Three test tubes contain an identical solution of unknown pH.
- The first one is tested with thymolphthalein and turns colourless
- 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:
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}
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}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:
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.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?