What weight of lead nitrate in grams would need to be dissolved in 400 mL of water to prepare a 270 ppm solution of Pb2+?
1 ppm = 1 mg/L
270 ppm = 270 mg/L
Convert that to mg/400 mL.
270 x (400/1000) = ? mg in 400 mL
Change to grams.
0.108
To determine the weight of lead nitrate required to prepare a 270 ppm solution of Pb2+, we can follow these steps:
1. Understand the problem:
- Pb2+ concentration in the solution is given as 270 ppm (parts per million).
- We need to calculate the weight of lead nitrate required.
- The volume of water given is 400 mL.
2. Convert the given concentration to mg/L:
- ppm is equivalent to mg/L (miligrams per liter).
- Therefore, 270 ppm = 270 mg/L.
3. Calculate the amount (in moles) of Pb2+ ions required:
- The molar mass of Pb2+ is 207.2 g/mol (atomic mass of Pb = 207.2 g/mol).
- Since the concentration is given in mg/L, we need to convert it to g/L:
- 270 mg/L = 270/1000 g/L = 0.27 g/L.
- Now, we can calculate the moles of Pb2+ ions required using the molarity formula:
- Moles = Concentration (in mol/L) x Volume (in L).
- Moles = (0.27 g/L) / (207.2 g/mol).
4. Convert the moles of Pb2+ ions to grams of lead nitrate:
- The molar mass of lead nitrate (Pb(NO3)2) is 331.2 g/mol (207.2 g/mol for Pb2+ + 62.0 g/mol for NO3-).
- Multiply the moles of Pb2+ ions by the molar mass of lead nitrate:
- Weight = Moles x Molar mass.
- Weight = (moles Pb2+) x (331.2 g/mol).
By following these steps, you can calculate the weight of lead nitrate in grams needed to prepare the desired solution.