Understanding colour changes of systems in dynamic equilbrium
In the general case of $\ce{A + B &lt;=&gt;C}$ if  there are initially $a_0, b_0,c_0$ amounts present , then if an amount x of A and B react to produce C then the equilibrium constant $K_e$ is given by
$$ K_e=\frac{k_1}{k_{2}}=\frac{c_0+x_e}{(a_0-x_e)(b_0-x_e)}$$
where $k_1, k_2$ are the rate constants for the forward and reverse reaction and $x_e$ is the amount reacted at equilibrium.
Solving the equation and  for (algebraic) simplicity let $c_0=0$, gives the amount at equilibrium
$$x_e=\frac{1}{2}\left(a_0+b_0+K_e^{-1}-\sqrt{(a_0-b_0)^2+2(a_0+b_0)/K_e +K_e^{-2}}\right)$$
Using this the amount of each species can be obtained as for example the equilibrium amount of A is $a_e=a_0-x_e$. However, in your question you ask about the colour not the concentrations and this alters things a little. The quantity that is measured optically as in a spectrophotometer is the optical density of for example species A is $D=\epsilon_\lambda[A]l$ where $\epsilon_\lambda$ is the extinction coefficient of the molecule at wavelength $\lambda $ and is effectively how strongly the molecule absorbs at that wavelength, $\ce{[A]}$ is the concentration and l the path-length of the sample through which the light passes.  

The total absorption is the sum of that for all species so depending on the values of $\epsilon$ and how the concentrations change the absorption may go up, down or remain constant at any given wavelength. 

Furthermore, if you do not use a spectrophotometer and measure optical density but you are looking through the sample you will observe the transmitted light which is by the Beer-Lambert law given by $I_{trans}=I_0e^{-D}$ for each species present at each wavelength, so what you may see becomes pretty complicated.