Given Kd find equilibrium concentrations of the reactants


> 
> $$\ce{Biotin + Avidin → Complex}$$
> 
> 
> Biotin is a vitamin derivative that binds extremely tightly to avidin (a protein found in egg white) with a $K\_\mathrm{d}$ of $\pu{1e-15 M}$.
> 
> 
> If you make a $\pu{2 mL}$ solution of $\pu{10 μL}$ biotin and $\pu{2 μL}$ avidin, what fraction of biotin molecules will remain free at equilibrium?
> 
> 
> 


### Answer Key


The TA found the


$$K\_\mathrm{d,new} = \frac{[\text{Biotin}][\text{Avidin}]}{[\text{Complex}]}$$


using the given initial reactant concentration values and making an ICE table to find the equilibrium concentrations and found $K\_\mathrm{d,new} = \pu{1e-9}.$


$$K\_\mathrm{d} = \frac{(10 - δ)(2 - δ)}{δ} = 10^{-9}$$


Where does the TA get that $K\_\mathrm{d} = \pu{1e-9 μL}$? Don't you need to either know the delta term or have the value of $ΔG$ of reaction? Neither of which was given in the problem statement. Also, why did TA find another $K\_\mathrm{d,new}$? Doesn't the provided $K\_\mathrm{d}$ already describe the dissociation, which gives the concentration of reactants on top?


I understand that $K\_\mathrm{d}$ is the equilibrium constant which describes dissociation.


### My solution


My solution was to find


$$f\_\mathrm{empty} = \frac{[\ce{RL}]}{[\ce{R}] + [\ce{RL}]}$$


I rewrote the $f\_\mathrm{empty}$ relation to be in terms of the known values, the provided $K\_\mathrm{d}$ and the concentration values.


I found $f\_\mathrm{empty}$ to be $0.0009999$ which is also essentially zero, meaning that $f\_\mathrm{bound}$ is essentially $1$, which is also what the TA found.


The TA's approach wasn't intuitive for me, so if someone could help clarify it?


