A student prepares an equilibrium solution by mixing 5.00 mL0.00200 M Fe^3+ 2.00
ID: 523836 • Letter: A
Question
A student prepares an equilibrium solution by mixing 5.00 mL0.00200 M Fe^3+ 2.00 mL 0.00200 M SCN^-, and 3.00 mL 0.1 M HNO_3. What are the initial concentrations of Fe^3+ and SCN^- in this solution? c) After allowing the system in problem (b) above to come to equilibrium, the absorbance of the solution was measured. Using the Beer's law plot the student found that [FeSCN^2+] = 8.70 times 10^- 5. Determine the equilibrium concentrations Fe^3+, SCN^-, and FeSCN^2_ in this solution and calculate Keq.Explanation / Answer
Fe3+ (aq) + SCN– (aq) <----> FeSCN2+ (aq)
[FeSCN2+ ] / [Fe3+ ][SCN- ] = Kc
5 ml 0.002 M Fe3+ is mixed with 2 mL 0.002 M SCN-
The initial concentrations of the reactants can be found from the following relation:
initial concentration of reactants in mixture = concentration of stock solution × volume of stock solution used / total volume of mixture
initial concentration of Fe3+ = 2.00 × 10–3 M × 0.005 L / (0.005 L + 0.002 L) = 1.43× 10–3M
initial concentration of SCN– = 2.00 × 10–3 M × 0.002 L / (0.005 L + 0.002 L ) = 5.71 × 10-4 M
initial concentration of FeSCN2+ = 0.00 M, because we did not add any to the mixture.
B :
Fe3+ (aq) + SCN– (aq) <----> FeSCN2+ (aq)
initial conc 1.43× 10–3M 5.71 × 10-4 M 0
equilb conc 1.43× 10–3M - 8.7*10-5 M 5.71 × 10-4 M - 8.7*10-5 M 8.7*10-5 M
1.3 * 10-3 M 0.484 * 10-3 M 8.7*10-5 M
[FeSCN2+ ] / [Fe3+ ][SCN- ] = Kc
Kc = 8.7*10-5 M / ( 1.3 * 10-3 M) (0.484 * 10-3 M )
Kc = 138.27
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