Use Figure 19.5 to find an indicator for these titrations:

(a) $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\text{M}}{\mathbf{\left(}{\mathbf{\text{CH}}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\text{NH\#000A64}}$(Appendix C) with$\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{HBr}}$.

(b) $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{KOH}}$ with$\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{\text{M}}\mathbf{ }\mathbf{ }{\mathbf{\text{HNO}}}_{\mathbf{3}}$.

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

This is a weak base-strong acid titration. The following chemical reactions occur.

${\left({\mathrm{CH}}_3\right)}_2\mathrm{NH}+\mathrm{HBr}\to {\left({\mathrm{CH}}_3\right)}_2\mathrm{N}{{\mathrm{H}}_2}^{+}+{\mathrm{Br}}^{-}$

${\left({\text{CH}}_3\right)}_2\text{N}{{\text{H}}_2}^{+}+{\text{H}}_2\text{O}\rightleftharpoons {\left({\text{CH}}_3\right)}_2\text{NH}+{\text{H}}_3{\text{O}}^{+}$

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

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

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

${\mathrm{K}}_{\mathrm{a}}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}{\left[{\left({\text{CH}}_3\right)}_2\text{N}{{\text{H}}_2}^{+}\right]}$

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

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \\ =\frac{1\times {10}^{14}}{5.9\times {10}^{-4}} \\ {\mathrm{K}}_{\mathrm{a}}=1.69\times {10}^{11} \end{gathered}$$

Note that all ${\left({\text{CH}}_3\right)}_2\text{NH}$is turned to ${\left({\text{CH}}_3\right)}_2{\text{NH}}_2$at equivalent point. Therefore, $\left[{\left({\text{CH}}_3\right)}_2{\text{NH}}_2\right]=0.5\text{M}$.

Solve for $\left[{\left({\text{CH}}_3\right)}_2{\text{NH}}_2\right]=0.5\text{M}$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}{\left[{\text{CH}}_3\text{N}{{\text{H}}_2}^{+}\right]} \\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\sqrt{{\text{K}}_{\mathrm{a}}\times \left[{\text{CH}}_3\text{N}{{\text{H}}_2}^{+}\right]} \\ =\sqrt{1.69\times {10}^{11}\times 0.5} \\ \left[{\text{H}}_3{\text{O}}^{+}\right]=2.91\times {10}^6 \end{gathered}$$

Lastly, solve for the pH.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ =-\log \left(2.91\times {10}^6\right) \\ \mathrm{pH}=5.53 \end{gathered}$$

Therefore, the $\mathrm{pH}=5.53$is the equivalent point. On the equivalence point or beyond the equivalence point, the most appropriate indication must change colour. Methyl red changes colour from orange to yellow near this pH, as shown in Figure 19.5 in the book.

$0.2 \text{MKOH}$with $0.2 {\text{MHNO}}_3$.

This is a titration with a strong base and a strong acid. Since they will neutralise each other to generate water, the 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.