How to prove at what pH the color of an indicator will change based on its pKa? The equivalence point of an acid-base titration can be determined by choosing the appropriate colored indicator. This indicator is a weak acid $\ce{HA}$. The undissociated acid has a different color than it's ion $\ce{A-}$. The $\mathrm{p}K_\mathrm{a}$ value of a colored indicator is given to be $\mathrm{p}K_\mathrm{a}=6$. At what value of the $\mathrm{pH}$  will the color change? Please explain your answer. (Assume that both colors are equally well recognizable).


I've had this question on an exam and I can't seem to figure out what exactly the answer could be.

I would intuitively say that the color change happens at $\mathrm{pH}=6$. If the $\mathrm{pH}$ is lower than 6 then there are less $\ce{A-}$ ions and more $\ce{HA}$ (acid-indicator) in the solution. Hence the solution will take on the color of the undissociated acid-indicator. If the pH is higher than 6 then the $\ce{HA}$ starts to dissociate more to $\ce{H3O+}$ hence there being a higher $\ce{A-}$ concentration and the solution takes the color of the indicator's conjugate base. But I'm not sure if my thought process is right or how to 'prove' it's right.
Thank you guys for your answers! I am fairly confident I have found the correct answer, or at least one of the possible correct answers. Here's it goes:

The indicator reacts with water according to the following reaction equation:
$$\ce{HA + H2O &lt;=&gt; H3O+ + A-}$$
Then using the Henderson-Hasselbach equation, I arrive at the following point:
 $$ K_s = \frac{[\ce{H3O+}]*[\ce{A-}]}{[\ce{HA}]} $$
Now, at the equivalence point, the concentration of the indicator an it's conjugate base are equal, hence:
 $$\require{cancel} K_s = \frac{[\ce{H3O+}]*\cancel{[\ce{A-}]}}{\cancel{[\ce{HA}]}} $$
$K_s = [\ce{H3O+}]$. After taking the negative log of both sides I arrive at the equation: $pK_s = pH$ (which was given to be 6).

Now, when the solution is acidic(pH lower than 6 and high [$\ce{H3O+}$] concentration), Le Chatelier sais then the balance will shift to the left of the top most equation. Favoring the indicator $\ce{HA}$ as opposed to it's conjugate base. Naturally, if the solution contains more indicator molecules, then it will also take on the indicators color. 


 Analogue to the previous reasoning, when the pH reaches values above those of 6, Le Chatleier sais that the reaction will favor the formation of the indicators conjugate base [$\ce{A-}$](the right side) and the solution takes on the color accordingly.