calculate: initial and equilibrium[SCN-], initial and equilibrium [Fe+3], equili

Question

calculate: initial and equilibrium[SCN-], initial and equilibrium [Fe+3], equilibrium [FeNCS+2]

Lets see if you can replicate the calculations that you did on your data sheet. Gi students data for the sarme lab experiment for thertest student's data for the same lab experiment for their test tube #3. sheet G Concentration of Fe(NO3)3 in 0.10 M HNO3 solution 0.002 M Concentration of NasCN in 0.10 M HNO3 solution 0.002 M Volume of Fe(NO3)3 solution Volume of NaSCN solution Absorbance Equation of the trendline in their graph 5.00 mL 2.00 mL 0.346 y 2278.6x +0.0121 Calculate the initial [SCN] in the test tube.

Explanation / Answer

Initial concentration calculations:

[SCN-]o=(Concentration of stock*volume taken)/total volume=0.002M*(2ml/7ml)=0.000571M=5.71*10^-4M

[Fe3+]o=0.002M*(5ml/7ml)=0.00143M=1.43*10^-3M

Equilibrium concentration of SCN=[SCN-]eq

Explanation: Fe3+ (aq) +SCN-(aq)<--->[FeSCN]2+ (aq)

[SCN-]o=initial concentration <<<[Fe3+]o (initial),so it can be assumed that SCN- is completely used up in the reaction and at equilibrium [SCN-]eq=[FeSCN2+]eq

Moreover ,[FeSCN2+]eq is the absorbing species,so its absorbance can be used to calculate its concentration using callibration curve for standard [FeSCN2+] solutions.

[Using Beer's law ,

Absorbance=e*l*C

where e=absorptivity of species

l=path length of light(1 cm)]

The equation for callibration curve ,is y=2278.6x+0.0121

y=Absorbance ,x=[FeSCN2+]

given Abs=0.346

x=(y-0.0121)/2278.6=(0.346-0.0121)/2278.6=1.465*10^-4M

[FeSCN2+]eq=[SCN-]eq=1.465*10^-4M

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