Use Figure 19.5 to find an indicator for these titrations:

(a) $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}{\mathbf{\text{MCH}}}_{\mathbf{3}}{\mathbf{\text{NH}}}_{\mathbf{2}}$(Appendix C) with$\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{\text{MHCl}}$.

(b) $\mathbf{\text{0.50M}}$HI with$\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{\text{MKOH}}$.

pH is a measure of hydrogen ion concentration, which indicates whether a solution is acidic or alkaline.

This is a titration with a weak base and a strong acid. Write down the reaction that occurs first. There will be no${\mathrm{CH}}_3{\mathrm{NH}}_2$remaining because we are at the equivalence point. As a result, the reaction will be$\mathrm{CH}{ }_3{\text{NH}}_3^{+}$reacting with water.

${\text{CH}}_3{\text{NH}}_3^{+}+{\text{H}}_2\text{O}\rightleftharpoons {\text{CH}}_3{\text{NH}}_2+{\text{H}}_3{\text{O}}^{+}$

Write the${\mathrm{K}}_{\mathrm{a}}$expression.

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\text{CH}}_3{\text{NH}}_2\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[{\text{CH}}_3{\text{NH}}_3^{+}\right]}$

We know that $\left[{\text{CH}}_3{\text{NH}}_2\right]=\left[{\text{H}}_3{\text{O}}^{+}\right]$because they are produced with 1 mole each.

${\mathrm{K}}_{\mathrm{a}}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}{\left[{\text{CH}}_3{\text{NH}}_3^{+}\right]}$

Solve for the ${\mathrm{K}}_{\mathrm{a}}$of the equation using the ${\text{K}}_b$of ${\text{CH}}_3{\text{NH}}_2$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \\ =\frac{1\times {10}^{-14}}{4.4\times {10}^{-4}} \\ {\mathrm{K}}_{\mathrm{a}}=2.27\times {10}^{-11} \end{gathered}$$

Note that all ${\text{CH}}_3{\text{NH}}_2$is turned to ${\text{CH}}_3{\text{NH}}_3$at equivalent point Therefore, $\left[{\text{CH}}_3{\text{NH}}_3\right]=0.10\text{M}$. Solve for $\left[{\text{H}}_3\text{O}\right]$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{{\left[{\mathrm{H}}_3\mathrm{O}\right]}^2}{\left[{\mathrm{CH}}_3{\mathrm{NH}}_3\right]} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{\top }\right]=\sqrt{{\mathrm{K}}_{\mathrm{a}}\times \left[{\mathrm{CH}}_3{\mathrm{NH}}_3\right]} \\ =\sqrt{2.27\times {10}^{-11}\times 0.10} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{-}\right]=1.51\times {10}^{-6} \end{gathered}$$

Lastly, solve for the pH.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\text{H}}_3\text{O}\right] \\ =-\log \left(1.51\times {10}^{-6}\right) \\ \mathrm{pH}=5.82 \end{gathered}$$

Therefore, the equivalence point is at $\mathrm{pH}=5.82$. The most appropriate indicator must change color on the equivalence point or after the equivalence point. From Figure 19.5 in the book, methyl red transitions its color from orange to yellow near this pH.

This is a titration with a strong acid and a strong base. Since they will neutralise each other to generate water, the pH at the equivalence point is $7.00$. On the equivalence point or beyond the equivalence point, the most appropriate indication must change colour.

Therefore, Bromthymol blue changes its color from yellow green to blue in Figure 19.5 of the book.