How do I calculate 25.00mL of 5.0*10^-4 M of FeSCN^2+ using 0.002 M Fe(NO3)3, 0.002 M KSCN, 0.1 M of HNO3 and water?
To calculate the required volume of Fe(NO3)3 and KSCN solutions to make 25.00 mL of a 5.0*10^-4 M FeSCN^2+ solution, you need to know the stoichiometry of the reaction between Fe(NO3)3, KSCN, and FeSCN^2+. Here is the balanced chemical equation for the reaction:
Fe(NO3)3 + 3KSCN → Fe(SCN)3 + 3KNO3
From the balanced equation, you can see that the stoichiometric ratio between Fe(NO3)3 and Fe(SCN)3 is 1:1. This means that you need the same molar amount of Fe(NO3)3 as Fe(SCN)3.
First, calculate the moles of FeSCN^2+ needed:
Moles of FeSCN^2+ = Molarity * Volume
= (5.0*10^-4 M) * (25.00 mL / 1000 mL/1 L)
= 1.25*10^-5 moles
Since the stoichiometric ratio of Fe(NO3)3 to FeSCN^2+ is 1:1, you need 1.25*10^-5 moles of Fe(NO3)3.
Next, you need to calculate the volume of 0.002 M Fe(NO3)3 solution containing this amount of moles:
Volume of Fe(NO3)3 solution = Moles / Molarity
= 1.25*10^-5 moles / 0.002 M
= 6.25 mL
Therefore, you need to measure 6.25 mL of 0.002 M Fe(NO3)3 solution.
Similarly, since the stoichiometric ratio of KSCN to FeSCN^2+ is 1:1, you need 1.25*10^-5 moles of KSCN. You can calculate the volume of 0.002 M KSCN solution containing this amount of moles:
Volume of KSCN solution = Moles / Molarity
= 1.25*10^-5 moles / 0.002 M
= 6.25 mL
Therefore, you also need to measure 6.25 mL of 0.002 M KSCN solution.
Finally, to prepare the 25.00 mL solution, you can mix 6.25 mL of 0.002 M Fe(NO3)3, 6.25 mL of 0.002 M KSCN, and 12.50 mL of water. Add the solutions to a container and dilute with 12.50 mL of 0.1 M HNO3 and water to reach a total volume of 25.00 mL.