Chemistry 142 Lab Lab partner Poarte ve Experiment 10: Determination of the Equi

Question

Chemistry 142 Lab Lab partner Poarte ve Experiment 10: Determination of the Equilibrium Constant for a Complex Ion Report Sheet Data and Calculations Table 1: SOLUTION OF "KNOWN" CONCENTRATION: Procedure 17 Beer's Law constam Volume 0.200 M Volume 0.00200 M NaSCN (ml.) FeSCN AbsorbanceAbsorbance molarity mt o 0 000200M 5 SSG5 Initial Iswl (by dilution) Initial [Fe 000Mo 000200 Table II: EQUILIBRIUM ABSORBANCE DATA: Procedure B Volume Volume Volume (mL.) TotalAbsorbance Transmittance (mL) Volume(mL) Cuvette 2.00x10 M 2.00x10'M Deionized 10 AL 10 mL

Explanation / Answer

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

From Table II

Sample #1

a. [Fe3+]initial = 0.002 M x 10 ml/20 ml = 0.001 M

[SCN-]initial = 0.001 M x 2 ml/20 ml = 0.0002 M

b. [FeSCN]2+ equilibrium = 0.279/5565 = 5.0 x 10^-5 M

c. change in [Fe3+] and [SCN-] = 5.0 x 10^-5 M

d. [Fe3+] equilibrium = 0.001 - 5.0 x 10^-5 M = 9.5 x 10^-4 M

[SCN-] equilibrium = 0.0002 - 5.0 x 10^-5 M = 1.5 x 10^-4 M

e. equilibrium constant Kc

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

     = 5 x 10^-5/(9.5 x 10^-4 x 1.5 x 10^-4)

     = 350.88

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1. With all of SCN- converted

[FeSCN2+] = 0.0002 M

[SCN-] equilibrium = 0.0002 M

2. when reaction occurs,

tube #1,

[SCN-] equiilibrium = 1.5 x 10^-4 M

percent [SCN-] remained = 1.5 x 10^-4 x 100/0.0002 = 75%

So the assumption of all [SCN-] forming complex is incorrect as not all SCN- is consumed in the reaction.

3. [Fe3+] = 0.002 M x 25 ml/55 ml = 9.1 x 10^-4 M

[SCN-] = 0.002 M x 17 ml/55 ml = 6.2 x 10^-4 M

Keq = 351

So,

[FeSCN]2+ = 351 x 9.1 x 10^-4 x 6.2 x 10^-4 = 2.0 x 10^-4 M

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