Identifying unknown solution with indicators. why is one pKa value ignored and how to treat negatives values?

Question

Three test tubes contain an identical solution of unknown pH.

1. The first one is tested with thymolphthalein and turns colourless
2. The second is tested with a-naphtholphthalein and turns blue

What colour would thmyol blue appear in the third tube?

Note: If the ratio of one form to the other form (e.g. acid to base form or vice versa) is greater than $10:1$, the colour observed will be that of the predominant form. However, if between $10$ and $0.1$, the colour will be a mixture.

\begin{array}{lccc} \text{Indicator} & \mathrm{p}K_\mathrm{a} & \text{Colour of acid form} & \text{Colour of base form} \\ \hline \text{$\alpha$-naphtholphthalein} & 8 & \text{yellow} & \text{blue} \\ \text{thymol blue} & 2 & \text{red} & \text{yellow} \\ & 8.8 & \text{yellow} & \text{blue} \\ \text{thymolphthalein} & 10 & \text{colourless} & \text{blue} \\ \hline \end{array}

Steps to solve this question:

1. Find pH of solution via Henderson-Hasselbalch equation: \begin{align} \mathrm{p}K_\mathrm{a} &= \ce{pH} - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]} \end{align}

2. For α-naphtholphthalein, as it appears blue, its basic form dominates and so
   \begin{align} \frac{[\ce{In-}]}{[\ce{HIn}]} &> 10:1 \\ \mathrm{p}K_\mathrm{a} &= 8 = \ce{pH} - \log_{10} (10) = \ce{pH} - 1\\ \ce{pH} &= 9 \end{align}

3. For thymol blue \begin{align} \mathrm{p}K_\mathrm{a} &= \ce{pH} - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}\\ 8.8 &= 9 - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}\\ \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]} &= -0.2 \end{align}

Where I get stuck:

1. How do I handle the negative sign in negative 0.2?
   If it was positive 0.2, this would tell me that the ratio of $\log_{10} ([\ce{In-}]/[\ce{HIn}])$ is between $10$ and $0.1$, hence a mix between the acid colour and basic colour would result, giving a green solution. But I don't know how to treat the negative.

2. In my solution manual they ignore the $\mathrm{p}K_\mathrm{a}=2.0$ value of the thymol blue and only use the $\mathrm{p}K_\mathrm{a}=8.8$ value. Why is this the case?

Answer 1

As you noted, the $\mathrm{pH}$ is fairly close to 9.

It's probably quite different from the $\mathrm{p}K_{\mathrm{a}}$ value of $\alpha$-naphtholphthanlein because if you have any of the yellow acidic form around, the solution will take on a greenish tinge. One order of magnitude seems sufficient so, $\mathrm{pH} \geq 9$

The sample with thymolphtalein follows the same logic in that since we don't see any blue, there must be little of the base form, so the $\mathrm{pH} \leq 9$.

9 is very close to 8.8, the $\mathrm{pH}$ where we would expect equal forms of thymol blue. Indeed, $10^{0.2}\approx 1.05$, so both forms are present in solution in roughly equal amounts.

Further assuming that intensity of the two colors is similar, you would have a mix of yellow and blue, also known as green.