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 follow these steps:
Step 1: Calculate the concentration of the stock solution.
- We are given that 5.00 g of stannic nitrate is dissolved in enough water to make 250.0 mL of stock solution.
- The molecular weight of stannic nitrate (Sn(NO3)4) is 366.71 g/mol.
- First, we need to convert grams to moles: 5.00 g / 366.71 g/mol = 0.0136 mol.
- Now, we can calculate the concentration of the stock solution: 0.0136 mol / 0.250 L = 0.0544 mol/L.
Step 2: Calculate the number of moles in the 15.0 mL aliquot.
- We are provided with a 15.0 mL aliquot of the stock solution.
- To calculate the number of moles, we need to use the concentration of the stock solution.
- Concentration (mol/L) = Number of moles (mol) / Volume (L).
- Rearranging the equation, Number of moles (mol) = Concentration (mol/L) x Volume (L).
- Number of moles = 0.0544 mol/L x 0.015 L = 0.00082 mol.
Step 3: Calculate the final volume of the solution.
- We are told that the 15.0 mL aliquot is added to 75.0 mL of water.
- The final volume is the sum of the aliquot volume and the volume of water: 15.0 mL + 75.0 mL = 90.0 mL.
Step 4: Calculate the final concentration of the ions in the solution.
- Since the 15.0 mL aliquot was added to 75.0 mL of water, the final solution has a volume of 90.0 mL.
- We know that the number of moles is still 0.00082 mol, and the final volume is 90.0 mL.
- Concentration (mol/L) = Number of moles (mol) / Volume (L).
- Concentration (mol/L) = 0.00082 mol / 0.090 L = 0.0091 mol/L.
Step 5: Breakdown the stannic nitrate into its ions.
- The formula for stannic nitrate is Sn(NO3)4.
- This dissociates into one Sn4+ ion and four NO3- ions per formula unit.
So, in the final solution, the concentration of Sn4+ ion is 0.0091 mol/L, and the concentration of NO3- ions is 4 times that, which is 4 x 0.0091 mol/L = 0.0364 mol/L.