You made 100.0 ml of a lead (II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 ml left. In addition, you forgot the initial concentration of the solution. You decide to take 2.00 ml of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.407g. What was the concentration ofthe original lead (II) nitrate solution?
Convert 3.407g PbCl2 to mols. That will be mols Pb(NO3)2 in 2.00 mL solution. That times 80/2 will be the mols in 80 mL and mols in the original 100 mL. Calculate molarity from there.Post your work if you get stuck. We will need that to know where you are having a problem.
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i got the answer 4.9M
To find the concentration of the original lead (II) nitrate solution, we can follow these steps:
Step 1: Convert 3.407g of PbCl2 to moles.
The molar mass of PbCl2 is calculated as follows:
Pb = 207.2 g/mol
Cl = 35.45 g/mol (since there are 2 Cl atoms in PbCl2, we multiply by 2)
Total molar mass = (207.2 g/mol) + (2 * 35.45 g/mol) = 278.1 g/mol
Using the given mass:
moles of PbCl2 = mass / molar mass = 3.407g / 278.1 g/mol = 0.0123 mol
Step 2: Calculate the moles of Pb(NO3)2 in the 2.00 mL solution.
Since PbCl2 reacts with NaCl to form Pb(NO3)2, we can assume that the moles of PbCl2 and Pb(NO3)2 are equal.
moles of Pb(NO3)2 in 2.00 mL = 0.0123 mol
Step 3: Convert the moles in 2.00 mL to the moles in 80.0 mL and the original 100.0 mL.
We know that the moles of Pb(NO3)2 are directly proportional to the volume of the solution. Therefore:
moles of Pb(NO3)2 in 80.0 mL = (moles in 2.00 mL) * (80.0 mL / 2.00 mL) = 0.0123 mol * 40.0 = 0.492 mol
moles of Pb(NO3)2 in 100.0 mL = (moles in 80.0 mL) * (100.0 mL / 80.0 mL) = 0.0123 mol * 1.25 = 0.615 mol
Step 4: Calculate the molarity of the original lead (II) nitrate solution.
Molarity (M) is defined as moles of solute per liter of solution.
Given that we started with 100.0 mL of solution, which is equal to 0.1000 L:
Molarity of Pb(NO3)2 = moles of Pb(NO3)2 / volume of solution in liters
Molarity = 0.615 mol / 0.1000 L = 6.15 M
Therefore, the concentration of the original lead (II) nitrate solution is 6.15 M.
To find the concentration of the original lead (II) nitrate solution, we can follow these steps:
Step 1: Convert 3.407g PbCl2 to moles.
- To do this, we need the molar mass of PbCl2. The molar mass of lead (II) chloride (PbCl2) is approximately 278.10 g/mol (207.2 g/mol for Pb + 2 * 35.45 g/mol for Cl).
- Divide the given mass (3.407g) by the molar mass to find the number of moles: 3.407g / 278.10 g/mol = 0.01225 mol.
Step 2: Determine the moles of Pb(NO3)2 in the 2.00 mL solution.
- Since we are assuming there was an excess of sodium chloride (NaCl) in the reaction, the formation of lead (II) chloride (PbCl2) can be used to determine the moles of lead (II) nitrate (Pb(NO3)2) in the original solution.
- From the balanced equation: Pb(NO3)2 + 2 NaCl → PbCl2 + 2 NaNO3, we can see that 1 mole of Pb(NO3)2 forms 1 mole of PbCl2.
- Therefore, the number of moles of Pb(NO3)2 is the same as the number of moles of PbCl2: 0.01225 mol.
Step 3: Calculate the moles of Pb(NO3)2 in the original 80 mL solution and the original 100 mL solution.
- Using the given information that there were only 80.0 mL left and assuming no evaporation or loss of solute during the transfer, we can set up a proportion:
(moles in 2.00 mL) / (2.00 mL) = (moles in 80 mL) / (80 mL)
- Solving for the moles in 80 mL:
(0.01225 mol) / (2.00 mL) = (x mol) / (80 mL)
- Cross-multiplying and solving for x:
x = (0.01225 mol * 80 mL) / (2.00 mL) = 0.49 mol
- Similarly, we can calculate the moles in the original 100 mL solution:
(0.01225 mol) / (2.00 mL) = (y mol) / (100 mL)
y = (0.01225 mol * 100 mL) / (2.00 mL) = 0.6125 mol
Step 4: Calculate the molarity of the original lead (II) nitrate solution.
- Molarity (M) is defined as moles of solute divided by the volume of solution in liters.
- Since we have found the moles of Pb(NO3)2 in the original 100 mL solution (0.6125 mol), we can now calculate the molarity:
Molarity = moles / volume = 0.6125 mol / 0.100 L = 6.125 M
Therefore, the concentration of the original lead (II) nitrate solution is 6.125 M.