Q19.53P - Chemistry: Molecular Nature Of Matter And Change

Question

Use Figure 19.5 to find an indicator for these titrations:

(a) $\mathbf{0}\mathbf{.25}{\mathbf{\text{M\hspace{0.17em}\hspace{0.17em}C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}\mathbf{\text{COOH}}$  (Appendix C) with  $\mathbf{0.25}\mathbf{\text{M\hspace{0.17em}\hspace{0.17em}KOH}}$.

(b) $\mathbf{0.50}{\mathbf{\text{M\hspace{0.17em}\hspace{0.17em}NHNH}}}_{\mathbf{4}}\mathbf{\text{Cl}}$  (Appendix  $\mathbf{C}$  ) with $\mathbf{0.50}\mathbf{\text{M\hspace{0.17em}\hspace{0.17em}NaOH}}$ .

Step-by-Step Solution

Answer

(a) An indicator for $0.25{\text{M\hspace{0.17em}\hspace{0.17em}C}}_6{\text{H}}_5\text{COOH}$  (Appendix C) with $0.25\text{M\hspace{0.17em}\hspace{0.17em}KOH}$  is thymol blue or phenolphthalein.

(b) An indicator for  $0.50{\text{M\hspace{0.17em}\hspace{0.17em}NHNH}}_4\text{Cl}$ (Appendix C) with $0.50\text{M\hspace{0.17em}\hspace{0.17em}NaOH}$ is Alizarin or Alizarin Yellow R.

Step 1: Definition of pH

A solution's $\mathbf{\text{pH}}$  value that measures the concentration of hydrogen ions, reveal whether a solution is acidic or alkaline.

Step 2: Find an indicator for 0 .25 M  C 6 H 5 COOH (Appendix C) with 0.25 M  KOH

a)

This is a weak acid-strong base titration. Write down the first reaction. Since this is in the equivalence point, there will be no ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$  left. So, the reaction will be  ${\text{C}}_6{\text{H}}_5\text{COO}$ reacting  with water.

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COO}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{OH}$

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

 ${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right]\left[\mathrm{OH}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COO}\right]}$

We know that $\left[{\text{C}}_6{\text{H}}_5\text{COOH}\right]=\left[\text{OH}\right]$  because they are produced with 1 mole each.

 ${\mathrm{K}}_{\mathrm{b}}=\frac{{\left[\mathrm{OH}\right]}^2}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COO}\right]}$

Solve for the  ${\mathrm{K}}_{\mathrm{b}}$ of the equation using the ${\mathrm{K}}_{\mathrm{a}}$  of  $\mathrm{HCOOH}$.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}}\\ {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}\\ =\frac{1\times {10}^{14}}{6.3\times {10}^{-5}}\\ =1.59\times {10}^{10}\end{array}$

Note that all ${\text{C}}_6{\text{H}}_5\text{COOH}$  is turned to ${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$at an equivalence point. Therefore, $\left[{\text{C}}_6{\text{H}}_5{\text{COO}}^{-}\right]=0.25\text{M}$ . Now, solve for  $\left[\text{OH}\right]$.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{b}}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\right]}\\ \left[{\mathrm{OH}}^{-}\right]=\sqrt{{\mathrm{K}}_{\mathrm{b}}\times \left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\right]}\\ =\sqrt{1.59\times {10}^{-10}\times 0.25}\\ =6.30\times {10}^{-6}.\end{array}$

Now, solve for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ .

 $\begin{array}{c}{\mathrm{K}}_{\mathrm{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[\text{OH}\right]}\\ =\frac{1\times {10}^{14}}{6.3\times {10}^{-6}}\\ =1.59\times {10}^{-9}.\end{array}$

Lastly, solve for the  $\mathrm{pH}$.

$\begin{array}{c}\mathrm{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]\\ =-\log (1.59\times {10}^{-9})\\ =\mathrm{8.80.}\end{array}$

Therefore, the equivalence point is at $\mathrm{pH}=\mathrm{8.80.}$ The most appropriate indicator must change color at the equivalence point or after the equivalence point. From Figure $19.5$  in the book, thymol blue transitions its color from yellow-green to blue or phenolphthalein transitions from clear to pink near this pH.

Step 3: Find an indicator for 0 .50 M  NHNH 4 Cl (Appendix C ) with 0.50 M  NaOH

b)

This is a weak acid-strong base titration. First, write the reaction that occurs. The  ${\text{NH}}_4\text{Cl}$ is a salt and will dissolve first.

${\text{NH}}_4\text{Cl}\to {\text{NH}}_4^{+}+{\text{Cl}}^{-}$

Since this is in the equivalence point, no  ${\text{NH}}_4^{+}$ will be left. So, the reaction will be the ${\text{NH}}_3$  reacting with water.

 ${\text{NH}}_3+{\text{H}}_2\text{O}\rightleftharpoons {\text{NH}}_4^{+}+{\text{OH}}^{-}$

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

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\text{NH}}_4^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\mathrm{N}{\text{H}}_3\right]}$

We know that  $\left[{\text{NH}}_4^{+}\right]=\left[{\text{OH}}^{-}\right]$ because they are produced with 1 mole each.

${\mathrm{K}}_{\mathrm{b}}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[{\mathrm{NH}}_3\right]}$

Note that all  ${\text{NH}}_4^{+}$ is turned to ${\text{NH}}_3$  at an equivalence point. Therefore, $\left[{\text{NH}}_3\right]=0.50\text{M}$ . Now, solve for  $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{b}}=\frac{{\left[{\text{OH}}^{-}\right]}^2}{\left[{\text{NH}}_3\right]}\\ \left[{\text{OH}}^{-}\right]=\sqrt{{\mathrm{K}}_{\mathrm{b}}\times \left[{\text{NH}}_3\right]}\\ =\sqrt{1.76\times {10}^{-5}\times 0.50}\\ =2.97\times {10}^{-3}.\end{array}$

Now, solve for  $\left[{\text{H}}_3{\text{O}}^{+}\right]$.

 $\begin{array}{c}{\mathrm{K}}_{\mathrm{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[\text{OH}\right]}\\ =\frac{1\times {10}^{14}}{2.97\times {10}^{-3}}\\ =3.37\times {10}^{-12}\end{array}$

Lastly, solve for the $\mathrm{pH}$ .

$\begin{array}{c}\mathrm{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]\\ =-\log (3.37\times {10}^{-12})\\ =\mathrm{11.47.}\end{array}$

The equivalence point is at $\mathrm{pH}=11.47$ . The most appropriate indicator must change color on the equivalence point or after the equivalence point. From Figure   in the book, Alizarin transitions its color from pink to blue or Alizarin Yellow  $\mathrm{R}$ transitions from yellow to pink near the pH 11.47.