Question

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.

Answer 1

2.

(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

2. (a)

Fe⁺³ + 2HSCN = Fe(SCN)₂⁺ + 2H⁺

(b)

Keq = [Fe(SCN)₂⁺ ] * [ H⁺]² / [Fe⁺³]* [HSCN]²