A beaker contain 300.ml of a 0,20 M Pb(NO3) solution. If 200 ml of 0,20M solution og MgCl2 is added to the beaker, What will be the concentration of Pb2+ ions in the resulting solution
I'm sure you meant Pb(NO3)2
Pb(NO3)2 + MgCl2 ==> PbCl2 + Mg(NO3)2
The key to this problem is knowing that PbCl2 is a ppt. First you must determine which is the limiting reagent.
mols Pb(NO3)2 = M x L = approx 0.06
mols MgCl2 = M x L = approx 0.04
If we used 0.06 mols Pb(NO3)2 and all of the MgCl2 we needed, now much PbCl2 would ppt? That's 0.06 x (1 mol PbCl2/1 mol Pb(NO3)2) = 0.06 x 1/1 = 0.06 mols PbCl2.
How much PbCl2 would ppt if we used 0.04 mols MgCl2 and all of the Pb(NO3)2 we needed. That's 0.04 x (1 mol PbCl2/1 mol MgCl2) = 0.04 x 1/1 = 0.04
So MgCl2 is the limiting reagent and that makes it an easier problem.
Thus 0.04 mols PbCl2 will be formed (the problem doesn't ask for that), you will have no MgCl2 left over, but you will have 0.06-0.04 - 0.02 mols Pb(NO3)2 left over. What will the concn be? That's M Pb^2+ = mols Pb^2+/L solution which should be 0.02 mols/0.5L = ?
To find the concentration of Pb2+ ions in the resulting solution, we first need to determine the amount of Pb(NO3)2 that is present in the initial solution, and then calculate the final volume of the resulting solution.
Step 1: Calculate the initial amount of Pb(NO3)2:
Initial volume = 300 ml
Initial concentration of Pb(NO3)2 = 0.20 M
Amount of Pb(NO3)2 = Initial concentration × Initial volume
= 0.20 M × 300 ml
= 60 mmol
Step 2: Calculate the final volume of the resulting solution:
Initial volume of Pb(NO3) solution = 300 ml
Volume of MgCl2 solution added = 200 ml
Final volume = Initial volume + Volume added
= 300 ml + 200 ml
= 500 ml
Step 3: Calculate the final concentration of Pb2+ ions:
Final amount of Pb(NO3)2 = Initial amount + Amount of Pb(NO3)2 in MgCl2 solution
= 60 mmol + 0 mmol (Pb(NO3)2 is not present in MgCl2)
= 60 mmol
Final concentration of Pb2+ ions = Final amount / Final volume
= 60 mmol / 500 ml
= 0.12 M
Therefore, the concentration of Pb2+ ions in the resulting solution is 0.12 M.
To calculate the concentration of Pb2+ ions in the resulting solution, we need to use the concept of moles and the volume of the solution. Here's how you can calculate it:
Step 1: Find the moles of Pb(NO3)2 initially present in the beaker.
Since the volume of the Pb(NO3)2 solution is 300 ml and the concentration is 0.20 M, we can find the moles using the formula:
moles = concentration (M) × volume (L)
Convert the volume to liters: 300 ml ÷ 1000 ml/L = 0.3 L
Moles of Pb(NO3)2 = 0.20 M × 0.3 L = 0.06 moles
Step 2: Find the moles of MgCl2 added to the beaker.
The volume of the MgCl2 solution added is 200 ml, and the concentration is 0.20 M. Using the same formula as before:
moles = concentration (M) × volume (L)
Convert the volume to liters: 200 ml ÷ 1000 ml/L = 0.2 L
Moles of MgCl2 = 0.20 M × 0.2 L = 0.04 moles
Step 3: Determine the limiting reagent.
The limiting reagent is the one with fewer moles, which in this case is MgCl2 since it has 0.04 moles compared to Pb(NO3)2 with 0.06 moles.
Step 4: Find the moles of Pb2+ ions in the resulting solution.
Moles of Pb(NO3)2 will react with an equal number of moles of MgCl2 to form PbCl2. Therefore, the moles of Pb(NO3)2 remaining in the solution can be calculated as follows:
Moles of Pb(NO3)2 remaining = moles of Pb(NO3)2 initially - moles of MgCl2 used
Moles of Pb(NO3)2 remaining = 0.06 moles - 0.04 moles = 0.02 moles
Step 5: Calculate the concentration of Pb2+ ions in the resulting solution.
To find the concentration, we divide the moles of Pb2+ ions by the new volume of the resulting solution (500 ml = 300 ml + 200 ml) in liters:
Concentration of Pb2+ ions = Moles of Pb2+ ions ÷ Volume (L)
Concentration of Pb2+ ions = 0.02 moles ÷ 0.5 L = 0.04 M
Therefore, the concentration of Pb2+ ions in the resulting solution, after adding the MgCl2 solution, is 0.04 M.