol + 2 * 126.9 g/mol = 460.0 g/mol.
- Now, divide the mass of the precipitate (0.545 g) by the molar mass of PbI2 to get the number of moles: 0.545 g / 460.0 g/mol ≈ 0.00119 mol.
2. Calculate the number of moles of lead(II) ion in the precipitate:
- Since each mole of PbI2 contains one mole of Pb2+ ions, the number of moles of lead(II) ions in the precipitate is also approximately 0.00119 mol.
3. Determine the volume of the original lead(II) nitrate solution:
- The volume of the lead(II) nitrate solution is given as 32.0 mL.
4. Convert the volume of the lead(II) nitrate solution to liters:
- There are 1000 mL in 1 L, so divide 32.0 mL by 1000 to get the volume in liters: 32.0 mL / 1000 mL/L = 0.0320 L.
5. Calculate the molarity of the lead(II) ion in the original solution:
- Molarity (M) is defined as moles of solute per liter of solution.
- Divide the number of moles of lead(II) ions (0.00119 mol) by the volume of the solution in liters (0.0320 L): 0.00119 mol / 0.0320 L ≈ 0.0372 M.
Therefore, the molarity of the lead(II) ion in the original solution is approximately 0.0372 M.