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.
millimols Fe(NO3)3 = mL x M = 150 x 3.5 = 525
mmols NO3^- from Fe(NO3)3 = 3*525 = approx 1500 but you need it more accurately.
mmols Mg(NO3)2 = 760 x 1.6 = 1216
mmols NO3^- from Mg(NO3)2 = 2*1216 = approx 2400
Total NO3^- = approx 2400 + 1500 = approx 3900 mmols.
Total volume assuming volumes are additive = 150 + 760 = 910 mL
You took 20 of that so you have how many millimols NO3^- in that 20?
That's approx 3900 x (20/910) = approx 86. So now you have 85 mmols NO3^- in 250 mL and concn is M = millimols/mL = ?
To calculate the nitrate concentration in the final solution, we need to determine the amount of nitrate ions present in the solution after the two solutions are mixed.
First, let's calculate the moles of nitrate ions in the iron(III) nitrate solution and the magnesium nitrate solution separately.
The moles of nitrate ions in the iron(III) nitrate solution can be calculated using the formula:
moles = concentration (M) x volume (L)
Converting the volume of the iron(III) nitrate solution from mL to L:
Volume of iron(III) nitrate solution = 150.00 mL / 1000 = 0.150 L
Moles of nitrate ions in the iron(III) nitrate solution:
moles of nitrate = 3.500 M x 0.150 L = 0.525 mol
Similarly, for the magnesium nitrate solution:
Volume of magnesium nitrate solution = 760.00 mL / 1000 = 0.760 L
Moles of nitrate ions in the magnesium nitrate solution:
moles of nitrate = 1.600 M x 0.760 L = 1.216 mol
Now, since the two solutions were mixed together, the moles of nitrate ions are additive.
Total moles of nitrate ions in the final solution = moles of nitrate ions from iron(III) nitrate solution + moles of nitrate ions from magnesium nitrate solution
= 0.525 mol + 1.216 mol = 1.741 mol
Next, we need to determine the concentration of nitrate ions in the final solution.
We know that the final solution has a volume of 20.00 mL. However, it was diluted to a total volume of 250.00 mL.
The moles of nitrate ions stays the same after dilution. So, we can calculate the concentration using the formula:
concentration (M) = moles / volume (L)
Converting the volume of the final solution from mL to L:
Volume of final solution = 250.00 mL / 1000 = 0.250 L
Nitrate concentration in the final solution:
concentration = 1.741 mol / 0.250 L = 6.964 M
Therefore, the nitrate concentration in the final solution is 6.964 M.