A student mixes 5.0 mL of 0.00200 M Fe(NO3)3 with 5.0 mL 0.00200 KSCN. She finds
ID: 1030332 • Letter: A
Question
A student mixes 5.0 mL of 0.00200 M Fe(NO3)3 with 5.0 mL 0.00200 KSCN. She finds that the concentration of FeSCN2 in the equilibrium mixture is 0.000125 M. Follow these steps to determine the corresponding experimental value of Kc for the reaction of Fe and SCN to produce this complex ion. Show your calculations for each step below and then place the appropriate value(s) in the equilibrium (or 'ICE table near the bottom of the page. Step 1. Calculate the molarity of Fe, sCN, and FesSCN2 initially present after mixing the two solutions, but prior to any reaction taking place. (M,V M-V2) Step 2. Determine the expression and initial value for Qc. Then give the appropriate signs of the concentration changes for each species in terms of the reaction's shift, or x, into the 'ICE' table. Step 3. Fill in the equilibrium value for the molarity of FeSCN2. From this, you can determine the value of x. Step 4. Given the value of x, determine the equilibrium molarities of Fe and SCN ICE' Table Fe" (aq) SCN (aq)FeSCN (aq) Step 5. Give the correct expression for Ke for this equation. Then calculate the value of Kc for the reaction from the equilibrium concentrations. Use correct significant figures. Step 6. On the reverse side, complete an 'ICE' table using this same procedure, but using a different reaction stoichiometry: Fe 2 SCNFeSCN2 Assume that the equilibrium concentration of FeSCN2 is 0.0000625 M, or one-half its previous value. Remember how the reaction stoichiometry affects the expression for KeExplanation / Answer
Molarity of Fe(NO3)3 = 0.00200M
Volume of Fe(NO3)3 = 5 mL
No. of moles of Fe(NO3)3 added is
n = MV
where M = molarity
V = volume in L
n = 0.00200 mol/L x (5 x 10-3)L
n = 1 x 10-5 mol
Molarity of Fe3+ initially present after mixing is
M = (no. of moles)/total volume of the solution in L
Total volume = 5 mL + 5 mL = 10 mL
M = (1 x 10-5 mol)/ (10 x 10-3L)
M = 1 x 10-3 mol/L = 0.00100 mol/L
Molarity and volume of FSCN added is the same as that of Fe(NO3)3
Therefore, Molarity of SCN- will be the same as the molarity of Fe3+ = 1 x 10-3 mol/L = 0.00100 mol/L
Before the reaction started, the concentration of FeSCN2+ will be zero because the reaction has not happened yet.
Fe3+ (aq) + SCN- (aq) (equilibrium arrow) FeSCN2+ (aq)
The ICE table is
The equilibrium concentration of FeSCN2+ is 0.000125
The equilibrium constant for the above equilibrium is
Kc = [FeSCN2+]/[Fe3+][SCN-]
Substituting the values,
Kc = 0.000125 / (0.000875 x 0.000875)
Kc = 0.000125 / (7.6563 x 10-7)
Kc = 1.6 x 10-5 x 107
Kc = 1.6 x 102
Kc = 160
Therefore, the Kc of the equilibrium is 160.
For the equilibrium
Fe3+ + 2SCN- (equilibrium arrow) FeSCN2+
The ICE table is
The equilibrium concentration of FeSCN2+ is 0.0000625M
[SCN]eq = [SCN]initial - 2[FeSCN2+]
This is because, in the equilibrium 2SCN- is taken which can react to form 2FeSCN2+
2[FeSCN2+] = 2 x 0.0000625 = 0.000125
Therefore, for [SCN-]eq 0.000125 is subtracted from [SCN-]initial
The Kc is given by the expression
Kc = [FeSCN2+]/[Fe3+][SCN-]2
The concentration of SCN- is squared in the denominator because the coefficient of SCN- is 2.
Kc = 0.0000625 / (0.0009375 x 0.0008752)
Kc = 0.0000625 / (0.0009375 x 7.656 x 10-7)
Kc = 0.0000625 / (7.1775 x 10-10)
Kc = 8.71 x 10-6 x 1010
Kc = 8.71 x 104
For the second equilibrium with different stoichiometry, the Kc value is 8.71 x 104
Fe3+ (aq) SCN- (aq) FeSCN2+(aq) Initial 0.00100 0.00100 0 Change 0.00100-0.000125 0.00100-0.000125 0.000125 Equilibrium 0.000875 0.000875 0.000125Related Questions
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