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=
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. ....
To calculate the equilibrium concentration of [ML]2+ in the trial solution, you need to use the following formula:
[ML]2+ equilibrium = (abs[ML]2+ at equilibrium) / (abs [ML]2+std * (concentration [ML]2+ std))
Using the given data:
abs[ML]2+ at equilibrium = 0.523
abs[ML]2+std = 0.842
[ML]2+std = 0.0500 M
Now, substitute the values into the formula:
[ML]2+ equilibrium = (0.523) / (0.842 * 0.0500)
Calculate the result:
[ML]2+ equilibrium = 12.43 M
To calculate Kc, we need to complete the ICE table using the given data:
Rows of table: [M3+], [L-], [ML]2+
Columns of table: I, C, E
I = Initial concentration
C = Change in concentration
E = Equilibrium concentration
[M3+], initial = 0.210 M
[L-], initial = 0.150 M
[ML]2+, initial = 0 M
For the change in concentration, we need to consider the stoichiometry of the reaction. From the balanced equation, the stoichiometry tells us that for every 1 mole of [M3+], 1 mole of [ML]2+ is formed. Similarly, for every 1 mole of [L-], 1 mole of [ML]2+ is formed. So, the changes in concentrations are:
[M3+], change = -x (since 1 mole of [ML]2+ is formed for every mole of [M3+])
[L-], change = -x (since 1 mole of [ML]2+ is formed for every mole of [L-])
[ML]2+, change = +x (stoichiometry consideration)
Now, calculate the equilibrium concentrations:
[M3+], equilibrium = [M3+], initial - x
[L-], equilibrium = [L-], initial - x
[ML]2+, equilibrium = [ML]2+, initial + x
Using the given data, substitute the initial concentrations into the equilibrium expressions:
[M3+], equilibrium = 0.210 - x
[L-], equilibrium = 0.150 - x
[ML]2+, equilibrium = 0 + x
Now, substitute the values into the equation:
Kc = ([ML]2+, equilibrium) / ([M3+], equilibrium * [L-], equilibrium)
Kc = (x) / ((0.210 - x) * (0.150 - x))
However, we cannot determine the value of Kc with just a single trial run of the equilibrium reaction. To calculate Kc accurately, multiple trials are required with different initial concentrations.