Chapter 13, Problem 13.12QAP

The equilibrium constant for the reaction $2{\text{CrO}}_4^{2-}+2{\text{H}}^{{}^{+}}\underset{}{\rightleftharpoons }{\text{Cr}}_2{\text{O}}_7^{2-}+{\text{H}}_2\text{O}$ is 4.2 × 10¹⁴. The molar absorptivities for the two principal species in a solution of K₂ Cr₂ O₂ are

Four solutions were preparedby dissolving 4.00 × 10-4, 3.00 × 10-4, 2.00 × 10-4,and 1.00 × 10-4 moles of K₂ Cr₂ O₇ in water and diluting to 1.00 L with a pH 5.60 buffer. Derive theoretical absorbance values (1.00-cm cells) for each solution and plot the data for (a) 345 nm, (b) 370 nm, and (c) 400 nm.

Interpretation Introduction

(a)

Interpretation:

The theoretical absorbance value for 345 nm should be derived and the data should be plotted.

Concept introduction:

The relationship between absorbance and concentration of absorbance is linear. If the incident light $I_o$ falls on the sample solutions then only $I$ amount of light gets transmitted through the solution. And the relationship is defined as below:

$\text{Absorbance(A)}=\frac{I_o}{I}$

$\text{Transmission(T)}=\frac{I_{}}{I_o}$

In the case of no absorbing sample, all the light gets passed and the value of $T=100\%$. But in the case of absorbing sample, the absorbance is given by $A=\log \left(\frac{1}{T}\right)$.

Explanation of Solution

Given:

The equilibrium constant is $4.2\times {10}^{14}$ and the pH is $5.6$. $\lambda =345,{\epsilon }_1(Cr{O}_4{}^{2-})=1.84\times {10}^3,{\epsilon }_2(Cr{O}_7{}^{2-})=10.7\times {10}^2$.

The given reaction is $2Cr{O}_4{}^{2-}+2H^{+}\rightleftharpoons C{r}_2{O}_7{}^{2-}+H_{2}O$

The equilibrium constant for the given reaction is $4.2\times {10}^{14}$ and the pH is $5.6$.

The formula to determine pH is: $pH=-\log \left[H^{+}\right]$.

Therefore

$\begin{array}{l}5.6=-\log \left[H^{+}\right]\\ \left[H^{+}\right]=2.51\times {10}^{-6}M\end{array}$

For the given reaction the expression for the equilibrium constant can be written as

$K=\frac{\left[C{r}_2{O}_7{}^{2-}\right]}{{\left[C{r}_2{O}_4{}^{2-}\right]}^2{\left[H^{+}\right]}^2}$

The equilibrium concentration of dichromate is $4.00\times {10}^{-4}$. Now, substitute the values and find out the equilibrium concentration of $C{r}_2{O}_4{}^{2-}$ at the given pH.

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[4.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[4.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=3.88\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(1.84\times {10}^3 M^{-1}c{m}^{-1})\times (3.88\times {10}^{-4} M)+(10.7\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 =1.14192\approx 1.14\end{array}$

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $3.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[3.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[3.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=2.91\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(1.84\times {10}^3 M^{-1}c{m}^{-1})\times (2.91\times {10}^{-4} M)+(10.7\times {10}^2 M^{-1}c{m}^{-1})\times (3.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 =0.8564\approx 0.85\end{array}$

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $2.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[2.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[2.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=1.94\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(1.84\times {10}^3 M^{-1}c{m}^{-1})\times (1.94\times {10}^{-4} M)+(10.7\times {10}^2 M^{-1}c{m}^{-1})\times (2.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 0.57\end{array}$

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $1.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[1.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[1.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=0.97\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below:

$\begin{array}{l}{A}_1 =\left[(1.84\times {10}^3 M^{-1}c{m}^{-1})\times (0.97\times {10}^{-4} M)+(10.7\times {10}^2 M^{-1}c{m}^{-1})\times (1.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 0.28\end{array}$

Now take the values in excel and plot to get the graph:

Interpretation Introduction

(b)

Interpretation:

The theoretical absorbance value for 370 nm should be derived and the data should be plotted.

Concept introduction:

The relationship between absorbance and concentration of absorbance is linear. If the incident light $I_o$ falls on the sample solutions then only $I$ amount of light gets transmitted through the solution. And the relationship is defined as below:

$\text{Absorbance(A)}=\frac{I_o}{I}$

$\text{Transmission(T)}=\frac{I_{}}{I_o}$

In the case of no absorbing sample, all the light gets passed and the value of $T=100\%$. But in the case of absorbing sample, the absorbance is given by $A=\log \left(\frac{1}{T}\right)$.

Explanation of Solution

Given:

The equilibrium constant is $4.2\times {10}^{14}$ and the pH is $5.6$. $\lambda =370,{\epsilon }_1(Cr{O}_4{}^{2-})=4.81\times {10}^3,{\epsilon }_2(Cr{O}_7{}^{2-})=7.28\times {10}^2$.

The given reaction is $2Cr{O}_4{}^{2-}+2H^{+}\rightleftharpoons C{r}_2{O}_7{}^{2-}+H_{2}O$

The equilibrium constant for the given reaction is $4.2\times {10}^{14}$ and the pH is $5.6$.

The formula to determine pH is: $pH=-\log \left[H^{+}\right]$.

Therefore

$\begin{array}{l}5.6=-\log \left[H^{+}\right]\\ \left[H^{+}\right]=2.51\times {10}^{-6}M\end{array}$

For the given reaction the expression for the equilibrium constant can be written as

$K=\frac{\left[C{r}_2{O}_7{}^{2-}\right]}{{\left[C{r}_2{O}_4{}^{2-}\right]}^2{\left[H^{+}\right]}^2}$

The equilibrium concentration of dichromate is $4.00\times {10}^{-4}$. Now substitute the values and find out the equilibrium concentration of $C{r}_2{O}_4{}^{2-}$ at the given pH.

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[4.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[4.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=3.88\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(4.81\times {10}^3 M^{-1}c{m}^{-1})\times (3.88\times {10}^{-4} M)+(7.28\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 1.93\end{array}$

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $3.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[3.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[3.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=2.91\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(4.81\times {10}^3 M^{-1}c{m}^{-1})\times (2.91\times {10}^{-4} M)+(7.28\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 1.47\end{array}$

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $2.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[2.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[2.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=1.94\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(4.81\times {10}^3 M^{-1}c{m}^{-1})\times (1.94\times {10}^{-4} M)+(7.28\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 1.00\end{array}$

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $1.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[1.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[1.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=0.97\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(4.81\times {10}^3 M^{-1}c{m}^{-1})\times (0.97\times {10}^{-4} M)+(7.28\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 0.54\end{array}$

Now take the values in excel and plot to get the graph:

Interpretation Introduction

(c)

Interpretation:

The theoretical absorbance value for 400 nm should be derived and the data should be plotted.

Concept introduction:

The relationship between absorbance and concentration of absorbance is linear. If the incident light $I_o$ falls on the sample solutions then only $I$ amount of light gets transmitted through the solution. And the relationship is defined as below:

$\text{Absorbance(A)}=\frac{I_o}{I}$

$\text{Transmission(T)}=\frac{I_{}}{I_o}$

In the case of no absorbing sample, all the light gets passed and the value of $T=100\%$. But in the case of absorbing sample, the absorbance is given by $A=\log \left(\frac{1}{T}\right)$.

Explanation of Solution

Given:

The equilibrium constant is $4.2\times {10}^{14}$ and the pH is $5.6$. $\lambda =400,{\epsilon }_1(Cr{O}_4{}^{2-})=1.88\times {10}^3,{\epsilon }_2(Cr{O}_7{}^{2-})=1.89\times {10}^2$.

The given reaction is $2Cr{O}_4{}^{2-}+2H^{+}\rightleftharpoons C{r}_2{O}_7{}^{2-}+H_{2}O$

The equilibrium constant for the given reaction is $4.2\times {10}^{14}$ and the pH is $5.6$.

The formula to determine pH is: $pH=-\log \left[H^{+}\right]$.

Therefore

$\begin{array}{l}5.6=-\log \left[H^{+}\right]\\ \left[H^{+}\right]=2.51\times {10}^{-6}M\end{array}$

For the given reaction the expression for the equilibrium constant can be written as

$K=\frac{\left[C{r}_2{O}_7{}^{2-}\right]}{{\left[C{r}_2{O}_4{}^{2-}\right]}^2{\left[H^{+}\right]}^2}$

The equilibrium concentration of dichromate is $4.00\times {10}^{-4}$. Now substitute the values and find out the equilibrium concentration of $C{r}_2{O}_4{}^{2-}$ at the given pH.

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[4.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[4.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=3.88\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(1.88\times {10}^3 M^{-1}c{m}^{-1})\times (3.88\times {10}^{-4} M)+(1.89\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 0.74\end{array}$

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $3.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[3.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[3.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=2.91\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(1.88\times {10}^3 M^{-1}c{m}^{-1})\times (2.91\times {10}^{-4} M)+(1.89\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 0.57\end{array}$

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $2.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[2.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[2.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=1.94\times {10}^{-4}\end{array}$

The theoretical absorbance value of the first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(1.88\times {10}^3 M^{-1}c{m}^{-1})\times (1.94\times {10}^{-4} M)+(1.89\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 0.38\end{array}$

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $1.00\times {10}^{-4}$.

So, the new required concentration will be as

$\begin{array}{l}4.2\times {10}^{14}=\frac{\left[1.00\times {10}^{-4}\right]}{{\left[Cr{O}_4{}^{2-}\right]}^2{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=\frac{\left[1.00\times {10}^{-4}\right]}{\left[4.2\times {10}^{14}\right]{\left[2.51\times {10}^{-6}\right]}^2}\\ {\left[Cr{O}_4{}^{2-}\right]}^2=0.97\times {10}^{-4}\end{array}$

Theoretical absorbance value of first solution can be calculated as below

$\begin{array}{l}{A}_1 =\left[(1.88\times {10}^3 M^{-1}c{m}^{-1})\times (0.97\times {10}^{-4} M)+(1.89\times {10}^2 M^{-1}c{m}^{-1})\times (4.00\times {10}^{-4} M)]\times \left(1.00cm\right)\right]\\ A_1 \approx 0.20\end{array}$

Now take the values in excel and plot to get the graph: