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 84.8 mL left (the rest had evaporated). 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 1.650 g. What was the concentration of the original lead(II) nitrate solution?
mols PbCl2 = grams/molar mass = approx 6E-3 but you need a better number than that. I've estimated as well as all of the calculations that follow.
That's mols in 2.00 mL out of 84.8 mL. Convert to mols in 100 mL (the original volume) by 6E-3 x (100/84.8) = approx 7E-3 mols in the original solution.
Then M = 7E-3/0.100 L = about 0.07M
To determine the concentration of the original lead(II) nitrate solution, we can use the concept of stoichiometry and the data provided.
First, let's break down the steps we will follow to find the answer:
1. Calculate the change in volume of the solution.
2. Determine the moles of lead(II) nitrate in the remaining solution.
3. Use the mole ratio to find the moles of sodium chloride reacted.
4. Calculate the molar mass of lead(II) chloride.
5. Calculate the concentration of the original lead(II) nitrate solution.
Let's go through each step in detail:
1. Calculate the change in volume of the solution:
Change in volume = Initial volume - Final volume
= 100.0 mL - 84.8 mL
= 15.2 mL
2. Determine the moles of lead(II) nitrate in the remaining solution:
To find the moles, we need to know the initial concentration of the solution. Since it is not specified, we cannot directly determine the moles of lead(II) nitrate. However, we can calculate the moles of lead (Pb) ions in the remaining solution because lead(II) nitrate dissociates into one lead ion and two nitrate ions in solution.
Since the lead(II) nitrate solution is completely dissociated, we can use the equation:
moles of lead (Pb2+) = volume (in liters) x concentration (in mol/L)
In this case, the volume is (84.8 mL - 15.2 mL) = 69.6 mL = 0.0696 L (converted from mL to L).
3. Use the mole ratio to find the moles of sodium chloride reacted:
Based on the balanced chemical equation, 1 mole of lead(II) nitrate reacts with 2 moles of sodium chloride to form 1 mole of lead(II) chloride and 2 moles of sodium nitrate. Therefore, the moles of sodium chloride reacted would be half the moles of lead(II) nitrate.
4. Calculate the molar mass of lead(II) chloride:
To find the molar mass, we divide the mass of the solid obtained (1.650 g) by the moles of lead in it (which we will determine in the next step).
5. Calculate the concentration of the original lead(II) nitrate solution:
Now, we can use the moles of lead in the remaining solution and the initial volume (100.0 mL) to calculate the concentration of the original lead(II) nitrate solution. Note that we will express the concentration in mol/L (M).
By following these steps, you can determine the concentration of the original lead(II) nitrate solution using the provided information.