Consider the following Reaction

Consider the following Reaction
M+ + L- ↔ ML

A reaction was initiated with 0.010 M M^+ and 0.0200 M L^- with no ML present. The equilibrium concentration of ML was determined to be 0.00500 M via a Beer’s Law experiment. Determine the equilibrium constant for this reaction.


a. 25.0

b. 4444

c. 66.7

d. 0.0150

e. 0.00400

Well, I think setting up an ICE table would be the easiest thing show how to do this, but that would be hard to do on yahoo answers, so I'll just set up the equation.

In equilibrium, K = products/reactants
Luckily enough, neither M+ or L- have a coefficient so no exponents are needed for the following equation,

K = [ML]/ ( [M+][L-] )

If you formed .005M of product in the reaction, that means that it took .005M of both M+ and L- to form the product, since every ratio in the equation is one to one. So, simply subtract .005 from both M+ and L- and stick them in the equilibrium equation.

[M+]=.01M-.005M=.005M
[L-]=.02M-.005M=.015M

K=[.005]/ ( [.005][.015] )
K=[.005]/(.000075)
K=66.6666=66.7

I believe your answer is C. I'm pretty sure this is the right way to do this.

Hope this helps.

Since everything in the reaction occurs in a 1:1:1 ratio, Keq = [ML]/[M+][L-]

Since [ML] = 0.00500 M, equilibrium concentration of M+ = 0.010 - 0.005 = 0.005. Also, [L-] = 0.0200 - 0.005 = 0.015

So, Keq = 0.005 / 0.005X0.015 = 66.7