# How to predict the color of a pH indicator using Le Chatelier's principle The question I have is above. Could someone explain why the answer is (A)?

What I think so far:

• Does the sodium hydroxide turn the indicator blue because it is a strong base and hence it will react completely to form water, shifting equilibrium really far to the right?

• Does the HCl turn the indicator red for a similar reason?

• I am unsure of the sodium acetate effect
• Please note that the homework policy on this site requires you to show some personal effort. Mar 7 '15 at 6:40
• IMHO, the equations in the question have been incorrectly written. This is not chemistry, it's rather confusing and playing mean tricks on students. In the alkaline, H3O+ does not exist. It should read H2Z + OH- <--> HZ- + H2O etc. Apr 26 '20 at 7:13

You're absolutely on the right track!

The keys to the question are

1. to order $\ce{HCl}$, $\ce{NH4Cl}$, $\ce{NaOAc}$ and $\ce{NaOH}$ in terms of strong acid, weak acid, weak base and strong base

2. to realize that in the presence of $\ce{HCl}$, the indicator exists as $\ce{H2Z}$, while it is fully deprotonated to $\ce{Z^{2-}}$ in the presence of $\ce{NaOH}$

3. in the presence of weak acids or weak bases, the indicator exists in form of each two species. Consequently, a mixed colour is observed.

In the presence of $\ce{NaOAc}$, a weak base, $\ce{H2Y}$ is completely deprotonated, thus no red colour is observed. In addition, a part of $\ce{HY-}$ is deprotonated too: the second equilibrium isn't completely on the left side. Some amount of $\ce{Y^{2-}}$ is present too. As a result, the green colour (yellow + blue) is observed.

It is all about the position of the equilibria.

• Alright! I understand points 1 and 2, but why does the indicator exist as two species in the presence of weak base? I think I may have some sort of idea, but I would prefer it if you gave a definite explanation. Thankyou! Mar 7 '15 at 6:57
• @chemistryyo I've tried to address the last point too. Mar 7 '15 at 7:06
• Sorry, I just have another question: How do we know that H2Y is completely deprotonated. If this was a single equilibrium reaction, there should be some H2Y present Mar 7 '15 at 7:13