Question

The equilibrium constant for the reaction

is 4.2 * 10¹⁴. The molar absorptivities for the two principal species in a solution of K₂Cr₂O₇ are

Short answer

a)

b)

c)

Step-by-step solution

Part (a) Step 1: Given Information

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

$$\begin{gathered} \lambda =345 \\ {\epsilon }_1\left(Cr{O_4}^{2-}\right)=1.84\times {10}^3 \\ {\epsilon }_2\left(Cr{O_7}^{2-}\right)=10.7\times {10}^2 \end{gathered}$$

Part (a) Step 2: Explanation

The given reaction is $2{{\mathrm{CrO}}_4}^{2-}+2{\mathrm{H}}^{+}\rightleftharpoons {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+{\mathrm{H}}_2\mathrm{O}$

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

Therefore

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

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

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

The equilibrium concentration of dichromate is $4.0\times {10}^{-4}$. Now, substitute the values and find out the equilibrium concentration of ${\mathrm{Cr}}_2{{\mathrm{O}}_4}^{2-}$at the given pH.

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

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

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

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to data-custom-editor="chemistry" $3.0\times {10}^{-4}$.

So, the new required concentration will be as

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

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

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

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to data-custom-editor="chemistry" $2.0\times {10}^{-14}$.

So, the new required concentration will be as

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

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

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

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to $1.0\times {10}^{-14}$.

So, the new required concentration will be as

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

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

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

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

Part (b) Step 1: Given Information

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

$$\begin{gathered} \lambda =370 \\ {\epsilon }_1\left(Cr{O_4}^{2-}\right)=4.81\times {10}^3 \\ {\epsilon }_2\left(Cr{O_7}^{2-}\right)=7.28\times {10}^2 \end{gathered}$$

Part (b) Step 2: Explanation

The given reaction is $2{{\mathrm{CrO}}_4}^{2-}+2{\mathrm{H}}^{+}\rightleftharpoons {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+{\mathrm{H}}_2\mathrm{O}$

The equilibrium constant for the given reaction is 4.2 * 10 ¹⁴and the pH is 5.6.

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

Therefore

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

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

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

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

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

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

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

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 3.00 * 10 ^-4.

So, the new required concentration will be as

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

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

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

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 2.00 * 10 ^-14.

So, the new required concentration will be as

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

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

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

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 1.00 * 10 ^-14.

So, the new required concentration will be as
$$\begin{gathered} 4.2\times {10}^{-14}=\frac{[1.00\times {10}^{-4}]}{{\left[{{\mathrm{CrO}}_4}^{2-}\right]}^2{[2.51\times {10}^{-6}]}^2} \\ {\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]}^2=\frac{[1.00\times {10}^{-4}]}{[4.2\times {10}^{-14}]{[2.51\times {10}^{-6}]}^2} \\ {\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]}^2=0.97\times {10}^{-4} \end{gathered}$$

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

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

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

Part (c) Step 1: Given Information

$$\begin{gathered} \lambda =400 \\ {\epsilon }_1\left(Cr{O_4}^{2-}\right)=1.88\times {10}^3 \\ {\epsilon }_2\left(Cr{O_7}^{2-}\right)=1.89\times {10}^2 \end{gathered}$$

Part (c) Step 2: Explanation

The given reaction is $2{{\mathrm{CrO}}_4}^{2-}+2{\mathrm{H}}^{+}\rightleftharpoons {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+{\mathrm{H}}_2\mathrm{O}$

The equilibrium constant for the given reaction is 4.2 * 10 ¹⁴and the pH is 5.6.

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

Therefore

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

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

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

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

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

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

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

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 3.00 * 10 ^-4.

So, the new required concentration will be as

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

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

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

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 2.00 * 10 ^-14.

So, the new required concentration will be as

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

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

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

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 1.00 * 10 ^-14.

So, the new required concentration will be as

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

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

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

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