If 8 cm3 of 9.9 mol m-3 Fe(NO3)3 are mixed with 10 cm3 of 1.0 mol m-3 KSCN solution in a 20 cm3 volumetric flask and water is added to bring the total volume up to 20 cm3, what is the concentration of the Fe3+ ions (in mol m-3)?
You know Fe^3+ forms the FeSCN^2+ complex. What are the units for cc x m^-3? millimols/m^3?
8 x 9.9 = 79.2
10 x 1 = 10
.........Fe^3+ + SCN^- ==> FeSCN^2+
I.......79.2.....10.........0
C......-10......-10........10
E.......69.2......0........10
Keq for this reaction is relatively large; we assume all of the "free" Fe^3+ is due to that which has not reacted. Thus the concn Fe^3+ is
69.2 mmols x (8 cc/20 cc) = ? mol/m^3
To find the concentration of Fe3+ ions in the solution, we need to understand the reaction that occurs when Fe(NO3)3 and KSCN are mixed. Fe(NO3)3 dissociates in water to produce Fe3+ ions, while KSCN dissociates to produce SCN- ions.
The balanced equation for the reaction between Fe(NO3)3 and KSCN is:
Fe(NO3)3 + 3KSCN → Fe(SCN)3 + 3KNO3
From the equation, we can see that 1 mole of Fe(NO3)3 reacts with 3 moles of KSCN to produce 1 mole of Fe(SCN)3. This means that the number of moles of Fe3+ ions formed is equal to the number of moles of Fe(NO3)3 used.
Given that the volume of the Fe(NO3)3 solution is 8 cm3 and its concentration is 9.9 mol m-3, we can calculate the number of moles of Fe(NO3)3 using the formula:
Moles = Concentration × Volume
Moles of Fe(NO3)3 = 9.9 mol m-3 × 0.008 m3 (since the volume is given in cm3, we need to convert it to m3)
Moles of Fe(NO3)3 = 0.0792 mol
Since 1 mole of Fe(NO3)3 produces 1 mole of Fe3+ ions, the concentration of Fe3+ ions in the solution is also 0.0792 mol m-3.
Therefore, the concentration of Fe3+ ions in the solution is 0.0792 mol m-3.