Calculate the concentration of FeSCN2+ in a standard solution prepared by combining 10.00 mL of 0.200 M Fe(NO3)3 and 2.50 mL of 0.00200M KSCN and diluting to a total volume of 25.00 mL with deionized water.
Well, let's dive into some chemistry calculations! Don't worry, I'll try to keep it fun.
To calculate the concentration of FeSCN2+, we'll use the principle of "moles of solute divided by liters of solution."
First, let's calculate the moles of Fe(NO3)3:
Moles of Fe(NO3)3 = (0.200 mol/L) × (0.01000 L) = 0.00200 mol
Now, let's calculate the moles of KSCN:
Moles of KSCN = (0.00200 mol/L) × (0.00250 L) = 0.00000500 mol
Since KSCN is the limiting reagent (it's present in the least amount), we'll base our calculations on it.
Now, let's calculate the total volume of the solution:
Total volume = 0.01000 L + 0.00250 L + 0.01250 L = 0.01250 L
Finally, to determine the concentration of FeSCN2+, we'll divide the moles of KSCN by the total volume:
Concentration of FeSCN2+ = (0.00000500 mol) / (0.01250 L) = 0.0004 mol/L
So, the concentration of FeSCN2+ in the solution is 0.0004 mol/L.
Now that the chemistry is done, let the laughter begin!
To calculate the concentration of FeSCN2+ in the standard solution, we can use the concept of stoichiometry and the dilution equation.
First, let's determine the moles of Fe(NO3)3 and KSCN used in the solution:
1. Moles of Fe(NO3)3:
Moles = Volume x Concentration
Moles = 10.00 mL x 0.200 M
Moles = 2.0 mmol (millimoles)
2. Moles of KSCN:
Moles = Volume x Concentration
Moles = 2.50 mL x 0.00200 M
Moles = 0.0050 mmol
Next, we need to determine the limiting reactant in the reaction between Fe(NO3)3 and KSCN. The reaction is as follows:
Fe(NO3)3 + 3KSCN -> Fe(SCN)3 + 3KNO3
Since the stoichiometric ratio is 1:3 between Fe(NO3)3 and KSCN, we can calculate the theoretical moles of Fe(SCN)2+ formed. It is three times the moles of KSCN used since each mole of KSCN forms three moles of Fe(SCN)2+.
The moles of Fe(SCN)2+ = 3 x Moles of KSCN = 3 x 0.0050 mmol = 0.015 mmol.
Now, let's determine the concentration of FeSCN2+ in the final solution after dilution:
1. Volume of final solution = 25.00 mL
2. Moles of FeSCN2+ = 0.015 mmol (calculated above)
Using the dilution equation, we can relate the moles, volume, and concentration:
Moles = Concentration x Volume
Concentration (FeSCN2+) = Moles / Volume
Concentration (FeSCN2+) = 0.015 mmol / 25.00 mL
To express the concentration in molarity, divide the moles by the volume in liters:
Concentration (FeSCN2+) = 0.015 mmol / (25.00 mL x 0.001 L/mL)
Finally, calculate the concentration of FeSCN2+:
Concentration (FeSCN2+) = 0.015 mmol / 0.025 L = 0.60 M