PRE-LAB In the lab, five (6.0) mL of 0.0200 M Fc(NO,), are mixed with 5.0 ml of
ID: 1041963 • Letter: P
Question
PRE-LAB In the lab, five (6.0) mL of 0.0200 M Fc(NO,), are mixed with 5.0 ml of 0.00200 M NaSCN to form the blood-red FeSCN2 ion. The equilibrium concentration of the FeSCN2 ion, detcrmined spectrophotometrically, is 7.0 x 104 mol/L a. Moles Fe, initial b. Moles SCN, initial c. Moles FeSCN, at equilibriumm d. Moles Fe reacted e. Moles SCN reacted f Moles Fe remaining unreacted, at equilibrium g. Moles SCN remaining unrcacted, at equilibrium h. [Fe3"] at equilibrium i. [SCN] at equilibrium j. Ke of FeSCN Type here to searchExplanation / Answer
(a)
5 mL of 0.02 M Fe(NO3)3
So, moles of Fe(NO3)3 = 0.02 M x 0.005 L = 0.0001 mol
So, Mole Fe3+ initial = 0.0001 mol
(b)
5 mL of 0.002 M NaSCN
So, moles of NaSCN = 0.002 M x 0.005 L = 0.00001 mol
So, Mole SCN- initial = 0.00001 mol
(c)
[FeSCN2+]equi = 7.0 x 10-4 mol /L
Total volume = 5 mL + 5 mL = 10 mL = 0.010 L
Equilibrium moles of FeSCN2+ = (7.0 x 10-4 mol /L) x 0.010 L
=7.0 x 10-6 mol
Fe3+(aq) + SCN- <=====> FeSCN2+
I: 0.0001 0.00001 0
C: - x - x + x
E: 0.0001 – x 0.00001 – x x = 7.0 x 10-6
So, x = 7.0 x 10-6
(d) Moles of Fe3+ reacted = x =7.0 x 10-6 mol
(e) Moles of SCN- reacted = 7.0 x 10-6 mol
(f) Moles of Fe3+ reacted = initial moles of Fe3+ – equilibrium moles of Fe3+
= 0.0001 mol – (7.0 x 10-6) mol
= 9.3 x 10-5 mol
(g) Moles of SCN- reacted = initial moles of SCN- – equilibrium moles of SCN-
= 0.00001 mol – (7.0 x 10-6) mol
= 3.0 x 10-6 mol
(h) [Fe3+]equil = (9.3 x 10-5 mol) / 0.010 L = 0.0093 M
(i) [SCN-]equil = (3.0 x 10-6 mol) / 0.010 L = 0.0003 M
(j) Kc = [FeSCN2+]equil / {[Fe3+]equil x [SCN-]equil }
= (7.0 x 10-4) / { (0.0093) x (0.0003)}
= 250.9
= 251
= 2.51 x 102
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