If we assume that Le Châtelier\'s Principle has pushed the reaction entirely to
ID: 1030139 • Letter: I
Question
If we assume that Le Châtelier's Principle has pushed the reaction entirely to completion in the Part I solutions, then we can calculate the concentration of Fe(SCN)2+ formed as a result of all SCN reacting simply by accounting for the dilutions occurring in the mixing process. M1V1=M2V2, and in this case MFE(SCN)2+ = (MSCN- * VSCN-)/total volume. Calculate the equilibrium concentration of Fe(SCN)2+ in each Part I solution, 1-4. Show at least one example calculation.
Solution [Fe(SCN)^+2] M
1
2
3
4
5 0.000
This is Part 1: Label five 100 mL beakers 1–5. Obtain ~35 mL of each of 0.200 M Fe(NO3)3, 0.00200 M NaSCN, and distilled water. Prepare four solutions according to the mixing schedule below (the fifth beaker is a blank). Use a 0-10 mL autopipette to transfer each solution to a 50 mL volumetric flask. Fill each solution up to the mark with deionized water and mix each solution thoroughly. Transfer the solution to the appropriately labeled beaker and re-use the 50 mL volumetric to make the next solution. Measure and record the temperature of one of the above solutions.
Mixing: Beaker1 0.200 M Fe(NO3)3, mL=5.000. 0.00200 M NaSCN, mL = 2.00. total volume 50ml
Beaker 2 0.200 M Fe(NO3)3, mL=5.000. 0.00200 M NaSCN, mL=3.000. total volume 50ml.
Beaker 3 0.200 M Fe(NO3)3, mL=5.000. 0.00200 M NaSCN, mL=4.000. total volume 50ml
Beaker 4 0.200 M Fe(NO3)3, mL=5.000. 0.00200 M NaSCN, mL= 5.000. total volume 50ml
Beaker 5 (blank) 5.000. 0.000. total volume 50ml
Explanation / Answer
Fe3+ (aq)+ SCN- (aq)--------------> Fe(SCN)2+ (aq)
Beaker 1: The molarity of Fe(SCN)2+ can be calculated using the formula M1V1 = M2V2
Where M1 is [SCN-] and V1 is volume of [SCN-], M2 is [Fe(SCN)2+] and V2 is the total volume.
[Fe(SCN)2+ ](M2) = {[SCN-] * VolSCN-}/Total vol
= {2.00x10-3 moles/L*2.00 x 10-3 L}/50.00 x 10-3 L = 8.00 x 10-5 M
Beaker 2: The molarity of Fe(SCN)2+ can be calculated using the formula M1V1 = M2V2
Where M1 is [SCN-] and V1 is volume of [SCN-], M2 is [Fe(SCN)2+] and V2 is the total volume.
[Fe(SCN)2+ ](M2) = {[SCN-] * VolSCN-}/Total vol
= {2.00x10-3 moles/L*3.00 x 10-3 L}/50.00 x 10-3 L = 1.20 x 10-4 M
Beaker 3: The molarity of Fe(SCN)2+ can be calculated using the formula M1V1 = M2V2
Where M1 is [SCN-] and V1 is volume of [SCN-], M2 is [Fe(SCN)2+] and V2 is the total volume.
[Fe(SCN)2+ ](M2) = {[SCN-] * VolSCN-}/Total vol
= {2.00x10-3 moles/L*4.00 x 10-3 L}/50.00 x 10-3 L = 1.60 x 10-4 M
Beaker 4: The molarity of Fe(SCN)2+ can be calculated using the formula M1V1 = M2V2
Where M1 is [SCN-] and V1 is volume of [SCN-], M2 is [Fe(SCN)2+] and V2 is the total volume.
[Fe(SCN)2+ ](M2) = {[SCN-] * VolSCN-}/Total vol
= {2.00x10-3 moles/L*5.00 x 10-3 L}/50.00 x 10-3 L = 2.00 x 10-4 M
Beaker 5: The molarity of Fe(SCN)2+ can be calculated using the formula M1V1 = M2V2
Where M1 is [SCN-] and V1 is volume of [SCN-], M2 is [Fe(SCN)2+] and V2 is the total volume.
[Fe(SCN)2+ ](M2) = {[SCN-] * VolSCN-}/Total vol
= {0 moles/L*5.00 x 10-3 L}/50.00 x 10-3 L = 0 M (No Fe(SCN)2+ is formed)