suppose 100 g of lead nitrate salt is added to enough water to make 1 L of solution, what is the concentration of the lead ion in the moles per liter in each of the solutions?
To find the concentration of the lead ion in the solution, we need to calculate the number of moles of lead ions present in the solution.
To start, we need to determine the molar mass of lead (Pb) and nitrate (NO3-) ions. The molar mass of lead is approximately 207.2 g/mol, and the molar mass of nitrate is approximately 62 g/mol (14 g/mol for nitrogen + 48 g/mol for oxygen).
Next, we calculate the moles of lead nitrate used. Since the mass of lead nitrate is given as 100 g, we divide this value by its molar mass:
moles of lead nitrate = mass / molar mass = 100 g / 331.2 g/mol (207.2 g/mol for Pb + 62 g/mol for NO3-)
moles of lead nitrate = 0.302 mol
Since lead nitrate dissociates into one lead ion (Pb2+) and two nitrate ions (NO3-) in solution, the number of moles of lead ions is equal to the number of moles of lead nitrate.
Thus, the moles of lead ions = 0.302 mol.
Finally, to find the concentration of lead ions in moles per liter (mol/L), we divide the moles of lead ions by the volume of the solution.
Concentration of lead ions = moles of lead ions / volume of solution
Remember, the volume of the solution is given as 1 L.
Concentration of lead ions = 0.302 mol / 1 L = 0.302 mol/L
So, the concentration of the lead ion in the solution is 0.302 moles per liter (0.302 M).