# Question #fdd69

##### 1 Answer
Aug 19, 2015

The answer is (a) yellow, green, blue.

#### Explanation:

Bromothymol blue acts as a weak acid in solution.
You can write the equilibrium reaction for the ionization of any acid indicator like this

${\text{HIn"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq]) + "In}}_{\textrm{\left(a q\right]}}^{-}$

The color of the solution will depend on whether the weak acid form, $\text{HIn}$, or the conjugate base form, ${\text{In}}^{-}$, is the dominant form.

In other words, when the acid is mostly undissociated, i.e. you have more $\text{HIn}$ than ${\text{In}}^{-}$ present, the color of the solution will be yellow.

When the acid id mostly dissociated, i.e. you have more ${\text{In}}^{-}$ than $\text{HIn}$ present, the color of the solution will be blue.

To determine what pH will cause the solution to change colors, take a look at the expression of the acid dissociation constant, ${K}_{a}$

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["In"^(-)])/(["HIn}\right]\right)$

When this equilibrium is established, the pH of the solution will be equal to the $p {K}_{a}$. Now, when the pH drops below $p {K}_{a}$, the equilibrium will shift to the left.

That happens because a lower pH means a higher concentration of hydronium ions, ${\text{H"_3"O}}^{+}$. The excess hydronium ions will protonate the conjugate base form and form the weak acid, $\text{HIn}$.

The dominant species in the solution will thus be $\text{HIn}$, which implies that the color of the solution will be yellow.

On the other hand, when pH exceeds $p {K}_{a}$, the concentration of the hydronium ions will decrease.

This means that the equilibrium will shift right so that more acid molecules will give off their proton to form hydronium ions and the conjugate base ${\text{In}}^{-}$.

If the pH is high enough, like you have here (a few units over $p {K}_{a}$), the conjugate base will become the dominant form, so the color of the solution will be blue.

Depending on the indicator, at $p {H}_{\text{sol}} = p {K}_{a} \pm 1$ the solution will be a mix of the two colors associated with acid form and conjugate base form. In your case, the solution will have a green color at $p {H}_{\text{sol}} = p {K}_{a}$.

Here's a video illustrating the bromothymol blue solution color changes