Question

Given the following data for a hypothetical reaction: M3+ (aq)+ L-(aq) ⇋ [ML]2+(aq); calculate the equilibrium concentration of [ML]2+in the trial solution and Kc based on this single trial run of the equilibrium reaction.

Standard solution data:
[ML]2+std = 0.0500 M
abs [ML]2+std = 0.842

Trial solution data:
[M3+]initial= 0.210
[L-]initial= 0.150 M
abs[ML]2+at equilibrium = 0.523

Ratio (conc std/abs std) ______ (this is your absorbance to concentration “conversion factor”)
The equilibrium concentration of ML2+in the sample __________
Complete the ICE table below using the data above to determine Kc based on this data.
Rows of table: [M3+], [L-], [ML]2+
Columns of table: I, C, E

Kc=

Answer 1

Note: Kevin, you should tell us what you don't understand and what you understand about a problem like this and don't leave us up in the air. You don't show any work at all so I'm giving you a heads up that you can't gret another freebie like this again.
Abs = A stand = 0.842 for 0.05M [ML]^2+
A for trial 1 run is 0.523 so concn [ML]^2+ is 0.05 x (0.523/0.842) = 0.0311 M. I think this is way is easier to see than to go through the cnversion ratio. Then plug this into the ICE chart.

................M^3+ + L- ==> [ML]^2+
I..............0.210..0.150........0
C (fill in last)................................
E......................................0.0311

Now you can fill in the C line as
..............-0.0311...-0.0311.....+0.0311
Now the rest of the E line becomes:
E..........0.179.....0.119.............0.0311

Plug the E line into the Kc expression and solve for Kc.
Post your work if you get stuck. ....