Ion Concentrations
1.) A solution is prepared by dissolving 5.00 g of stannic nitrate in enough water to make 250.0 mL of stock solution. A 15.0 mL aliquot (portion) of this stock solution is then removed and added to 75.0 mL of water. Calculate the concentrations of all ions in the final solution.
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To calculate the concentrations of all ions in the final solution, we need to determine the number of moles of each ion present in the solution.
First, let's calculate the number of moles of stannic nitrate (Sn(NO3)4) in the stock solution:
1. Determine the molar mass of stannic nitrate (Sn(NO3)4):
Sn: 118.7 g/mol (from periodic table)
N: 14.0 g/mol
O: 16.0 g/mol
Total molar mass = (118.7 + (14.0 + 3 * 16.0) * 4) g/mol = 435.7 g/mol
2. Calculate the number of moles of stannic nitrate:
Moles = Mass / Molar mass
Moles of Sn(NO3)4 = 5.00 g / 435.7 g/mol = 0.0115 mol
Since we are taking a 15.0 mL aliquot of the stock solution and diluting it to a final volume of 75.0 mL, we can assume that the number of moles of stannic nitrate (Sn(NO3)4) remains constant.
Let's now calculate the concentrations of all ions in the final solution:
1. Calculate the concentration of stannic nitrate (Sn(NO3)4):
Concentration = Moles / Volume
Concentration of Sn(NO3)4 = 0.0115 mol / 0.075 L = 0.153 M
2. Since stannic nitrate dissociates into one tin ion (Sn4+) and four nitrate ions (NO3-), the concentrations of tin and nitrate ions are the same as that of the stannic nitrate:
Concentration of tin ion (Sn4+) = Concentration of Sn(NO3)4 = 0.153 M
Concentration of nitrate ion (NO3-) = Concentration of Sn(NO3)4 = 0.153 M
Therefore, the concentrations of all ions in the final solution are as follows:
- Concentration of tin ion (Sn4+): 0.153 M
- Concentration of nitrate ion (NO3-): 0.153 M