what mass of Pb^2+ ion is left in solution when 50 ml of 0.2M PbCl2 is added to 50ml of 1.5 NaCl?[Ksp for PbCl2= 1.7* 10^(-4)]
To find the mass of Pb^2+ ion left in solution, we need to calculate the concentration of Pb^2+ ion in the final solution first. Here's how you can do it:
Step 1: Calculate the moles of PbCl2 added:
Moles of PbCl2 = Volume of PbCl2 solution (in L) × Molarity of PbCl2
Given:
Volume of PbCl2 solution = 50 mL = 50/1000 L = 0.05 L
Molarity of PbCl2 = 0.2 M
Moles of PbCl2 = 0.05 L × 0.2 mol/L = 0.01 mol
Step 2: Calculate the moles of NaCl in the solution:
Moles of NaCl = Volume of NaCl solution (in L) × Molarity of NaCl
Given:
Volume of NaCl solution = 50 mL = 50/1000 L = 0.05 L
Molarity of NaCl = 1.5 M
Moles of NaCl = 0.05 L × 1.5 mol/L = 0.075 mol
Step 3: Calculate the moles of chloride ion (Cl-) in the solution:
Since PbCl2 dissociates into one Pb^2+ ion and two Cl- ions, the moles of chloride ions are twice the moles of PbCl2.
Moles of chloride ion = 2 × Moles of PbCl2 = 2 × 0.01 mol = 0.02 mol
Step 4: Calculate the concentration of chloride ion (Cl-) in the final solution:
Total volume of the final solution = Volume of PbCl2 solution + Volume of NaCl solution = 0.05 L + 0.05 L = 0.1 L
Concentration of chloride ion = Moles of chloride ion / Total volume of the solution
Concentration of chloride ion = 0.02 mol / 0.1 L = 0.2 M
Step 5: Use the Ksp expression to find the concentration of Pb^2+ ion in the final solution:
Ksp = [Pb^2+][Cl-]^2
Given:
Ksp for PbCl2 = 1.7 × 10^(-4)
Concentration of chloride ion (Cl-) = 0.2 M (from Step 4)
[Pb^2+] = Ksp / [Cl-]^2
[Pb^2+] = (1.7 × 10^(-4)) / (0.2)^2
[Pb^2+] = (1.7 × 10^(-4)) / 0.04
[Pb^2+] ≈ 0.00425 M
Step 6: Calculate the mass of Pb^2+ ion in the final solution:
Mass of Pb^2+ ion = Moles of Pb^2+ ion × Molar mass of Pb^2+
Molar mass of Pb^2+ = Molar mass of Pb = 207.2 g/mol
Moles of Pb^2+ ion = Concentration of Pb^2+ ion × Volume of the solution (in L)
Moles of Pb^2+ ion = 0.00425 M × 0.1 L = 0.000425 mol
Mass of Pb^2+ ion = 0.000425 mol × 207.2 g/mol ≈ 0.088 g
Therefore, approximately 0.088 grams of Pb^2+ ion is left in the solution.
To find the mass of Pb^2+ ion left in solution, we first need to calculate the moles of Pb^2+ ions present in the solution.
Step 1: Calculate the moles of Pb^2+ ions in PbCl2 solution:
Moles of Pb^2+ ions = concentration (M) x volume (L)
Moles of Pb^2+ ions = 0.2 M x 0.050 L (concentration in PbCl2 solution)
Step 2: Calculate the moles of Cl- ions in NaCl solution:
Moles of Cl- ions = concentration (M) x volume (L)
Moles of Cl- ions = 1.5 M x 0.050 L (concentration in NaCl solution)
Step 3: Since the stoichiometry of the reaction is 1:2 (Pb^2+:Cl-), the number of moles of Pb^2+ ions that will react with Cl- ions is twice the number of moles of Cl- ions.
Moles of Pb^2+ ions that react = 2 x Moles of Cl- ions = 2 x (1.5 M x 0.050 L)
Step 4: The moles of Pb^2+ ions remaining in solution is obtained by subtracting the moles of Pb^2+ ions that reacted from the initial moles of Pb^2+ ions.
Moles of Pb^2+ ions remaining = Moles of Pb^2+ ions - Moles of Pb^2+ ions that react
Substituting the values:
Moles of Pb^2+ ions remaining = [0.2 M x 0.050 L] - [2 x (1.5 M x 0.050 L)]
Step 5: Calculate the mass of Pb^2+ ions remaining in solution using the molar mass of Pb^2+ (207.2 g/mol).
Mass of Pb^2+ ions remaining = Moles of Pb^2+ ions remaining x Molar mass of Pb^2+
Substituting the values:
Mass of Pb^2+ ions remaining = [0.2 M x 0.050 L - 2 x (1.5 M x 0.050 L)] x 207.2 g/mol
Now you can perform the calculations to find the mass of Pb^2+ ions remaining in solution.
50 mL x 0.2M = 10 millimols Pb^2+
50 mL x 1.5M = 75 mmols Cl^-
........Pb^2+ + 2Cl^- ==> PbCl2
I.......10......75.........0
C......-10.....-20........10
what's left.0....55.......10
So we have left a PbCl2 solid in a solution with an excess of 55 mmoles Cl^- in a volume of 100 mL.
(Cl^-) = 55/100 = 0.55M
Ksp = (Pb^2+)(Cl^-)^2
Ksp = 1.7E-4 = (Pb^2+)(0.55)^2
Solve for (Pb^2+).