To tackle this problem, we need to break it down into manageable parts and use stoichiometry to find the mass percentages of PbO2 and Pb3O4 in the original sample. Let's go through the steps systematically.
Understanding the Components
We have a mixture of lead compounds: Pb3O4 and PbO2, along with some inert impurities. When this mixture is treated with dilute nitric acid (HNO3), all lead is converted to Pb2+. The addition of sodium oxalate (Na2C2O4) helps in precipitating lead as lead oxalate (PbC2O4), and any excess oxalate can be titrated with potassium permanganate (KMnO4).
Step 1: Determine the amount of excess oxalate
From the titration, we know that a 10 mL portion of the solution required 8.0 mL of 0.02 M KMnO4. First, we calculate the moles of KMnO4 used:
- Molarity (M) = moles/volume (L)
- Volume of KMnO4 = 8.0 mL = 0.008 L
- Moles of KMnO4 = 0.02 M × 0.008 L = 0.00016 moles
Since KMnO4 reacts with oxalate in a 1:5 ratio, the moles of oxalate that reacted can be calculated as:
- Moles of oxalate = 5 × moles of KMnO4 = 5 × 0.00016 = 0.0008 moles
Step 2: Calculate the total moles of oxalate added
Next, we need to find the total moles of Na2C2O4 added. The molar mass of Na2C2O4 is approximately 110 g/mol. Thus, for 2.7 g of Na2C2O4:
- Moles of Na2C2O4 = mass/molar mass = 2.7 g / 110 g/mol ≈ 0.0245 moles
Step 3: Determine moles of Pb2+
The moles of oxalate that reacted with Pb2+ can be calculated by subtracting the moles of excess oxalate from the total moles added:
- Moles of oxalate that reacted with Pb2+ = 0.0245 - 0.0008 = 0.0237 moles
Since each mole of Pb2+ reacts with one mole of oxalate, the moles of Pb2+ present in the solution is also 0.0237 moles.
Step 4: Calculate the mass of lead
The molar mass of lead (Pb) is approximately 207.2 g/mol. Therefore, the mass of Pb2+ in the solution is:
- Mass of Pb = moles × molar mass = 0.0237 moles × 207.2 g/mol ≈ 4.91 g
Step 5: Analyze the second experiment
In the second experiment, we know that 10 mL of permanganate solution oxidized Pb2+ to Pb4+. This was equivalent to 4.48 mL of 5 V H2O2 solution. The molarity of H2O2 can be calculated as follows:
- 5 V H2O2 means 5 g of H2O2 in 100 mL solution, so molarity = (5 g / 34 g/mol) / 0.1 L = 1.47 M
Thus, the moles of H2O2 in 4.48 mL (0.00448 L) is:
- Moles of H2O2 = 1.47 M × 0.00448 L ≈ 0.00656 moles
Since H2O2 reacts with Pb2+ in a 1:1 ratio, the moles of Pb2+ oxidized to Pb4+ is also 0.00656 moles.
Step 6: Calculate the mass of Pb in the original sample
Now, we can find the total mass of lead in the original sample:
- Total mass of Pb = mass from oxalate + mass from H2O2 = 4.91 g + (0.00656 moles × 207.2 g/mol) ≈ 4.91 g + 1.36 g ≈ 6.27 g
Step 7: Determine the mass percentages of PbO2 and Pb3O4
Next, we need to find the individual contributions of PbO2 and Pb3O4. The molar masses are:
- PbO2: 207.2 + 2 × 16 = 239.2 g/mol
- Pb3O4: 3 × 207.2 + 4 × 16 = 686.6 g/mol
Let x be the mass of PbO2 and y be the mass of Pb3O4. We have the following equations:
- x + y = 5.0 g (total mass of the sample)
- (x / 239.2) + (3y / 686.6) = 6.27 g (total mass of Pb)
Solving these equations simultaneously will give us the values of x and y, which represent the masses of PbO2 and Pb3O4, respectively. After solving, we can find the mass percentages:
- Mass % of PbO2 = (x / 5.0 g) × 100
- Mass % of Pb3O4 = (y / 5.0 g) × 100
By following these steps and calculations, you can determine the mass percentages of PbO2 and Pb3O4 in the original sample. If you need help with the algebraic solution or any specific calculations, feel free to ask!
