Chem110 Don’t get stressed lab

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Chemistry

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Apr 3, 2024

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Experiment 5 Don’t Get Stressed in the Lab Total Points: 43 Student’s Name: Parker Gould Lab Section: 503 NOTE: You must show your work for all calculations; no work, no credit. Equilibrium Reactions of Iron with Thiocyanate Table Test Tube Initial [Fe 3+ ] Initial [SCN - ] Absorban ce Equilibrium [Fe 3+ ] Equilibrium [SCN - ] Equilibrium [FeSCN 2+ ] 1 0.001 0.0004 0.175 9.65x10^-4 3.65x10^-4 3.5x10^-5 2 0.001 0.0006 0.200 9.60x10^-4 5.60x10^-4 4.00x10^-5 3 0.001 0.0008 0.216 9.57x10^-4 7.57x10^-4 4.32x10^-5 4 0.001 0.001 0.225 9.55x10^-4 9.55x10^-4 4.50x10^-5 5 0.18 0.0002 0.999 0.1798 0 2.0x10^-4 Initial Temperature = 21.0 °C (5 pts) 1. (2 pts) Express the equilibrium constant (K c ) for the iron complex formed in this investigation. (For instance: K c = x/y) Kc= [FeSCN^2+](aq)/[Fe3+(aq)][SCN^-(aq)] 2. (2 pts) Calculate the initial concentration of Fe 3+ ([Fe 3+ ] i ) for all the test tubes . This is based on the dilution that results from adding the KSCN and H 2 O to the original 0.0020 M Fe(NO 3 ) 3 solution 1- (0.002)(5)=M(10)=0.001m 2- (0.002)(5)=M(10)=0.001m 3- (0.002)(5)=M(10)=0.001m 4- (0.002)(5)=M(10)=0.001m 5- (0.2)(18)=M(20)= 0.18m 3. (4 pts) Calculate the initial concentration of SCN - ([SCN - ] i ) for all the test tubes . It was diluted by Fe(NO 3 ) 3 and H 2 O. 1- (0.002)(2)=M(10) = 0.0004m 2- (0.002)(3)=M(10)= 0.0006m 3- (0.002)(4)=M(10)= 0.0008m
Experiment 5 Don’t Get Stressed in the Lab Total Points: 43 4-(0.002)(5)=M(10)= 0.001m 5-(0.002)(2)=M(20)= 0.0002m 4. (2 pts) Calculate the concentration of FeSCN 2+ in the standard solution , test tube 5. (The conversion of SCN - to FeSCN 2+ is essentially 100% because of the large excess of Fe 3+ ; thiocyanate is the limiting reagent.) M1V1=M2V2 (0.002)(2)=M2(20) = 0.002M SCN = 0.002M FeSCN^2+ 5. (4 pts) Using the following formula, calculate the [FeSCN 2+ ] eq for test tubes 1-5 . [FeSCN 2+ ] eq = (A eq /A std ) x [FeSCN 2+ ] std A eq = absorbance values for the equilibrium solutions A std = absorbance values for the standard test tube 1- Xeq= (0.175/0.999)(0.002)= 3.5x10^-5 2- Xeq=(0.200/0.999)(0.002)= 4.00x10^-5 3- Xeq= (0.216/0.999)(0.002)= 4.32x10^-5 4- Xeq=(0.255/0.999)(0.002)= 4.50x10^-5 5- Xeq=(0.999/0.999)(0.002)= 2.0x10^-4 6. (4 pts) Calculate the concentration of Fe 3+ at equilibrium ([Fe 3+ ] eq ) for test tubes 1-5 . (Hint: [Fe 3+ ] eq = [Fe 3+ ] I – [FeSCN 2+ ] eq ) 1- 0.001-3.5x10^-5= 9.65x10^-4 2- 0.001-4.00x10^-5= 9.60x10^-4 3- 0.001-4.32x10^-5= 9.57x10^-4 4- 0.001-4.50x10^-5= 9.55x10^-4 5- 0.18-2.0x10^-4= 0.1798 7. (4 pts) Calculate the concentrations of SCN - at equilibrium ([SCN - ] eq ) for test tubes 1-5 , including the standard. (Hint: [SCN - ] eq = [SCN - ] I – [FeSCN 2+ ] eq ) 1- 0.0004- 3.5x10^-5= 3.65x10^-4 2- 0.0006- 4.00x10^-5= 5.60x10^-4 3- 0.0008- 4.32x10^-5= 7.57x10^-4 4- 0.001- 4.50x10^-5= 9.55x10^-4 5- 0.0002- 2.0x10^-4= 0 8. (4 pts) Calculate the K c values for test tubes 1-4 ; show their expression before you calculate.
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