The following two solutions were combined and mixed well: 150.00 mL of a 3.500 M iron(¡¡¡) nitrate solution and 760.00mL of a 1.600M magnesium nitrate solution. 20.00mL of the final solution was transferred with a pipette into an empty 250.00mL volumetric flask. It was made up to the mark with distilled water. Calculate the nitrate concentration in the final solution.
To calculate the nitrate concentration in the final solution, we need to determine how many moles of nitrate ions (NO3-) are present in both the iron(III) nitrate and magnesium nitrate solutions.
First, let's calculate the moles of nitrate ions in the iron(III) nitrate solution:
Moles of nitrate ions = volume of solution (in L) x molarity
Moles of nitrate ions in iron(III) nitrate solution = 0.150 L x 3.500 mol/L = 0.525 mol
Next, let's calculate the moles of nitrate ions in the magnesium nitrate solution:
Moles of nitrate ions in magnesium nitrate solution = 0.760 L x 1.600 mol/L = 1.216 mol
Now, we need to calculate the total moles of nitrate ions in the final solution by summing up the moles of nitrate ions from both solutions:
Total moles of nitrate ions = moles of nitrate ions in iron(III) nitrate solution + moles of nitrate ions in magnesium nitrate solution
Total moles of nitrate ions = 0.525 mol + 1.216 mol = 1.741 mol
Since 20.00 mL of the final solution was transferred to the volumetric flask and made up to 250.00 mL with distilled water, the final solution has a volume of 250.00 mL (0.250 L). Therefore, to find the nitrate concentration in the final solution, we divide the total moles of nitrate ions by the final volume of the solution:
Nitrate concentration in the final solution = total moles of nitrate ions / final volume of solution
Nitrate concentration in the final solution = 1.741 mol / 0.250 L = 6.964 mol/L
Therefore, the nitrate concentration in the final solution is 6.964 mol/L.
To calculate the nitrate concentration in the final solution, we need to determine the amount of nitrate ions present in the combined solutions and then calculate the concentration in the final solution after dilution.
Step 1: Calculate the moles of nitrate ions in each original solution.
For the iron(III) nitrate solution:
moles of nitrate ions = volume (L) x concentration (M)
moles of nitrate ions = 0.150 L x 3.500 M
moles of nitrate ions = 0.525 mol
For the magnesium nitrate solution:
moles of nitrate ions = volume (L) x concentration (M)
moles of nitrate ions = 0.760 L x 1.600 M
moles of nitrate ions = 1.216 mol
Step 2: Calculate the total moles of nitrate ions in the combined solutions.
total moles of nitrate ions = moles of nitrate ions from iron(III) nitrate + moles of nitrate ions from magnesium nitrate
total moles of nitrate ions = 0.525 mol + 1.216 mol
total moles of nitrate ions = 1.741 mol
Step 3: Calculate the concentration in the final solution after dilution.
The final solution volume is 250.00 mL, and a 20.00 mL aliquot was transferred into the flask. This means that the 20.00 mL aliquot represents 1/12.5 (250.00 mL / 20.00 mL) of the final solution.
Concentration in the final solution = concentration of nitrate ions in the combined solutions x dilution factor
Concentration in the final solution = total moles of nitrate ions / final solution volume (L)
Concentration in the final solution = 1.741 mol / (1/12.5) L
Concentration in the final solution = 1.741 mol / 0.08 L
Concentration in the final solution = 21.76 M
Therefore, the concentration of nitrate ions in the final solution is 21.76 M.