A solution from Part 1 (used to make the Beer\'s law plot) contains 1.5 ml of 2.

Question

A solution from Part 1 (used to make the Beer's law plot) contains 1.5 ml of 2.0 times 10^-2 M KSCN, 5 00 mL of 0.30 M FeCI_3, and enough 0.10 M HNO_3 to make a total volume of 25 mL. What is the ratio of moles of Fe^3+ to moles of KSCN? Assuming that the concentration of Fe^3+ is sufficient to drive the reaction virtually to completion, what is the concentration of FeSCN^2+ in this solution? A student prepares an equilibrium solution (from Part 2) by mixing 5.00 mL of 2.00 times 106-3 M Fe^3+, 2.00 mL of 2.00 times 10^-3 M SCN^-, and 3.00 mL of 0.100 M HNO_3. What are the initial concentrations of Fe^3+ and SCN^- m this solution? After allowing the system in problem 2 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. Set up an ICE table for this system and determine the equilibrium concentrations of Fe^3+ SCN^-, and FeSCN^2+ in this solution. Show your work and use the correct number of significant figures.

Explanation / Answer

For the given reaction,

Fe3+ + SCN- <===> [FeSCN]2+

1. a. ratio (Fe3+/KSCN) moles = (0.3 x 5)/(2 x 10^-3 x 1.5) = 500

b. [FeSCN]2+ concentration = 2 x 10^-3 x 1.5/25 = 0.00012 mmol

2. [Fe3+] initial = 2 x 10^-3 M x 5 ml/25 ml = 0.0004 M

[SCN-] initial = 2 x 10^-3 M x 2 ml/25 ml = 0.00016 M

3. ICE chart

                    Fe3+    +   SCN- <===> [FeSCN]2+

I                0.0004        0.00016                   -

C           -8.7x10^-5    -8.7x10^-5         8.7x10^-5                     

E             0.003913     0.000073          8.7x10^-5

So,

Keq = [FeSCN]2+/[Fe3+][SCN-]

        = 8.7 x 10^-5/0.003913 x 0.000073

        = 304.57

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