Final answer:
Before any reaction occurs, the initial concentration of lead (II) ions (Pb²⁺) is 0.600 M, sodium ions (Na⁺) is 1.50 M, chloride ions (Cl⁻) is 1.50 M, and nitrate ions (NO3⁻) is 1.20 M, immediately after mixing the two solutions.
Step-by-step explanation:
Before the reaction between lead (II) nitrate (Pb(NO3)2) and sodium chloride (NaCl) to form the precipitate lead (II) chloride (PbCl2), the initial concentration of all ions right after the solutions are mixed can be calculated. To find the concentration of each ion in the combined volume, we can follow this procedure:
Calculate the number of moles of each solute in their respective solutions before mixing. The number of moles is given by the product of molarity and volume (in liters).
Sum the total volume of the solutions to get the combined volume.
Divide the number of moles of each solute by the total volume to find their concentration in the mixture.
For Pb(NO3)2, we have:
Number of moles = 1.50 M × 0.200 L
= 0.300 moles
The total volume after mixing = 0.200 L + 0.300 L
= 0.500 L
Initial concentration of Pb²⁺ = 0.300 moles / 0.500 L
= 0.600 M
For NaCl, we have:
Number of moles = 2.50 M × 0.300 L
= 0.750 moles
Initial concentration of Na⁺ = 0.750 moles / 0.500 L
= 1.50 M
Initial concentration of Cl⁻ = 0.750 moles / 0.500 L =
1.50 M
For Nitrate ions, considering that Pb(NO3)2 dissociates into one Pb²⁺ and two NO3⁻ ions:
Initial concentration of NO3⁻ = (0.300 moles × 2) / 0.500 L = 1.20 M