Posted by Hannah on Monday, March 5, 2012 at 1:51pm.
I completed a lab to find the determination of Kc. I have to find the concentrations of reactants at equilibrium using an ICE table. The equation that were are using is
Fe^3+(aq) + SCN^-(aq) -> Fe(SCN)^2+(aq)
I have to create 5 ICE tables because we used 5 different test tubes each with a different concentration of SCN. The values I already have so far are the initial concentrations of reactants and the concentration of products at equilibrium.
For the first table the initial value of Fe is 1.0M and SCN is 1.0M. The concentration of product at equilibrium is 5.7 X 1o^-5M.
I set up the first ICE table like this:
Equation: Fe^3+ SCN^- Fe(SCN)^2+
I 1.0M 1.0M 0
C -5.7e-5M -5.7e^-5 +5.7e^-5
E 0.99M 0.99M 5.7e^-5
So for the concentrations of reactants at equilibrium, Fe is 0.99M and SCN is 0.99M
I put it in my calculator as 1.0 - 5.7 X e-5 and got 0.999943. So besides the weird number I got I'm guessing that I set it up correctly???
The last step is to determine Kc.
so it would be FeSCN / [Fe][SCN]
Kc= 5.7e-5M / (0.99m)(0.99M) = 5.8e-5
Did I set this up correctly??
Thank you for your help!!!
- Chemistry(Please check) - Hannah, Monday, March 5, 2012 at 2:05pm
I think I know where I made a mistake.
For the initial concentrations of reactants Fe and SCN it says that I can use the table that was provided to determine how much the total volume of each solution was used.
The table that was provided provided the amount of Fe and SCN used in each test tube in mL.
So for test tube 1 it was 1.0mL of 0.005M Fe(NO3)3 and 1.0mL of 0.005M KSCN
For all 6 of the test tubes used 1.0mL of Fe was used. Do I need to use the c1v1=c2v2 equation or is the initial concentration for Fe 0.005M?
- Chemistry(Please check) - DrBob222, Monday, March 5, 2012 at 2:56pm
Test tube 1 is
(Fe) = 0.005M x (1.0 mL Fe/2 mL total) =?
(SCN) - 0.005M x (1.0 mL SCN/2.0 mL total) = ?
- Chemistry(Please check) - Hannah, Monday, March 5, 2012 at 3:30pm
Where did the 2mL come from?
- Chemistry(Please check) - Hannah, Monday, March 5, 2012 at 3:34pm
Oh I see thats the total of SCN and Fe but we also used 0.1M HNO3 so for the first test time we used 5mL of HNO3 with the Fe, SCN so the total would be 7mL.
Fe= (0.005M) (1.0mL) = x(7mL) = 0.000714 So 7.1 e-4 Correct.
The teacher said to put the answer in 2 sig figs.
- Chemistry(Please check) - DrBob222, Monday, March 5, 2012 at 4:29pm
The Fe will be (?M at start x (mL Fe/total volume)
The SCN will be (?SCN at start x (mL Fe/total volume).
- Chemistry(Please check) - Hannah, Monday, March 5, 2012 at 5:04pm
Right so for Fe I did (0.005M) (1.0mL Fe) / (7.0mL total of solutions in the tube) = 7.1e-4.
So then my ICE table would say 7.10e-4 for both Fe and SCN because SCN for tube 1 had the same amounts and then I would subtract 7.1e-4 by the concentration of product which was 5.7e-5. This would be 6.53e-4M for concentration of reactants for Fe and SCN correct?
Answer this Question
More Related Questions
- Chemistry(Urgent, please check) - I completed a lab to find the determination ...
- Chemistry(Please check, thank you!) - I completed a lab to find the ...
- chemistry - In a buffer pre-lab, we were asked to complete a table for 10 test ...
- science - At 472 deg C the reaction reaches equilibrium with the following ...
- Chemistry(Please check) - I need to calculate the valueof Kc for 5 test tubes ...
- chemistry - I have gotten the equilibrium concentrations of N202 to be 2M and ...
- math - Mike makees the following table of the distances he travels during the ...
- chemistry - The equilibrium constant Kc for the reaction C <--> D + E is 7...
- Dr. BOB chemistry help please! - hi At a particular temperature, K = 1.00 102 ...
- Chemistry(Please respond, thank you) - The ionization constant of HA is 4.0e-4. ...