A solution was made by dissolving 13.4 g of hydrated copper nitrate [Cu(NO3)2.3H20] in water and making it up to a volume of 250 mL. What is the concentration of NO3^-ions in the solution?
mol Cu(NO3)2.3H2O = grams/molar mass = ?
M = mols/L = ?mols/0.250 = ?
Then M NO3^- = twice that.
To find the concentration of NO3^- ions in the solution, we need to first calculate the number of moles of NO3^- ions present in the solution.
1. Calculate the molar mass of Cu(NO3)2.3H2O:
- Molar mass of Cu = 63.55 g/mol
- Molar mass of N = 14.01 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of H = 1.01 g/mol
- Molar mass of H2O = (2 * 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
- Molar mass of Cu(NO3)2.3H2O = (63.55 g/mol) + (2 * (14.01 g/mol)) + (6 * (16.00 g/mol)) + (3 * (18.02 g/mol))
= 249.57 g/mol
2. Calculate the number of moles of Cu(NO3)2.3H2O:
- Moles of Cu(NO3)2.3H2O = Mass / Molar mass
= 13.4 g / 249.57 g/mol
≈ 0.0536 mol
3. Since 1 mole of Cu(NO3)2 dissociates into 2 moles of NO3^- ions, we need to multiply the number of moles of Cu(NO3)2.3H2O by 2 to get the moles of NO3^- ions:
- Moles of NO3^- ions = 2 * Moles of Cu(NO3)2.3H2O
= 2 * 0.0536 mol
= 0.1072 mol
4. Finally, calculate the concentration of NO3^- ions in the solution:
- Concentration (molarity) = Moles / Volume
= 0.1072 mol / 0.250 L
= 0.429 M
Therefore, the concentration of NO3^- ions in the solution is 0.429 M.