2.) FeSCN2+ equilibrium concentration was found to be 3.20 x 10-5 M in a solution made by mixing 5.00 mL of 1.00 x 10-3 M Fe(NO3)3 with 5.00 mL of 1.00 x 10-3 M HSCN. The H+ concentration is maintained at 0.500 M at all times since the HSCN and Fe3+ solutions were prepared using 0.500 M HNO3 in place of distilled water. a. How many moles FeSCN2+ are present at equilibrium? b. How many moles each of Fe3+ and HSCN were initially present? c. How many moles Fe3+ remained unreacted in the equilibrium mixture? 10 d. How many moles of HSCN remained unreacted? e. What are [Fe3+] and [HSCN] in the equilibrium solution? f. Calculate Keq for the reaction.
3.) In this experiment, we assume that the complex ion formed is FeSCN2+. It would be possible, however, to form Fe(SCN)2+ under certain conditions. a. Write the equation for the reaction between Fe3+ and HSCN in which Fe(SCN)2+ is produced. b. Formulate the expression, analogous to equation 4, for the equilibrium constant, Keq, associated with the reaction in part a.
(a) Total volume of the solution: 5=5 ml = 10 ml = 0.01 L
0.01 L * 3.20 x 10-5 M = 3.20 * 10-7 moles
(b) Moles of Fe+3 at beginning = 0.005 L * 1.00 x 10-3 M = 5*10-6 moles
(c) Moles of HSCN at beginning = 0.005 L * 1.00 x 10-3 M = 5*10-6 moles
(d) Moles of Fe+3 unreacted = 5*10-6 moles - 3.20 * 10-7 moles = 0.00000468 mole
(e) Moles of HSCN unreacted = 5*10-6 moles - 3.20 * 10-7 moles = 0.00000468 mole
(f) [Fe+3 ] at equilibrium = 0.00000468 mole/ 0.01 L = 0.000468 M
[HSCN] at equilibrium = 0.00000468 mole/ 0.01 L = 0.000468 M
(g) Keq = [Fe+3 ]*[HSCN]/ [FeSCN+2] * [H+]= 0.000468 * 0.000468/ 3.20 x 10-5 * 0.5 = 0.0136
Fe+3 + 2HSCN = Fe(SCN)2+ + 2H+
Keq = [Fe(SCN)2+ ] * [ H+]2 / [Fe+3]* [HSCN]2
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