Chapter 13, Problem 13.11QAP

The equilibrium constant for the conjugate acid-base pair $\text{HIn}+{\text{H}}_2\text{O}\underset{}{\rightleftharpoons }{\text{H}}_3{\text{O}}^{+}+{{\displaystyle \text{In}}}^{-}$ is 8.00 × 10^-5. From the additional information in the following table,
(a) calculate the absorbance at 430 nmand 600 nm for the following indicator concentrations:
3.00 × 10^-4M,2.00 × 10^-4M, 1.00 × 10^-4M, 0.500 × 10-4 M, and 0.250 × 10^-4M.
(b) plot absorbance as a function of indicator concentration.

Interpretation Introduction

(a)

Interpretation:

Absorbance values of the conjugate acid-base pair solutions of different concentrations at $430\text{ nm and 600 nm}$ should be calculated.

Concept introduction:

The absorbance can be calculated using the following formula:

$\begin{array}{l}\text{A = }\epsilon \text{bc}\\ \text{Where, }\\ \text{A = absorbance}\\ \epsilon \text{ = molar absorptivity}\\ \text{b = path length in cm}\\ \text{c = concentration of the sample}\\ \text{The total absorbance of a conjugate acid-base solution can be calculated as below,}\\ \text{HIn }\iff {\text{ H}}^{+}{\text{ + In}}^{-}\\ \text{A = }{\epsilon }_{{\text{In}}^{\text{-}}}{\text{b[In}}^{\text{-}}\text{] + }{\epsilon }_{\text{HIn}}{\text{b[HIn}}^{\text{-}}\text{]}\end{array}$

For the given conjugate acid-base pair, equilibrium can be written as,

$\begin{array}{l}{\text{HIn + H}}_2\text{O}\iff {\text{ H}}_3{\text{O}}^{+}{\text{ + In}}^{-}\\ K_a=8.00\times {10}^{-5}=\frac{[{\text{H}}^{+}][{\text{In}}^{-}]}{[\text{HIn}]}\end{array}$

Answer to Problem 13.11QAP

$\begin{array}{lllll}{\text{c}}_{\text{HIn}}\text{, M}\hfill & \left[\text{HIn}\right]\hfill & \left[{\text{In}}^{\text{-}}\right]\hfill & {\text{A}}_{\text{430}}\hfill & {\text{A}}_{\text{600}}\hfill \\ \text{3}\text{.00}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.80}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.20}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.54}\hfill & \text{1}\text{.06}\hfill \\ \text{2}\text{.00}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.07}\times {\text{10}}^{\text{-4}}\hfill & \text{9}\text{.30}\times {\text{10}}^{\text{-5}}\hfill & \text{9}\text{.32}\times {\text{10}}^{\text{-1}}\hfill & \text{7}\text{.79}\times {\text{10}}^{\text{-1}}\hfill \\ \text{1}\text{.00}\times {\text{10}}^{\text{-4}}\hfill & \text{4}\text{.20}\times {\text{10}}^{\text{-5}}\hfill & \text{5}\text{.80}\times {\text{10}}^{\text{-5}}\hfill & \text{3}\text{.83}\times {\text{10}}^{\text{-1}}\hfill & \text{4}\text{.55}\times {\text{10}}^{\text{-1}}\hfill \\ \text{5}\text{.00}\times {\text{10}}^{\text{-5}}\hfill & \text{1}\text{.50}\times {\text{10}}^{\text{-5}}\hfill & \text{3}\text{.50}\times {\text{10}}^{\text{-5}}\hfill & \text{1}\text{.48}\times {\text{10}}^{\text{-1}}\hfill & \text{2}\text{.62}\times {\text{10}}^{\text{-1}}\hfill \\ \text{2}\text{.50}\times {\text{10}}^{\text{-5}}\hfill & \text{5}\text{.00}\times {\text{10}}^{\text{-6}}\hfill & \text{2}\text{.00}\times {\text{10}}^{\text{-5}}\hfill & \text{5}\text{.57}\times {\text{10}}^{\text{-2}}\hfill & \text{1}\text{.45}\times {\text{10}}^{\text{-1}}\hfill \end{array}$

Explanation of Solution

Let’s use the equilibrium equation to find the concentrations of HIn and In^-(denoted as [HIn] and [In^-]) at a situation in which the total concentration (c_HIn) is $3.00\times {10}^{-4}\text{ M}$.

$\begin{array}{l}\text{at equilibrium, }[{\text{H}}^{+}]=[{\text{In}}^{-}]\\ \text{also, }[\text{HIn}]+[{\text{In}}^{-}]=3.00\times {10}^{-4}\text{ M}\\ \text{Therefore, the equilibrium equation can be rearranged as follows,}\\ K_a=8.00\times {10}^{-5}=\frac{[{\text{H}}^{+}][{\text{In}}^{-}]}{[\text{HIn}]}\\ 8.00\times {10}^{-5}=\frac{{[{\text{In}}^{-}]}^2}{3.00\times {10}^{-4}-[{\text{In}}^{-}]}\\ {[}^{{\text{In}}^{-}}+8.00\times {10}^{-5}[{\text{In}}^{-}]-8.00\times {10}^{-5}\times 3.00\times {10}^{-4}=0\\ {\text{The positive solution for the above quadratic equation gives the concentration of In}}^{-}\text{ at this situation}\text{.}\\ [{\text{In}}^{-}]=1.20\times {10}^{-4}\text{ M}\\ \text{Therefore, }\\ [\text{HIn}]=3.00\times {10}^{-4}-1.20\times {10}^{-4}\text{ =1}\text{.80}\times {10}^{-4}\text{ M}\end{array}$

Now, the absorbance values of this solution at 430 and 600 nm can be calculated as follows,

The cell length (b) is not given in the question, therefore taken as 1.00cm.

$\begin{array}{l}{\text{A}}_{\text{430}}={\epsilon }_{{\text{In}}^{\text{-}}}{\text{b[In}}^{\text{-}}\text{] + }{\epsilon }_{\text{HIn}}{\text{b[HIn}}^{\text{-}}\text{]}\\ {\text{A}}_{\text{430}}=\text{ 8}\text{.04}\times {\text{10}}^3\times 1.00\times 1.20\times {10}^{-4}+0.775\times {\text{10}}^3\times 1.00\times 1.80\times {10}^{-4}\\ {\text{A}}_{\text{430}}=\text{ 1}\text{.54}\\ \text{also,}\\ {\text{A}}_{\text{600}}={\epsilon }_{{\text{In}}^{\text{-}}}{\text{b[In}}^{\text{-}}\text{] + }{\epsilon }_{\text{HIn}}{\text{b[HIn}}^{\text{-}}\text{]}\\ {\text{A}}_{\text{600}}=\text{ 1}\text{.23}\times {\text{10}}^3\times 1.00\times 1.20\times {10}^{-4}+6.96\times {\text{10}}^3\times 1.00\times 1.80\times {10}^{-4}\\ {\text{A}}_{\text{600}}=\text{ 1}\text{.06}\end{array}$

Similarly, other absorbance values can be calculated for solutions with different concentrations.

Interpretation Introduction

(b)

Interpretation:

Absorbance as a function of indicator concentration should be plotted.

Explanation of Solution

The data obtained is as follows:

$\begin{array}{lllll}{\text{c}}_{\text{HIn}}\text{, M}\hfill & \left[\text{HIn}\right]\hfill & \left[{\text{In}}^{\text{-}}\right]\hfill & {\text{A}}_{\text{430}}\hfill & {\text{A}}_{\text{600}}\hfill \\ \text{3}\text{.00}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.80}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.20}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.54}\hfill & \text{1}\text{.06}\hfill \\ \text{2}\text{.00}\times {\text{10}}^{\text{-4}}\hfill & \text{1}\text{.07}\times {\text{10}}^{\text{-4}}\hfill & \text{9}\text{.30}\times {\text{10}}^{\text{-5}}\hfill & \text{9}\text{.32}\times {\text{10}}^{\text{-1}}\hfill & \text{7}\text{.79}\times {\text{10}}^{\text{-1}}\hfill \\ \text{1}\text{.00}\times {\text{10}}^{\text{-4}}\hfill & \text{4}\text{.20}\times {\text{10}}^{\text{-5}}\hfill & \text{5}\text{.80}\times {\text{10}}^{\text{-5}}\hfill & \text{3}\text{.83}\times {\text{10}}^{\text{-1}}\hfill & \text{4}\text{.55}\times {\text{10}}^{\text{-1}}\hfill \\ \text{5}\text{.00}\times {\text{10}}^{\text{-5}}\hfill & \text{1}\text{.50}\times {\text{10}}^{\text{-5}}\hfill & \text{3}\text{.50}\times {\text{10}}^{\text{-5}}\hfill & \text{1}\text{.48}\times {\text{10}}^{\text{-1}}\hfill & \text{2}\text{.62}\times {\text{10}}^{\text{-1}}\hfill \\ \text{2}\text{.50}\times {\text{10}}^{\text{-5}}\hfill & \text{5}\text{.00}\times {\text{10}}^{\text{-6}}\hfill & \text{2}\text{.00}\times {\text{10}}^{\text{-5}}\hfill & \text{5}\text{.57}\times {\text{10}}^{\text{-2}}\hfill & \text{1}\text{.45}\times {\text{10}}^{\text{-1}}\hfill \end{array}$

From the data, the graph between absorbance and concentration can be plotted as follows:

Here, blue curve indicates the plot at 430 nm and orange curve indicates the same at 600 nm.