Question

The equilibrium constant for the conjugate acid-base pair

$\mathrm{HIn}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{In}}^{-}$

is $8.00\times {10}^{-5}$. From the additional information in the following table,

(a) calculate the absorbance at 430 nm and 600 nm for the following indicator concentrations:

$3.00\times {10}^{-4}\mathrm{M},2.00\times {10}^{-4}\mathrm{M},1.00\times {10}^{-4}\mathrm{M},0.500\times {10}^{-4}\mathrm{M},\mathrm{and}0.250\times {10}^{-4}\mathrm{M}$.

(b) plot absorbance as a function of indicator concentration.

Short answer

a)

b)

Step-by-step solution

Part (a) Step 1: Given Information

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

Part (a) Step 2: Explanation

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 (cHIn) is $3.00\times {10}^{-4}\mathrm{M}$.

$$\begin{gathered} \mathrm{At}\mathrm{equibrium},\left[{\mathrm{H}}^{+}\right]=\left[{\ln }^{-}\right] \\ \mathrm{also},\left[\mathrm{Hln}\right]+\left[{\ln }^{-}\right]=3.00\times {10}^{-4}\mathrm{M} \\ \mathrm{Thus},\mathrm{on}\mathrm{rearranging}, \\ {\mathrm{K}}_{\mathrm{a}}=8.00\times {10}^{-5}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\ln }^{-}\right]}{\left[\mathrm{Hln}\right]} \\ 8.00\times {10}^{-5}=\frac{{\left[{\ln }^{-}\right]}^2}{3.00\times {10}^{-4}-\left[{\ln }^{-}\right]} \\ {\left[{\ln }^{-}\right]}^2+8.00\times {10}^{-5}\left[{\ln }^{-}\right]=8.00\times {10}^{-5}\times 3.00\times {10}^{-4}=0 \\ \mathrm{The}\mathrm{positive}\mathrm{solution}\mathrm{for}\mathrm{the}\mathrm{above}\mathrm{quadratic}\mathrm{equation}\mathrm{gives}\mathrm{the}\mathrm{concentration}\mathrm{of}{\ln }^{-}]\mathrm{at}\mathrm{this}\mathrm{solution} \\ [{\mathrm{In}}^{-}]=1.20\times {10}^{-4}\mathrm{M} \\ \mathrm{Therefore}, \\ [\mathrm{Hln}]=3.00\times {10}^{-4}-1.20\times {10}^{-4}=1.80\times {10}^{-4}\mathrm{M} \end{gathered}$$

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.

$$\begin{gathered} {\mathrm{A}}_{430}={\mathrm{\epsilon }}_{\ln }.\mathrm{b}\left[{\mathrm{In}}^{-}\right]+{\mathrm{\epsilon }}_{\mathrm{Hln}}.\mathrm{b}\left[{\mathrm{HIn}}^{-}\right] \\ {\mathrm{A}}_{430}=8.04\times {10}^3\times 1.00\times 1.20\times {10}^{-4}+0.775\times {10}^3\times 1.00\times 1.80\times {10}^{-4} \\ {\mathrm{A}}_{430}=1.54 \\ \mathrm{also}, \\ {\mathrm{A}}_{600}={\mathrm{\epsilon }}_{\ln }.\mathrm{b}\left[{\mathrm{In}}^{-}\right]+{\mathrm{\epsilon }}_{\mathrm{Hln}}.\mathrm{b}\left[{\mathrm{HIn}}^{-}\right] \\ {\mathrm{A}}_{600}=1.23\times {10}^3\times 1.00\times 1.20\times {10}^{-4}+6.96\times {10}^3\times 1.00\times 1.80\times {10}^{-4} \\ {\mathrm{A}}_{600}=1.06 \end{gathered}$$

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

The data obtained is as follows:

Part (b) Step 1: Explanation

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

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