Q19.50P - Chemistry: Molecular Nature Of Matter And Change

Question

Use Figure 19.5 to find an indicator for these titrations:

(a) $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{HCl}}$ with $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{NaOH}}$.

(b) $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{HCOOH}}$  (Appendix C) with $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{NaOH}}$.

Step-by-Step Solution

Answer

(a) An indicator for $0.10\text{M} \text{HCl}$  with $0.10\text{M} \text{NaOH}$ is titrations bromthymol blue.

(b) An indicator for $0.10\text{M} \text{HCOOH}$ with $0.10\text{MNaOH}$ is thymol blue or phenolphthalein.

Step 1: Definition of pH

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

Step 2: Subpart (a)

$0.10\text{MHCl}$ with $0.10\text{MNaOH}$.

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, Bromothymol blue changes hue from yellow green to blue in Figure 19.5 of the book.

Step 3: Subpart (b)

$0.10\text{M} \text{HCOOH}$ with $0.10\text{M} \text{NaOH}$.

This is a weak acid-strong base titration. First, write the reaction that occurs. Since this is in the equivalence point, there will be no H C O O H left. 

So, the reaction will be ${\text{HCOOO}}^{-}$ reacting with water.

${\text{HCOO}}^{-}+{\text{H}}_2\text{O}\rightleftharpoons \text{HCOOH}+{\text{OH}}^{-}$ 

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

 ${\mathrm{K}}_{\mathrm{b}}=\frac{\left[\mathrm{HCOOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\mathrm{HCOO}}^{-}\right]}$

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

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

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

${\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b} } \left(1\right)$

${\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}} \left(2\right)$

    $=\frac{1\times {10}^{-14}}{1.8\times {10}^{-4}} \left(3\right)$

${\mathrm{K}}_{\mathrm{b}}=5.56\times {10}^{-11} \left(4\right)$

Note that all $\mathrm{HCOOH}$ is turned to ${\mathrm{HCOO}}^{-}$ at equivalent point Therefore,  $\left[{\mathrm{HCOO}}^{-}\right]=0.10\text{M}$. Solve for  $\left[{\text{OH}}^{-}\right]$.

      $K_b=\frac{{\left[{\text{OH}}^{-}\right]}^2}{\left[{\text{HCOO}}^{-}\right]} \left(5\right)$

$\left[{\text{OH}}^{-}\right]=\sqrt{{\mathrm{K}}_{\mathrm{b}}\times \left[\text{HCOO}\right]} \left(6\right)$

          $=\sqrt{5.56\times {10}^{-11}\times 0.10} \left(7\right)$

$\left[{\text{OH}}^{-}\right]=2.36\times {10}^{-6} \left(8\right)$

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

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right] \left(9\right) \\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{K_w}{\left[{\text{OH}}^{-}\right]} \left(10\right) \\ =\frac{1\times {10}^{14}}{2.36\times {10}^{-6}} \left(11\right) \\ \left[{\text{H}}_3{\text{O}}^{+}\right]=4.24\times {10}^{-9} \left(12\right) \end{gathered}$$

Lastly, solve for the pH.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\text{O}}^{+}\right] \\ =-\log \left(4.24\times {10}^{-9}\right) \\ \mathrm{pH}=8.37 \end{gathered}$$

At pH=8.37, the equivalence point is reached. On the equivalence point or beyond the equivalence point, the most appropriate indication is it must change color. 

Therefore, near this pH, thymol blue transitions from yellow green to blue, while phenolphthalein transitions from clear to pink, as seen in Figure 19.5 in the book.