Question

The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution:

$$\begin{gathered} \mathrm{HIn}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{In}}^{-} \\ \mathrm{color}1\mathrm{color}2 \end{gathered}$$

The following absorbance data were obtained for a 5.00 x 10^-4 M solution of HIn in 0.1 M NaOH and 0.1 M HCl. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells.

In the NaOH solution, essentially all of the indicator is present as In^-; in the acidic solution, it is essentially all in the form of HIn.

(a) Calculate molar absorptivities for In2 and HIn at 485 and 625 nm.

(b) Calculate the acid dissociation constant for the indicator if a pH 5.00 buffer containing a small amount of the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells).

(c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of 0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)?

(d) A 25.00-mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the base was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the indicator under consideration, the absorbance was found to be 0.333 at 485 nm and 0.655 at 625 nm (1.00-cm cells). Calculate the pH of the solution and Ka for the weak acid.

(e) What would be the absorbance of a solution at 485 and 625 nm (1.50-cm cells) that was 2.00 3 1024 M in the indicator and was buffered to a pH of 6.000?

Short answer

(a) At 484 nm,

$$\begin{gathered} {\epsilon }_{I{n}^{-}}=150\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \\ {\mathrm{\epsilon }}_{\mathrm{HIn}}=974\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

At 625 nm,

$$\begin{gathered} {\epsilon }_{I{n}^{-}}=1808\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \\ {\mathrm{\epsilon }}_{\mathrm{HIn}}=362\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

(b) $K_a=1.88\times {10}^{-6}$

(c) pH=4.55

(d) pH=5.73

$K_a=1.85\times {10}^{-6}$

(e)role="math" localid="1650432922359" $$\begin{gathered} A_{485}=0.131 \\ {A}_{625}=0.391 \end{gathered}$$

Step-by-step solution

Part (a): Given Information

Molar absorptivities for In- and HIn at 485 and 625 nm should be calculated.

Part (a): Explanation

At 484 nm,

$$\begin{gathered} 0.075={\mathrm{\epsilon }}_{{\mathrm{In}}^{-}}\times 1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L} \\ {\mathrm{\epsilon }}_{{\mathrm{In}}^{-}}=\frac{0.075}{1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L}}=150\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \\ 0.487={\mathrm{\epsilon }}_{\mathrm{HIn}}\times 1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L} \\ {\mathrm{\epsilon }}_{H\mathrm{In}}=\frac{0.487}{1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L}}=974\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

At 625 nm,

$$\begin{gathered} 0.904={\mathrm{\epsilon }}_{{\mathrm{In}}^{-}}\times 1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L} \\ {\mathrm{\epsilon }}_{{\mathrm{In}}^{-}}=\frac{0.904}{1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L}}=1808\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \\ 0.181={\mathrm{\epsilon }}_{H\ln }\times 1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L} \\ {\mathrm{\epsilon }}_{H\mathrm{In}}=\frac{0.181}{1\mathrm{cm}\times 5\times {10}^{-4}\mathrm{mol}/\mathrm{L}}=362\mathrm{L}.{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

Part (b) : Given Information

The acid dissociation constant of the indicator should be calculated.

Part (b): Explanation

At 485 nm

$150\left[{\mathrm{In}}^{-}\right]+974\left[\mathrm{HIn}\right]=0.567\to \left(1\right)$

At 625 nm

$1808\left[{\mathrm{In}}^{-}\right]+362\left[\mathrm{HIn}\right]=0.395\to \left(2\right)$

Solving the above equations as:

$$\begin{gathered} \left(1\right)\times 1808-\left(2\right)\times 150 \\ 1760992\left[\mathrm{HIn}\right]-54300\left[\mathrm{HIn}\right]=1025.136-59.25 \\ 1760992\left[\mathrm{HIn}\right]=965.886 \\ \left[\mathrm{HIn}\right]=5.66\times {10}^{-4}\mathrm{mol}/\mathrm{L} \end{gathered}$$

Substituting to (1)

$$\begin{gathered} 150\left[{\mathrm{In}}^{-}\right]+974\times 5.66\times {10}^{-4}=0.567 \\ 150\left[{\mathrm{In}}^{-}\right]=0.567-0.551 \\ 150\left[{\mathrm{In}}^{-}\right]=0.016 \\ \left[{\mathrm{In}}^{-}\right]=1.06\times {10}^{-4}\mathrm{mol}/\mathrm{L} \end{gathered}$$

data-custom-editor="chemistry" $\mathrm{HIn}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{In}}^{-}$

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ 5.00=-\log \left[H^{+}\right] \\ \left[H^{+}\right]=1\times {10}^{-5}mol/L \end{gathered}$$

$$\begin{gathered} K_a=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]} \\ {K}_a=\frac{1\times {10}^{-5}\times 1.06\times {10}^{-4}}{5.66\times {10}^{-4}} \\ {K}_a=1.88\times {10}^{-6} \end{gathered}$$

Part (a): Given Information

pH of the solution should be calculated.

Part (b): Explanation

At 485 nm

$150\left[{\mathrm{In}}^{-}\right]+974\left[\mathrm{HIn}\right]=0.492\to \left(1\right)$

At 625 nm

$1808\left[{\mathrm{In}}^{-}\right]+362\left[\mathrm{HIn}\right]=0.245\to \left(2\right)$

$$\begin{gathered} \left(1\right)\times 1808-\left(2\right)\times 150 \\ 1760992\left[\mathrm{HIn}\right]-54300\left[\mathrm{HIn}\right]=889.536-36.75 \\ 1760992\left[\mathrm{HIn}\right]=852.786 \\ \left[\mathrm{HIn}\right]=5.00\times {10}^{-4}\mathrm{mol}/\mathrm{L} \end{gathered}$$

Substituting to (1)

$$\begin{gathered} 150\left[{\mathrm{In}}^{-}\right]+974\times 5.00\times {10}^{-4}=0.492 \\ 150\left[{\mathrm{In}}^{-}\right]=0.492-0.487 \\ 150\left[{\mathrm{In}}^{-}\right]=5\times {10}^{-3} \\ \left[{\mathrm{In}}^{-}\right]=3.33\times {10}^{-5}\mathrm{mol}/\mathrm{L} \end{gathered}$$

$$\begin{gathered} K_a=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]} \\ 1.88\times {10}^{-6}=\frac{\left[H^{+}\right]\times 3.33\times {10}^{-5}}{5.00\times {10}^{-4}} \\ \left[H^{+}\right]=\frac{1.88\times {10}^{-6}\times 5.00\times {10}^{-4}}{3.33\times {10}^{-5}} \\ \left[H^{+}\right]=2.82\times {10}^{-5}mol/L \end{gathered}$$

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ pH=-\log (2.82\times {10}^{-5}) \\ pH=4.55 \end{gathered}$$

Part (a): Given Information

The pH of the solution and Kafor the weak acid should be determined.

Part (a): Explanation

At 485 nm

$150\left[{\mathrm{In}}^{-}\right]+974\left[\mathrm{HIn}\right]=0.333\to \left(1\right)$

At 625 nm

$1808\left[{\mathrm{In}}^{-}\right]+362\left[\mathrm{HIn}\right]=0.655\to \left(2\right)$

$$\begin{gathered} \left(1\right)\times 1808-\left(2\right)\times 150 \\ 1760992\left[\mathrm{HIn}\right]-54300\left[\mathrm{HIn}\right]=602.064-98.25 \\ 1760992\left[\mathrm{HIn}\right]=503.814 \\ \left[\mathrm{HIn}\right]=2.95\times {10}^{-4}\mathrm{mol}/\mathrm{L} \end{gathered}$$

Substituting to (1)

$$\begin{gathered} 150\left[{\mathrm{In}}^{-}\right]+974\times 2.95\times {10}^{-4}=0.333 \\ 150\left[{\mathrm{In}}^{-}\right]=0.333-0.288 \\ 150\left[{\mathrm{In}}^{-}\right]=0.045 \\ \left[{\mathrm{In}}^{-}\right]=3\times {10}^{-4}\mathrm{mol}/\mathrm{L} \end{gathered}$$

$$\begin{gathered} K_a=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[HIn\right]} \\ 1.88\times {10}^{-6}=\frac{\left[H^{+}\right]\times 3\times {10}^{-4}}{2.95\times {10}^{-4}} \\ \left[H^{+}\right]=\frac{1.88\times {10}^{-6}\times 2.95\times {10}^{-4}}{3\times {10}^{-4}} \\ \left[H^{+}\right]=1.85\times {10}^{-6}mol/L \end{gathered}$$

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ pH=-\log (1.85\times {10}^{-6}) \\ pH=5.73 \end{gathered}$$

Since the HX solution is half neutralized $\left[\mathrm{HX}\right]=\left[{\mathrm{X}}^{-}\right]$

$$\begin{gathered} K_a=\frac{\left[H^{+}\right]\left[X^{-}\right]}{\left[HX\right]} \\ {K}_a=1.85\times {10}^{-6} \end{gathered}$$

Part (a) Given Information

The absorbance of a solution at 485 and 625 nm should be determined.

Part (a) Explanation

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ 6.00=-\log \left[H^{+}\right] \\ \left[H^{+}\right]=1.00\times {10}^{-6}mol/L \end{gathered}$$

So, $$\begin{gathered} \frac{\left[{\mathrm{In}}^{-}\right]}{\left[\mathrm{HIn}\right]}=\frac{1.86\times {10}^{-6}}{1.00\times {10}^{-6}}=1.858 \\ \left[{\mathrm{In}}^{-}\right]+\left[\mathrm{HIn}\right]=2.00\times {10}^{-4}\mathrm{M} \\ 1.858\left[\mathrm{HIn}\right]+\left[\mathrm{HIn}\right]=2.00\times {10}^{-4}\mathrm{M} \\ \left[\mathrm{HIn}\right]=7.00\times {10}^{-5}\mathrm{mol}/\mathrm{L} \\ \left[{\mathrm{In}}^{-}\right]=2.00\times {10}^{-4}-7.00\times {10}^{-5}=1.30\times {10}^{-4}\mathrm{mol}/\mathrm{L} \\ {\mathrm{A}}_{485}=150\times 1.50\times 1.30\times {10}^{-4}+974.1\times 1.50\times 7.00\times {10}^{-5}=0.131 \\ {\mathrm{A}}_{625}=1808\times 1.50\times 1.30\times {10}^{-4}+362\times 1.50\times 7.00\times {10}^{-5}=0.391 \end{gathered}$$