Dissociation constant for Indicators So I came across a question which asked to calculate the $\mathrm{pH}$ at which an indicator of given $\mathrm{K_{a}}$ would change colour. The solution was given in the following way
Since when the color change occurs $\ce{[HIn]=[In^{-}]}$,
As $\mathrm{K_a}= \left(\frac{[\ce{H^+} ][\ce{In^-}]}{[\ce{HIn}]}\right)$,
At $\mathrm{K_a}=[\ce{H+}]$ the colour change would occur.
And consequently from the $\ce{H+}$ concentration the $\mathrm{pH}$ was calculated, the solution imposes that during the colour change the equilibrium is attained, but how can that be since to observe colour change you add an acid or base which disturbs the initial equilibrium for which the $\mathrm{K_{a}}$ was defined and hence the now won't hold true.
First of all, the indicator does not change colour instantaneously at $\mathrm{pH = pK_a}$. The indicator is a different colour than it's conjugate base, and exists in an equilibrium with it in solution. For the colour change to be noticed, the concentration of the base has to be at least 10 times more than that of the conjugate acid, or vice versa. This is why there is a range of colour change(look at these pH charts), with the upper and lower bounds having a $\mathrm{pH}$ difference of $2$. As an example, here's a simplified reaction for Phenolphthalein ($\ce{Ph}$ is the phenolphthalein anion)
$$\ce{\underset{\text{colourless}}{H2Ph} &lt;=&gt; 2H+ + \underset{\text{violet}}{Ph^2-}}$$
If you take a simpler, weak monobasic acid indicator $\ce{HIn &lt;=&gt; H+ + In-}$, the $\mathrm{K_a}$ is
$$\mathrm{K_a} = \frac{\ce{[H+][In^-]}}{\ce{[HIn]}}$$
The relative concentration of the indicator and it's conjugate base is only determined by the concentration of the $\ce{H+}$ ion, as $\mathrm{K_a}$ is a constant.
Answering your second query, the author takes the condition of equilibrium, simply because the system is in equilibrium, that is $\mathrm{Q_a = K_a}$. It is assumed that the reaction is instantaneous or that titration is performed slowly enough so that the system is in equilibrium at every single instant.