Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for ${\mathbf{CO}}_{\mathbf{2}}$in the ocean was designed to operate without the need for calibration.33

The sensor compartment is separated from seawater by a silicone membrane through which ${\mathbf{CO}}_{\mathbf{2}}$, but not dissolved ions, can diffuse. Inside the sensor, ${\mathbf{CO}}_{\mathbf{2}}$equilibrates with ${\mathbf{HCO}}_{\mathbf{3}}$and ${{\mathbf{CO}}_{\mathbf{3}}}^{\mathbf{2}}$. For each

measurement, the sensor is flushed with fresh solution containingbromothymol blue indicator. All indicator is in the formnear neutral pH, so we can

write two mass balances:

$$\begin{gathered} \left[{\mathrm{HIn}}^{-}\right]\mathbf{+}\left[{\ln }^{2-}\right]\mathbf{=}{\mathbf{F}}_{\mathbf{In}}\mathbf{=}\mathbf{50}\mathbf{.0}\mathbf{\mu Mand} \\ \left[{\mathrm{Na}}^{+}\right]\mathbf{=}{\mathbf{F}}_{\mathbf{N\alpha }}\mathbf{=}\mathbf{50}\mathbf{.0}\mathbf{\mu M}\mathbf{+}\mathbf{42}\mathbf{.0}\mathbf{\mu M}\mathbf{=}\mathbf{92}\mathbf{.0}\mathbf{\mu M} \end{gathered}$$

has an absorbance maximum at 434 nm andhas a maximum at 620 nm. The sensor measures the absorbance ratio ${\mathbf{R}}_{\mathbf{A}}\mathbf{=}{\mathbf{A}}_{\mathbf{620}}\mathbf{/}{\mathbf{A}}_{\mathbf{434}}$reproducibly without need for calibration. From this ratio, we can findin the seawater as outlined here:

(a).From Beer’s law for the mixture, write equations forin terms of the absorbance at 620 and 434 nmThen show that

$\frac{\left[{\ln }^{2-}\right]}{\mathbf{[}\mathbf{Hln}\mathbf{-}\mathbf{]}}\mathbf{=}\frac{{\mathbf{R}}_{\mathbf{A}}{\mathbf{\epsilon }}_{\mathbf{434}}^{{\mathbf{HHn}}^{\mathbf{-}}}\mathbf{-}{\mathbf{\epsilon }}_{\mathbf{6}\mathbf{,}\mathbf{20}}^{\mathbf{Hln}\mathbf{-}}}{{\mathbf{\epsilon }}_{\mathbf{620}}^{{\mathbf{ln}}^{\mathbf{2}\mathbf{-}}}\mathbf{-}{\mathbf{R}}_{\mathbf{A}}{\mathbf{\epsilon }}_{\mathbf{434}}^{{\mathbf{ln}}^{\mathbf{2}\mathbf{-}}}}\mathbf{=}{\mathbf{R}}_{\mathbf{ln}}$ (A)

(b) From the mass balance (1) and the acid dissociation constant

, show that

$\left[{\mathrm{Hln}}^{-}\right]\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{1}\mathbf{n}}}{{\mathbf{R}}_{\mathbf{ln}}\mathbf{+}\mathbf{1}}$ (B)

$\left[{\ln }^{2-}\right]\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{ln}}{\mathbf{F}}_{\mathbf{ln}}}{\left[H^{+}\right]\left(R_{\ln }+1\right)}$ (C)

(c) Show that $\left[{\mathrm{H}}^{+}\right]={\mathbf{K}}_{\ln }/{\mathbf{R}}_{\ln }$ (D)

(d) From the carbonic acid dissociation equilibria, show that

$\begin{array}{r}\left[{\mathrm{HCO}}_3^{-}\right]\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{1}}\mathbf{\left[}\mathbf{CO}\mathbf{\right(}\mathbf{aq}\mathbf{\left)}\mathbf{\right]}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{E}}{\left[H^{+}\right]}\\ \left[{\mathrm{CO}}_3^{2-}\right]\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{1}}{\mathbf{K}}_{\mathbf{2}}\mathbf{\left[}\mathbf{CO}\mathbf{\right(}\mathbf{aq}\mathbf{\left)}\mathbf{\right]}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{F}}{{\left[H^{+}\right]}^{\mathbf{2}}}\end{array}$

(e) Write the charge balance for the solution in the sensor compartment. Substitute in expressions B, C, E, and F forHln,${\mathbf{In}}^{\mathbf{2}\mathbf{-}}\mathbf{,}\left[{\mathrm{HCO}}_3\right]\mathbf{\text{, and}}\left[{\mathrm{CO}}_3^{2-}\right]$

(f) Suppose that the various constants have the following values:

$\begin{array}{l}{\mathbf{\epsilon }}_{\mathbf{4344}}^{{\mathbf{HHn}}^{\mathbf{-}}}\mathbf{=}\mathbf{8}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}\mathbf{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}}{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\\ {\mathbf{\epsilon }}_{\mathbf{6620}}^{\mathbf{Hn}\mathbf{-}}\mathbf{=}\mathbf{0}\mathbf{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}}{\mathbf{K}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\\ {\mathbf{\epsilon }}_{\mathbf{434}}^{{\mathbf{ln}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{.90}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}\mathbf{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}}{\mathbf{K}}_{\mathbf{ln}}\mathbf{=}\mathbf{2}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\\ {\mathbf{\epsilon }}_{\mathbf{620}}^{{\mathbf{ln}}^{\mathbf{2}\mathbf{-}}}\mathbf{=}\mathbf{1}\mathbf{.70}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}\mathbf{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}}{\mathbf{K}}_{\mathbf{w}}\mathbf{=}\mathbf{6}\mathbf{.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\end{array}$

From the measured absorbance ratio=2.84, findin the seawater.

(g) Approximately what is the ionic strength inside the sensor compartment? Were we justified in neglecting activity coefficients in working this problem?

Consider,

A as absorbance,

e as molar absorptivity,

b as pathlength.

From the given question we can get,

$\begin{array}{c}{\mathrm{A}}_{620}={\epsilon }_{620}^{{\mathrm{HIn}}^{-}}\mathrm{b}\left[{\mathrm{HIn}}^{-}\right]+\epsilon {\ln }_{620}^{2-}\mathrm{b}\left[{\ln }^{2-}\right]\\ {\mathrm{A}}_{434}={\mathrm{e}}_{434}^{\mathrm{HIn}}\mathrm{b}\left[{\mathrm{HIn}}^{-}\right]+{\epsilon }_{434}^{\ln {\mathrm{nn}}^2}\mathrm{b}\left[{\ln }^{2-}\right]\end{array}$

By solving above equation we get,

$\begin{array}{r}\left[{\mathrm{HInn}}^{-}\right]=\frac{1}{\mathrm{D}}\left({\mathrm{A}}_{620}{\epsilon }_{434}^{{\ln }^{2-}}\mathrm{b}-{\mathrm{A}}_{434}{\mathrm{In}}_{620}^{{\ln }^{2-}}\mathrm{b}\right)\\ \left[{\ln }^{2-}\right]=\frac{1}{D}\left({\mathrm{A}}_{434}{\mathrm{HIn}}_{620}^{-}\mathrm{b}-{\mathrm{A}}_{620434}\underset{\mathrm{Hn}}{-1}\mathrm{b}\right)\end{array}$

Divide${\mathrm{In}}^{2-}$by Hln ,

Divide $A_{434}$to the numerator and denominator,

Hence proved that$\frac{\left[{\ln }^{2-}\right]}{\left[{\mathrm{HIn}}^{-}\right]}=\frac{{\mathrm{R}}_{\mathrm{A}}{\mathrm{s}}_{434}^{\mathrm{Hn}}-{\epsilon }_{6,20}^{\mathrm{HIn}}}{{\epsilon }_{620}^{{\ln }^{2-}}-{\mathrm{R}}_{\mathrm{A}}{\mathrm{s}}_{434}^{{\ln }^{2-}}}={\mathrm{R}}_{\ln }$

$\left[{\mathrm{HIn}}^{-}\right]=\frac{{\mathrm{F}}_{\ln }}{{\mathrm{R}}_{\mathrm{mn}+1}}$

To prove ${\mathrm{K}}_{\mathrm{In}}$,

For indicator, the mass balance can be given as,

$\left[{\mathrm{HIn}}^{-}\right]+\left[{\ln }^{2-}\right]={\mathrm{F}}_{\mathrm{In}}$

Divide Hln both side,

$\begin{array}{c}\frac{{\mathrm{Hln}}^{-}\mid }{\left[{\mathrm{Hln}}^{-}\right]}+\frac{\left[{\ln }^2\right]}{\left[{\mathrm{Hln}}^{-}\right]}=\frac{{\mathrm{F}}_{\ln }}{\left[{\mathrm{HIn}}^{-}\right]}\\ 1+{\mathrm{R}}_{\ln }=\frac{{\mathrm{F}}_{\ln }}{\left[\mathrm{HI}{\text{n}}^{-}]\right.}\end{array}$

Hence proved$\left[{\mathrm{HIn}}^{-}\right]=\frac{{\mathrm{F}}_{\ln }}{{\mathrm{R}}_{\ln +1}}$

The acid dissociation constant of the indicator is,

${\mathrm{K}}_{\ln }=\frac{\left[{\ln }^2\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HInn}}^{-}\right]}$

Substitute${\mathrm{F}}_{\text{In}}/\left({\mathrm{R}}_{\text{In}}+1\right)\text{for}\left[{\mathrm{HIn}}^{-}\right]$,

$\begin{array}{c}{\mathrm{K}}_{\ln }=\frac{\left[{\mathrm{In}}^{2-}\right]\left[{\mathrm{H}}^{+}\right]\left({\mathrm{R}}_{\ln }+1\right)}{{\mathrm{F}}_{\mathrm{In}}}\\ \left[{\mathrm{In}}^{2-}\right]=\frac{{\mathrm{K}}_{\ln }{\mathrm{F}}_{\ln }}{\left[{\mathrm{H}}^{+}\right]\left({\mathrm{R}}_{\ln }+\right)}\end{array}$

${\mathrm{R}}_{\ln }=\frac{\left[{\ln }^2\right]}{\left[\mathrm{H}{\ln }^{-}\right]}$

Hence,${\mathrm{K}}_{\ln }=\frac{\left[{\ln }^2\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HIn}}^{-}\right]}$

Therefore it has been proved, $\left[{\mathrm{H}}^{+}\right]={\mathrm{K}}_{\ln }/{\mathrm{R}}_{\mathrm{in}}$.

${\mathrm{K}}_1=\frac{\left[{\mathrm{HCO}}_3^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{CO}}_{2\left(\mathrm{aq}\right)}\right]}$ the acid dissociation reaction of Carbonic acid.

Hence proved $\left[{\mathrm{HCO}}_3\right]=\frac{{\mathrm{K}}_1\left[{\mathrm{CO}}_{2\left(\mathrm{aq}\right)}\right]}{\left[{\mathrm{H}}^{+}\right]}$

${\mathrm{K}}_2=\frac{\left[{\mathrm{CO}}_3^{2-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HOO}}_3^{-}\right]}$the acid dissociation reaction of Bicarbonate,

$\left[{\mathrm{CO}}_3^{2-}\right]=\frac{{\mathrm{K}}_2\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}^{+}\right]}$

Substitute$\left[{\mathrm{HCO}}_3\right]$ we get,

$\left[{\mathrm{CO}}_3^{2-}\right]=\frac{{\mathrm{K}}_1{\mathrm{K}}_2\left[{\mathrm{CO}}_{2\left(\mathrm{aq}\right)}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}$

$\begin{array}{r}\mathrm{Substitute}\mathrm{the}\mathrm{expression}{\mathrm{Hin}}^{-},{\mathrm{In}}^{2-},\left[{\mathrm{HCO}}_3\right],\text{and}\left[{\mathrm{CO}}_3^{2-}\right].\\ \left[{\mathrm{Na}}^{+}\right]+\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]+\left[{\mathrm{HIn}}^{-}\right]+2\left[{\mathrm{In}}^{2-}\right]+\left[{\mathrm{HCO}}_3^{-}\right]+2\left[{\mathrm{CO}}_3^{2-}\right]\\ \mathrm{The}\mathrm{solution}\mathrm{is}\\ {\mathrm{F}}_{\mathrm{Na}}+\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}^{+}\right]}+\frac{{\mathrm{F}}_{\ln }}{{\mathrm{R}}_{\ln +1}}+2\frac{{\mathrm{K}}_{\ln }{\mathrm{F}}_{\mathrm{in}}}{\left[{\mathrm{H}}^{+}\right]+\left({\mathrm{R}}_{\ln }+1\right)}+\frac{{\mathrm{K}}_1\left[{\mathrm{CO}}_{2\left(\mathrm{xq}\right)}\right]}{\left[{\mathrm{H}}^{+}\right]}+2\frac{{\mathrm{K}}_1{\mathrm{K}}_2\left[{\mathrm{CO}}_{2(\mathrm{mpq}}\right]}{{\left[{\mathrm{H}}^{+}\right]}_2}\end{array}$

$\begin{array}{r}\left[{\mathrm{H}}^{+}\right]={\mathrm{K}}_{\mathrm{in}}/{\mathrm{R}}_{\mathrm{in}}\\ \mathrm{For}{\mathrm{R}}_{\mathrm{in}}\\ {\mathrm{R}}_{\mathrm{in}}=\frac{{\mathrm{R}}_{\mathrm{A}}{{\epsilon }_{434}}^{\mathrm{Hn}}-{\epsilon }_{620}{\mathrm{H}}^{\mathrm{HIn}}}{{\epsilon }_{620}^{1^{2-}}-{\mathrm{R}}_{\mathrm{A}}{\epsilon }_{434}{\mathrm{In}}^2}\\ =\frac{\left(2.84\right)\left(8.00\times {10}^3\right)-\left(0\right)}{\left(1.70\times {10}^4\right)-\left(2.84\right)\left(1.90\times {10}^3\right)}\\ =1.958\\ \left[{\mathrm{H}}^{+}\right]=\left(2.0\times {10}^{-7}\right)/1.958\\ \left[{\mathrm{H}}^{+}\right]=1.02\times {10}^{-7}\mathrm{M}\end{array}$

To find$\left[{\mathrm{CO}}_2\left(\mathrm{aq}\right)\right]$, The value of $\left[{\mathrm{H}}^{+}\right]$is substituted into mass balance

${\mathrm{F}}_{\mathrm{Na}}+\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}^{+}\right]}+\frac{{\mathrm{F}}_{\mathrm{in}}}{{\mathrm{R}}_{\ln }}+2\frac{{\mathrm{K}}_{\ln }{\mathrm{F}}_{\mathrm{in}}}{\left[{\mathrm{H}}^{+}\right]\left({\mathrm{R}}_{\ln }+1\right)}+\frac{\left.{\mathrm{K}}_1\left[{\mathrm{CO}}_{2\mathrm{la}}\right)\right]}{\left[{\mathrm{H}}^{+}\right]}+2\frac{{\mathrm{K}}_1{\mathrm{K}}_2\left[{\mathrm{CO}}_{2\left(\mathrm{aq}\right)}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}$

$\begin{array}{r}92.0\times {10}^{-6}+1.02\times {10}^{-7}=\frac{\left(6.7\times {10}^{-15}\right)}{\left(1.02\times {10}^{-7}\right)}+\frac{\left(50.0\times {10}^6\right)}{1.958+1}+2\frac{\left(2.0\times {10}^7\right)\left(50.0\times {10}^6\right)}{\left(1.02\times {10}^{-7}\right)(1.958+1)}\\ +\frac{\left(3.0\times {10}^7\right)\left[{\mathrm{CO}}_{2(\mathrm{aj}}\right]}{\left(1.02\times {10}^{-7}\right)}+2\frac{\left(3.0\times {10}^7\right)\left(3.3\times {10}^{-11}\right)\left[{\mathrm{CO}}_{2\mathrm{x}}\right]}{{\left(1.02\times {10}^{-7}\right)}^1}\\ 9.21\times {10}^{-5}=6.56\times {10}^{-8}+1.69\times {10}^{-5}+6.62\times {10}^{-5}+2.94\left[{\mathrm{CO}}_{2\text{(aq)}}\right]+0.0019\left[{\mathrm{CO}}_{2\text{(aq)}}\right]\\ \mathrm{The}\mathrm{Solution}\mathrm{is}\left[{\mathrm{CO}}_{2\left(\text{(aq)}\right)}\right]=3.04\times {10}^{-6}\mathrm{M}\end{array}$

The ${\mathrm{Na}}^{+},{\mathrm{HIn}}^{-},{\mathrm{In}}^{2-},{{\mathrm{HCO}}_3}^{-},{{\mathrm{CO}}_3}^{2-}\text{and}{\mathrm{OH}}^{-}$are present in the solution. If the net cation charge is $92.1\mathrm{\mu M}$,the net charge on anion should be $92.1\mathrm{\mu M}$and the ionic strength must approximately $~92\mathrm{\mu M}\approx {10}^{-4}M$activity coefficients are close to 1.00 and hence an ionic strength of ${10}^{-4}$Mis low .

$\begin{array}{r}\left[{\mathrm{OH}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}^{+}\right]}=0.07\mu \mathrm{M}\\ \left[{\mathrm{HIn}}^{-}\right]=\frac{{\mathrm{F}}_{\ln }}{{\mathrm{R}}_{\ln }+\mathrm{I}}\\ =16.9\mu \mathrm{m}\\ \left[{\ln }^{2-}\right]=\frac{{\mathrm{K}}_{\ln }{\mathrm{F}}_{\ln }}{\left[{\mathrm{H}}^{+}\right]\left({\mathrm{R}}_{\ln }+1\right)}\\ =33\cdot 1\mu \mathrm{m}\\ \left[{{\mathrm{HCO}}_3}^{-}\right]=\frac{{\mathrm{K}}_1\left[{\mathrm{CO}}_{2\left(\mathrm{aq}\right)}\right]}{\left[{\mathrm{H}}^{+}\right]}\\ =2.94\left[{\mathrm{CO}}_{2\left(\mathrm{aq}\right)}\right]\\ =8.9\mu \mathrm{M}\\ \left[{{\mathrm{CO}}_3}^{2-}\right]=\frac{{\mathrm{K}}_1{\mathrm{K}}_2\left[{\mathrm{CO}}_{2\left(\mathrm{mq}\right)}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}\\ =0.0019\left[{\mathrm{CO}}_{2\text{(aq)}}\right]\\ =0.003\mu \mathrm{M}\end{array}$

$\begin{array}{c}\sum {\mathrm{c}}_{\mathrm{i}}{\mathrm{Z}}_{\mathrm{i}}^2=\\ \left\{\left[{\mathrm{Na}}^{+}\right]\cdot 1^2+\left[{\mathrm{H}}^{+}\right]\cdot 1^2+\left[{\mathrm{OH}}^{-}\right]\cdot 1^2+\left[{\mathrm{In}}^{2-}\right]\cdot 2^2+\left[{\mathrm{HCO}}_3^{-}\right]\cdot 1^2+\left[{\mathrm{CO}}_3^{2-}\right]\cdot 2^2+\left[{\mathrm{CO}}_3^{2-}\right]\cdot 2^2\right.\\ \}\end{array}$

The solution for ionic strength is 125$\mathrm{\mu M}$ .