When 0.046 moles of lead(II) nitrate are dissolved in enough water to make 485 milliliters of solution, what is the molar concentration of nitrate ions? Answer in units of M.
M = moles/L.
To find the molar concentration of nitrate ions (NO3-) in the solution, we need to first determine the number of moles of nitrate ions present and then divide it by the volume of the solution in liters.
The molar concentration (Molarity) is defined as moles of solute per liter of solution.
Given:
- Moles of lead(II) nitrate (Pb(NO3)2) = 0.046 mol
- Volume of solution = 485 mL = 0.485 L
We can start by calculating the number of moles of nitrate ions in 0.046 moles of lead(II) nitrate.
Lead(II) nitrate consists of one lead ion (Pb2+) and two nitrate ions (NO3-). Therefore, the number of moles of nitrate ions will be twice the number of moles of lead(II) nitrate.
Number of moles of nitrate ions = 2 × 0.046 mol
Now, we can calculate the molar concentration (Molarity) of nitrate ions.
Molar concentration of nitrate ions = (Number of moles of nitrate ions) / (Volume of solution in liters)
Molar concentration of nitrate ions = (2 × 0.046 mol) / 0.485 L
Dividing these values, we get:
Molar concentration of nitrate ions = 0.092 mol / 0.485 L
Calculating the quotient:
Molar concentration of nitrate ions ≈ 0.19 M
Therefore, the molar concentration of nitrate ions in the solution is approximately 0.19 M.