please help me the calculation of equilibrium concentration page with the three

Question

please help me the calculation of equilibrium concentration page with the three tables (I.C.E table A, B and C) and data analysis page with the table aswell

Revised 11/14/17 Calculating Equilibrium Concentrations A common method that is used to organize and calculate the concentrations of the species in an equilibrium system is colloquially known as an I.C.E, chart. "I.C,E" stands for Initial concentration, Change in concentration, and the Equilibrium concentration. The initial concentrations of the Fe" and the SCN ions can be calculated from the mixing chart in Part I I, 10. You have already determined the cquilibrium concentration of the FeSCN2 ions by completing the analysis in Part II. The rest is a little bit of math TEST TUBE A Fe* FeSCN 0.00 SCN Initial Change Equilibrium TEST TUBE B Fe3* SCN FeSCN Initial 0.00 Change Equilibrium TEST TUBE C Fe3" SCN FeSCN 0.00 Initial Change Equilibrium Page 2 of 3

Explanation / Answer

Calculate the initial concentration of [Fe3+], [SCN-] from the Table II

Test tube A

[Fe3+]i = volume of Fe(NO3)3 * concentration of Fe(NO3)3 / Total volume

= (5.00 mL) * (1.00*10-2 M) / 20 mL = 0.0025 M

[SCN-]i = volume of KSCN * concentration of KSCN / Total volume

= (5.00 mL) * (1.13*10-3 M) / 20 mL = 0.0002825 M

Similarly Calculate for test tube B and C

Test tube B

[Fe3+]i = (5.00 mL) * (1.00*10-2 M) / 20 mL = 0.0025 M

[SCN-]i = (10.00 mL) * (1.13*10-3 M) / 20 mL = 0.000565 M

Test tube C

[Fe3+]i = (10.00 mL) * (1.00*10-2 M) / 20 mL = 0.005 M

[SCN-]i = (5.00 mL) * (1.13*10-3 M) / 20 mL = 0.0002825 M

Now, the equilibrium concentration of FeSCN2+ is already determined in Part II, so now we can determine the equilibrium concentrations of Fe3+ and SCN- using ICE chart.

Test tube A

Fe3+ SCN- FeSCN2+

I 0.0025 M 0.0002825 M 0.00 M

C -0.0001512 -0.0001512 +0.0001512

E 0.0023488 M 0.0001313 M 0.0001512

Now that we have the equilibrium concentrations, we can calculate the equilibrium constant as follows:

Keq = [FeSCN2+] / [Fe3+] [SCN-]

= 0.0001512 M / (0.0023488 M) (0.0001313 M) = 490

Similarly calculate the equilibrium concentrations and then equilibrium constant for test tube B and C.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.