Answer
$pKa=4.68$
Work Step by Step
Suppose bromcresol green's chemical formula can be written conveniently as $HA$.
Acid-base reaction: $HA + H_{2}O -> H^{+} + HA^{-}$
According to the Henderson-Hasselbalch equation:
$pH=pKa + log(\frac{[A^{-}]}{[HA]})$, in which $HA$ is a monoprotic acid and $A^{-}$ is its conjugate base.
It has been given that "the yellow acid and blue base forms of the indicator are present in equal concentrations". Therefore, we know that: $[HA]=[A^{-}]$, or $\frac{[A^{-}]}{[HA]}=1$.
Now, we can substitute this value into the Henderson-Hasselbalch equation:
$pH=pKa + log(\frac{[A^{-}]}{[HA]})$
$pH=pKa+log(1)$
$pH=pKa+0$ (because $10^0=1$)
$pH=pKa$
Because bromcresol green's pH value at that moment is $4.68$, it is now known that its pKa is also $4.68$.
