Hi! I had a few questions and was hoping you could help out! I\'d love an equati

Question

Hi! I had a few questions and was hoping you could help out! I'd love an equation just to show the steps, and if its possible that there are tiny parts missing then please show me how to figure it out with the given parts! Thanks!

a. During the experiment you use a cell with a path length of 1.00cm. On the instrument the recorded absorbance is 0.246 for the first solution. Knowing the absorptivty constantof FeSCN2+ is 4700 L mol-1cm-1 at the wavelength used, calculate the concentration of FeSCN2+ in the first solution.

b. During the equilibrium experiment at room temperature, you prepare the standard solution of FeSCN2+ by taking 9.00 mL of 0.1925 M Fe(NO3)3 from the dispenser into a 20x150 mm test tube. You pipet 1.00mL of 0.001991 M KSCNinto same test tube and stir. The absorbance of this solution registers as 0.786 au on the spectrometer. Determine [FeSCN29]eq with units M.

c. For your first trial of an equilibrium experiment at room temperature: the standard solution of FeSCN2+ with an equilibrium concentration of 0.00200M has an absorbance in a 1.00cm parth length cell of 0.751 au. For your first trial, after preparing the sample as described, absorbance of tube #1 is 0.177au. What is the [FeSCN2+]eq of tube 1? M should be the units.

d. For one of your trials dyuring the equilibrium experiment at room temperature, you determine the following for Tube #1 while performing an equilibtrium experiment. All of the following apply to tube #1:

[Fe3+]0 = 0.00100359 M

[SCN-]o = 0.00040098 M

absorbance = 0.127 au

[FeSCN2+]eq = 0.00004517 M

determine Kc:

e. For a trial of an equilibrium experiment at room temperature, you determine the following for tube #1 while performing the experiment. For the standard solution in tube #5 the absorbance value is 0.789 and the [FeSCN2+]eq of the standard is 0.0002000M. The following results apply to tube#1:

[Fe3+]o = 0.00100546 M

[SCN-] = 0.00040071 M

absorbance = 0.1628 au

determine Kc (M-1 units)

Thanks again!

Explanation / Answer

Almost all the part you solve them with the following expression:

A = EbC

a) In this case, you have b, E and A, so solve for C:

C = A/Eb

C = 0.246 / 4700 * 1

C = 5.23x10-5 M

b) in this case, you can use the value of E of previous part, and the absorbance of this part, to get the concentration. The other way you have, is to know the value of Kc for this part, but you are not providing that value, so check this question and get back later.

c) With the first data, calculate E:

E = A/bC

E = 0.751 / 1 * 0.002 = 375.5 cm-1 M-1

With this value, calculate C with the second data:

C = 0.177 / 375.5 * 1 = 4.71x10-4 M

d) Let the overall reaction:

r: Fe3+ + SCN- <--------> FeSCN2+

At the equilibrium:

[Fe] = 0.00100359-x

[SCN] = 0.00040098-x

[FeSCN] = x = 0.00004517 M

This means that Kc:

Kc = 0.00004517 / (0.00100359-0.00004517) (0.00040098-0.00004517)

Kc = 132.45

PArt e) is solved in similar way as part d.

Hope this helps

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