The acid-base indicator methyl red has a $K_{a}$ = $1 \times 10^{-5}$. Its acidic form is red while its alkaline form is yellow. If methyl red is added in a colorless solution with a $pH = 7$, the color will be :





Seeing the $K_{a}$ = $1 \times 10^{-5}$, I concluded that it is a weak acid. Thus I think that when it is added to a neutral solution ($pH=7$), the resultant solution must be weakly acidic and thus its color must be red. However, contrary to my logic, the answer given in my book is option d) Yellow. Is there a flaw in my argument? Please Help!


A pH indicator is generally a weak acid and often work in a fashion similar to buffers in that most of the time both the acid and its conjugate base exist in significant concentration. From the Ka of the indicator you should therefore be able to find the "halfway" point of the indicator or the middle of the range of pH values over which the indicator operates.

Think about how indicators work. Indicators work simply by mixing colors. If I have an indicator which is black at pH = 0.0 and white at pH = 2.0, then I know that the protonated form of the indicator must be black-colored, and that the deprotonated form of the indicator must be white colored. This should make sense. As pH increase, hydroxide ion molarity increases, and hydroxide ion deprotonates other molecules.

Thus, at pH = 1.0, we should have a 50/50 mix of black and white molecules. In other words, the indicator shows a gray color at pH = 1.0. At pH = 1.0 we also have a buffer solution - we have the acid and conjugate base - and specifically in a 1:1 ratio.

By application of the Henderson-Hasselbach equation we will realize:

$\ce{pH = pK_{a} + log\frac{base}{conjugate~acid}}$ simplifies to ...

$\ce{pH = pK_{a}}$

I'll let you take the problem from here. Don't forget to apply common-sense considerations.


It's an indicator problem, so trying to balance the pH seems a bit off, since you typically add only a few drops of indicator to determine a solution's pH. Ideally I'd imagine you'd want the indicator to interfere as little as possible with whatever's going on.

Also, different indicators have different ranges. (They don't have to change color at exactly ${pH=7}$.) You're given that the acidic color is red, and the basic color is yellow. The ${K_a}$ helps to determine this switching point:

Regarding simple indicators, if the solution's pH is lower than the indicator's ${pK_a}$ you'll see the acidic color. Vice versa for the basic color.

The ${pK_a}$ that corresponds to ${K_a=1*10^{-5}}$ is simply ${5}$. As ${7 > 5}$, you'd expect the alkaline color to appear (in this case, yellow)


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