Chapter 14, Problem 14.10QAP

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

   $\underset{\text{color 1}}{{\displaystyle \text{HIn}}}\rightleftharpoons {\text{H}}^{+}+\underset{\text{color 2}}{{\displaystyle {\text{In}}^{-}}}$

The following absorbance data were obtained for a 5.00 × I0_-4 M solution of HIn in 0.1 M NaOH and 0.1 M HC1. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells.

0.1 M NaOH   A₄₈₅ = 0.075   A₆₂₅ = 0.904
0.1 M HC1  A₄₈₅ = 0.487   A₆₂₅ = 0.181

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 In_- and HIn at 485 and 625 nm.
(b) Calculate the acid dissociation constant for the indicator ¡fa 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 of0.492 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-cmcells). Calculate the pH of the solution and K_a 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 × 10-⁴ M in the indicator and was buffered to a pH of 6.000?

Interpretation Introduction

(a)

Interpretation:

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

Concept introduction:

The Beer-Lambert Law is:

$A=\epsilon lc$

A − absorbance

$\epsilon$ - molar absorptivity

l − length of the solution light passes through (cm)

c − concentration of solution (mol/L)

Answer to Problem 14.10QAP

At 484 nm,

${\epsilon }_{I{n}^{-}}=150\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}$

${\epsilon }_{HIn}=974\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}$

At 625 nm,

${\epsilon }_{I{n}^{-}}=1808\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}$

${\epsilon }_{HIn}=362\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}$

Explanation of Solution

At 484 nm,

$\begin{array}{l}0.075={\epsilon }_{I{n}^{-}}\times 1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}\\ {\epsilon }_{I{n}^{-}}=\frac{0.075}{1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}}=150\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}\end{array}$

$\begin{array}{l}0.487={\epsilon }_{HIn}\times 1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}\\ {\epsilon }_{HIn}=\frac{0.487}{1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}}=974\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}\end{array}$

At 625 nm,

$\begin{array}{l}0.904={\epsilon }_{I{n}^{-}}\times 1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}\\ {\epsilon }_{I{n}^{-}}=\frac{0.904}{1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}}=1808\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}\end{array}$

$\begin{array}{l}0.181={\epsilon }_{HIn}\times 1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}\\ {\epsilon }_{HIn}=\frac{0.181}{1\text{ cm }\times 5\times {10}^{-4}\text{ mol/L}}=362\text{ L}{\text{.mol}}^{-1}{\text{cm}}^{-1}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The acid dissociation constant of the indicator should be calculated.

Concept introduction:

The Beer-Lambert Law is:

$A=\epsilon lc$

A − absorbance

$\epsilon$ - molar absorptivity

l − length of the solution light passes through (cm)

c − concentration of solution (mol/L)

Answer to Problem 14.10QAP

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

Explanation of Solution

At 485 nm

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

At 625 nm

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

Solving the above equations as:

$\begin{array}{l}\left(1\right)\times 1808-\left(2\right)\times 150\\ 1760992\left[\text{HIn}\right]-54300\left[\text{HIn}\right]=1025.136-59.25\\ 1706692\left[\text{HIn}\right]=965.886\\ \left[\text{HIn}\right]=5.66\times {10}^{-4}\text{ mol/L}\end{array}$

$\begin{array}{l}\text{Substituting to }\left(1\right)\\ 150\left[{\text{In}}^{-}\right]+974\times 5.66\times {10}^{-4}=0.567\\ 150\left[{\text{In}}^{-}\right]=0.567-0.551\\ 150\left[{\text{In}}^{-}\right]=0.016\\ \left[{\text{In}}^{-}\right]=1.06\times {10}^{-4}\text{ mol/L}\end{array}$

$\text{HIn }\rightleftharpoons {\text{ H}}^{+}{\text{ + In}}^{-}$

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

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

Interpretation Introduction

(c)

Interpretation:

pH of the solution should be calculated.

Concept introduction:

The formula used to determine the pH is:

$pH=-\log \left[{\text{H}}^{+}\right]$

Where, $\left[{\text{H}}^{+}\right]$ represents concentration of ${\text{H}}^{+}$ ions.

Answer to Problem 14.10QAP

$pH=4.55$

Explanation of Solution

At 485 nm

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

At 625 nm

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

$\begin{array}{l}\left(1\right)\times 1808-\left(2\right)\times 150\\ 1760992\left[\text{HIn}\right]-54300\left[\text{HIn}\right]=889.536-36.75\\ 1706692\left[\text{HIn}\right]=852.786\\ \left[\text{HIn}\right]=5.00\times {10}^{-4}\text{ mol/L}\end{array}$

$\begin{array}{l}\text{Substituting to }\left(1\right)\\ 150\left[{\text{In}}^{-}\right]+974\times 5.00\times {10}^{-4}=0.492\\ 150\left[{\text{In}}^{-}\right]=0.492-0.487\\ 150\left[{\text{In}}^{-}\right]=5\times {10}^{-3}\\ \left[{\text{In}}^{-}\right]=3.33\times {10}^{-5}\text{ mol/L}\end{array}$

$\begin{array}{l}{K}_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}\\ 1.88\times {10}^{-6}=\frac{\left[{\text{H}}^{+}\right]\times 3.33\times {10}^{-5}}{5.00\times {10}^{-4}}\\ \left[{\text{H}}^{+}\right]=\frac{1.88\times {10}^{-6}\times 5.00\times {10}^{-4}}{3.33\times {10}^{-5}}\\ \left[{\text{H}}^{+}\right]=2.82\times {10}^{-5}\text{ mol/L}\end{array}$

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

Interpretation Introduction

(d)

Interpretation:

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

Concept introduction:

The formula used to determine the pH is:

$pH=-\log \left[{\text{H}}^{+}\right]$

Where, $\left[{\text{H}}^{+}\right]$ represents concentration of ${\text{H}}^{+}$ ions.

Answer to Problem 14.10QAP

$pH=5.73$

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

Explanation of Solution

At 485 nm

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

At 625 nm

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

$\begin{array}{l}\left(1\right)\times 1808-\left(2\right)\times 150\\ 1760992\left[\text{HIn}\right]-54300\left[\text{HIn}\right]=602.064-98.25\\ 1706692\left[\text{HIn}\right]=503.814\\ \left[\text{HIn}\right]=2.95\times {10}^{-4}\text{ mol/L}\end{array}$

$\begin{array}{l}\text{Substituting to }\left(1\right)\\ 150\left[{\text{In}}^{-}\right]+974\times 2.95\times {10}^{-4}=0.333\\ 150\left[{\text{In}}^{-}\right]=0.333-0.288\\ 150\left[{\text{In}}^{-}\right]=0.045\\ \left[{\text{In}}^{-}\right]=3\times {10}^{-4}\text{ mol/L}\end{array}$

$\begin{array}{l}{K}_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}\\ 1.88\times {10}^{-6}=\frac{\left[{\text{H}}^{+}\right]\times 3\times {10}^{-4}}{2.95\times {10}^{-4}}\\ \left[{\text{H}}^{+}\right]=\frac{1.88\times {10}^{-6}\times 2.95\times {10}^{-4}}{3\times {10}^{-4}}\\ \left[{\text{H}}^{+}\right]=1.85\times {10}^{-6}\text{ mol/L}\end{array}$

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

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

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

Interpretation Introduction

(e)

Interpretation:

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

Concept introduction:

The Beer-Lambert Law is:

$A=\epsilon lc$

A − absorbance

$\epsilon$ - molar absorptivity

l − length of the solution light passes through (cm)

c − concentration of solution (mol/L)

Answer to Problem 14.10QAP

$\begin{array}{l}{A}_{485}=0.131\\ A_{625}=0.391\end{array}$

Explanation of Solution

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

So, $\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}=\frac{1.86\times {10}^{-6}}{1.00\times {10}^{-6}}=1.858$

$\left[{\text{In}}^{-}\right]+\left[\text{HIn}\right]=2.00\times {10}^{-4}\text{ M}$

$\begin{array}{l}1.858\left[\text{HIn}\right]+\left[\text{HIn}\right]=2.00\times {10}^{-4}\text{ M}\\ \left[\text{HIn}\right]=7.00\times {10}^{-5}\text{ mol/L}\\ \left[{\text{In}}^{-}\right]=2.00\times {10}^{-4}-7.00\times {10}^{-5}=1.30\times {10}^{-4}\text{ mol/L}\end{array}$

$\begin{array}{l}{A}_{485}=150\times 1.50\times 1.30\times {10}^{-4}+974\times 1.50\times 7.00\times {10}^{-5}=0.131\\ A_{625}=1808\times 1.50\times 1.30\times {10}^{-4}+362\times 1.50\times 7.00\times {10}^{-5}=0.391\end{array}$