Chapter 19, Problem 19.10P

a)

Interpretation Introduction

Interpretation:

The equations have to be written for $\left[{\text{HIn}}^{\text{-}}\right]$ and $\left[{\text{In}}^{\text{2-}}\right]$ in terms of absorbance at given wavelength and also $\frac{\left[{\text{In}}^{\text{2-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\text{=}\frac{{\text{R}}_{\text{A}}{\text{ε}}_{\text{434}}{}^{\text{HIn}}{\text{-ε}}_{\text{620}}{}^{\text{HIn}}}{{\text{ε}}_{\text{620}}{}^{{\text{In}}^{\text{2-}}}{\text{-R}}_{\text{A}}{\text{ε}}_{\text{434}}{}^{{\text{In}}^{\text{2-}}}}{\text{=R}}_{\text{In}}$ has to be proved.

Concept Introduction:

Beer law:

$\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}$

$\begin{array}{l}\text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

To write the equations for $\left[{\text{HIn}}^{\text{-}}\right]$ and $\left[{\text{In}}^{\text{2-}}\right]$ in terms of absorbance at given wavelength

To prove $\frac{\left[{\text{In}}^{\text{2-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\text{=}\frac{{\text{R}}_{\text{A}}{\text{ε}}_{\text{434}}{}^{\text{HIn}}{\text{-ε}}_{\text{620}}{}^{{\text{In}}^{\text{2-}}}}{{\text{ε}}_{\text{620}}{}^{{\text{In}}^{\text{2-}}}{\text{-R}}_{\text{A}}{\text{ε}}_{\text{434}}{}^{\text{HIn}}}{\text{=R}}_{\text{In}}$

Explanation of Solution

$\begin{array}{l}{\text{A}}_{\text{620}}\text{=ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{620}\end{array}\text{b}\left[{\text{HIn}}^{\text{-}}\right]\text{+ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{620}\end{array}\text{b}\left[{\text{In}}^{\text{2-}}\right]\\ \\ {\text{A}}_{\text{434}}\text{=ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{434}\end{array}\text{b}\left[{\text{HIn}}^{\text{-}}\right]\text{+ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{434}\end{array}\text{b}\left[{\text{In}}^{\text{2-}}\right]\end{array}$

The solution obtained by solving the two above equations are,

$\begin{array}{l}\left[{\text{HIn}}^{\text{-}}\right]\text{=}\frac{\text{1}}{\text{D}}\left({\text{A}}_{\text{620}}\epsilon \begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{434}\end{array}{\text{b-A}}_{\text{434}}\epsilon \begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{620}\end{array}\text{b}\right)\\ \\ \left[{\text{In}}^{\text{2-}}\right]\text{=}\frac{\text{1}}{\text{D}}\left({\text{A}}_{\text{434}}\epsilon \begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{620}\end{array}{\text{b-A}}_{\text{620}}\epsilon \begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{434}\end{array}\text{b}\right)\\ \\ {\text{D=b}}^{\text{2}}\left(\text{ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{620}\end{array}\text{ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{434}\end{array}\text{ - ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{620}\end{array}\text{ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{434}\end{array}\right)\end{array}$

Divide the expression for $\left[{\text{In}}^{\text{2-}}\right]$ by expression for $\left[{\text{HIn}}^{\text{-}}\right]$ . The resultant equation is,

$\frac{\left[{\text{In}}^{\text{2-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\text{=}\frac{{\text{A}}_{\text{434}}\text{ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{620}\end{array}{\text{-A}}_{\text{620}}\text{ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{434}\end{array}}{{\text{A}}_{\text{620}}\text{ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{434}\end{array}{\text{-A}}_{\text{434}}\text{ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{620}\end{array}}$

Divide the numerator and denominator on right side by ${\text{A}}_{\text{434}}$ . The solution is

$\begin{array}{l}\frac{\left[{\text{In}}^{\text{2-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\text{=}\frac{\text{ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{620}\end{array}{\text{-R}}_{\text{a}}\text{ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{434}\end{array}}{{\text{R}}_{\text{a}}\text{ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{434}\end{array}\text{-ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{620}\end{array}}\\ \\ \frac{\left[{\text{In}}^{\text{2-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\text{=}\frac{{\text{R}}_{\text{a}}\text{ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{434}\end{array}\text{-ε}\begin{array}{c}{\text{HIn}}^{\text{-}}\\ \text{620}\end{array}}{\text{ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{620}\end{array}{\text{-R}}_{\text{a}}\text{ε}\begin{array}{c}{\text{In}}^{\text{2-}}\\ \text{434}\end{array}}\end{array}$

b)

Interpretation Introduction

Interpretation:

The equations

$\begin{array}{l}\left[{\text{HIn}}^{\text{-}}\right]\text{=}\frac{{\text{F}}_{\text{In}}}{{\text{R}}_{\text{In}}\text{+1}}\\ \\ \left[{\text{In}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{In}}{\text{F}}_{\text{In}}}{\left[{\text{H}}^{\text{+}}\right]\left({\text{R}}_{\text{In}}\text{+1}\right)}\end{array}$ Have to be proved for acid dissociation constant ${\text{K}}_{\text{In}}$ .

To prove ${\text{K}}_{\text{In}}$ = $\begin{array}{l}\left[{\text{HIn}}^{\text{-}}\right]\text{=}\frac{{\text{F}}_{\text{In}}}{{\text{R}}_{\text{In}}\text{+1}}\\ \\ \left[{\text{In}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{In}}{\text{F}}_{\text{In}}}{\left[{\text{H}}^{\text{+}}\right]\left({\text{R}}_{\text{In}}\text{+1}\right)}\end{array}$

Explanation of Solution

For indicator, the mass balance can be given as ,

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

Divide by $\left[{\text{HIn}}^{\text{-}}\right]$ on both sides ,

$\begin{array}{l}\frac{\left[{\text{HIn}}^{\text{-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\text{+}\frac{\left[{\text{In}}^{\text{2-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\text{=}\frac{{\text{F}}_{\text{In}}}{\left[{\text{HIn}}^{\text{-}}\right]}\\ \\ {\text{1+R}}_{\text{In}}\text{=}\frac{{\text{F}}_{\text{In}}}{\left[{\text{HIn}}^{\text{-}}\right]}\\ \\ \left[{\text{HIn}}^{\text{-}}\right]\text{=}\frac{{\text{F}}_{\text{In}}}{{\text{R}}_{\text{In}}\text{+1}}\end{array}$

The acid dissociation constant of the indicator is,

${\text{K}}_{\text{In}}\text{=}\frac{\left[{\text{In}}^{\text{2-}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}$

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

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

c)

Interpretation Introduction

Interpretation:

$\left[{\text{H}}^{\text{+}}\right]{\text{=K}}_{\text{In}}{\text{/R}}_{\text{In}}$ has to be proved.

To prove $\left[{\text{H}}^{\text{+}}\right]{\text{=K}}_{\text{In}}{\text{/R}}_{\text{In}}$

Explanation of Solution

${\text{R}}_{\text{In}}\text{=}\frac{\left[{\text{In}}^{\text{2-}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}$

Hence,

$\begin{array}{l}{\text{K}}_{\text{In}}\text{=}\frac{\left[{\text{In}}^{\text{2-}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HIn}}^{\text{-}}\right]}\\ \\ {\text{R}}_{\text{In}}\left[{\text{H}}^{\text{+}}\right]\end{array}$

Therefore,

$\left[{\text{H}}^{\text{+}}\right]{\text{=K}}_{\text{In}}{\text{/R}}_{\text{In}}$

d)

Interpretation Introduction

Interpretation:

With the help of Carbonic acid dissociation equilibria $\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{\left[{\text{H}}^{\text{+}}\right]}$ and $\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}$ have to be proved.

To prove $\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{\left[{\text{H}}^{\text{+}}\right]}$ and $\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}$

Explanation of Solution

From the acid dissociation reaction of Carbonic acid, it can be written as,

$\begin{array}{l}{\text{K}}_{\text{1}}\text{=}\frac{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{CO}}_{\text{2(aq)}}\right]}\\ \\ \left[{\text{HCO}}_{\text{3}}\right]\text{=}\frac{{\text{K}}_{\text{1}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\end{array}$

From the acid dissociation reaction of Bicarbonate, it can be written as,

$\begin{array}{l}{\text{K}}_{\text{2}}\text{=}\frac{\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]}\\ \\ \left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{2}}\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\end{array}$

Substitute in $\left[{\text{HCO}}_{\text{3}}\right]$ gives,

$\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}$

e)

Interpretation Introduction

Interpretation:

The charge balance for the solution has to be determined and substituted in expressions for $\left[{\text{HIn}}^{\text{-}}\right]\text{,}\left[{\text{In}}^{\text{2-}}\right]\text{,}\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\text{and}\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$ .

To determine the charge balance for the solution and substitute the expressions for $\left[{\text{HIn}}^{\text{-}}\right]\text{,}\left[{\text{In}}^{\text{2-}}\right]\text{,}\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\text{and}\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$

Explanation of Solution

The charge balance is given as,

$\left[{\text{Na}}^{\text{+}}\right]\text{+}\left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{+}\left[{\text{HIn}}^{\text{-}}\right]\text{+2}\left[{\text{In}}^{\text{2-}}\right]\text{+}\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\text{+2}\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$

${\text{F}}_{\text{Na}}\text{+}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{\text{+}}\right]}\text{+}\frac{{\text{F}}_{\text{In}}}{{\text{R}}_{\text{In}}\text{+1}}\text{+2}\frac{{\text{K}}_{\text{In}}{\text{F}}_{\text{In}}}{\left[{\text{H}}^{\text{+}}\right]\text{+}\left({\text{R}}_{\text{In}}\text{+1}\right)}\text{+}\frac{{\text{K}}_{\text{1}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\text{+2}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{{\left[{\text{H}}^{\text{+}}\right]}_{\text{2}}}$

f)

Interpretation Introduction

Interpretation:

The amount of $\left[{\text{CO}}_{\text{2(aq)}}\right]$ in sea water has to be found

To find the amount $\left[{\text{CO}}_{\text{2(aq)}}\right]$ in sea water

Explanation of Solution

$\left[{\text{H}}^{\text{+}}\right]{\text{=K}}_{\text{In}}{\text{/R}}_{\text{In}}$

${\text{R}}_{\text{In}}$ can be calculated from,

$\begin{array}{l}{\text{R}}_{\text{In}}\text{=}\frac{{\text{R}}_{\text{A}}{\text{ε}}_{\text{434}}{}^{\text{HIn}}{\text{-ε}}_{\text{620}}{}^{\text{HIn}}}{{\text{ε}}_{\text{620}}{}^{{\text{In}}^{\text{2-}}}{\text{-R}}_{\text{A}}{\text{ε}}_{\text{434}}{}^{{\text{In}}^{\text{2-}}}}\text{=}\frac{\left(\text{2}\text{.84}\right)\left(\text{8}\text{.00}{\text{×10}}^{\text{3}}\right)\text{-}\left(\text{0}\right)}{\left(\text{1}{\text{.70×10}}^{\text{4}}\right)\text{-}\left(\text{2}\text{.84}\right)\left(\text{1}{\text{.90×10}}^{\text{3}}\right)}\text{=1}\text{.958}\\ \\ \left[{\text{H}}^{\text{+}}\right]{\text{=K}}_{\text{In}}{\text{/R}}_{\text{In}}\\ \\ \left[{\text{H}}^{\text{+}}\right]\text{=}\left(\text{2}{\text{.0×10}}^{\text{-7}}\right)\text{/1}\text{.958}\\ \\ \left[{\text{H}}^{\text{+}}\right]\text{=1}{\text{.02×10}}^{\text{-7}}\text{M}\end{array}$

The value of $\left[{\text{H}}^{\text{+}}\right]$ is substituted into mass balance and an unknown equation of $\left[{\text{CO}}_{\text{2(aq)}}\right]$ is produced,

$\begin{array}{l}{\text{F}}_{\text{Na}}\text{+}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{\text{+}}\right]}\text{+}\frac{{\text{F}}_{\text{In}}}{{\text{R}}_{\text{In}}}\text{+2}\frac{{\text{K}}_{\text{In}}{\text{F}}_{\text{In}}}{\left[{\text{H}}^{\text{+}}\right]\left({\text{R}}_{\text{In}}\text{+1}\right)}\text{+}\frac{{\text{K}}_{\text{1}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\text{+2}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}\\ \\ \text{92}{\text{.0×10}}^{\text{-6}}\text{+1}{\text{.02×10}}^{\text{-7}}\text{=}\frac{\left(\text{6}{\text{.7×10}}^{\text{-15}}\right)}{\left(\text{1}{\text{.02×10}}^{\text{-7}}\right)}\text{+}\frac{\left(\text{50}{\text{.0×10}}^{\text{-6}}\right)}{\text{1}\text{.958+1}}\text{+2}\frac{\left(\text{2}{\text{.0×10}}^{\text{-7}}\right)\left(\text{50}{\text{.0×10}}^{\text{-6}}\right)}{\left(\text{1}{\text{.02×10}}^{\text{-7}}\right)\left(\text{1}\text{.958+1}\right)}\\ \text{+}\frac{\left(\text{3}{\text{.0×10}}^{\text{-7}}\right)\left[{\text{CO}}_{\text{2(aq)}}\right]}{\left(\text{1}{\text{.02×10}}^{\text{-7}}\right)}\text{+2}\frac{\left(\text{3}{\text{.0×10}}^{\text{-7}}\right)\left(\text{3}{\text{.3×10}}^{\text{-11}}\right)\left[{\text{CO}}_{\text{2(aq)}}\right]}{{\left(\text{1}{\text{.02×10}}^{\text{-7}}\right)}^{\text{1}}}\\ \\ \text{9}{\text{.21×10}}^{\text{-5}}\text{=6}{\text{.56×10}}^{\text{-8}}\text{+1}{\text{.69×10}}^{\text{-5}}\text{+6}{\text{.62×10}}^{\text{-5}}\text{+2}\text{.94}\left[{\text{CO}}_{\text{2(aq)}}\right]\text{+0}\text{.0019}\left[{\text{CO}}_{\text{2(aq)}}\right]\\ \\ \left[{\text{CO}}_{\text{2(aq)}}\right]\text{=3}{\text{.04×10}}^{\text{-6}}\text{M}\end{array}$

g)

Interpretation Introduction

Interpretation:

The ionic strength inside the sensor compartment has to be determined.

To determine the ionic strength inside the sensor compartment

Explanation of Solution

${\text{Na}}^{\text{+}}{\text{,HIn}}^{\text{-}}{\text{,In}}^{\text{2-}}{\text{,HCO}}_{\text{3}}{}^{\text{-}}{\text{,CO}}_{\text{3}}{}^{\text{2-}}\text{and}{\text{OH}}^{\text{-}}$ are present in the solution. If the net cation charge is $\text{92}\text{.1μM}$, the net charge on anion should be $\text{92}\text{.1μM}$ and the ionic strength must approximately $\sim \text{92μm}\approx {\text{10}}^{\text{-4}}\text{M}$ . The activity coefficients are close to $\text{1}\text{.00}$ and hence an ionic strength of ${\text{10}}^{\text{-4}}\text{M}$ is low.

The exact ionic strength can be calculated as,

$\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{\text{+}}\right]}\text{=0}\text{.07μM}$

$\begin{array}{l}\left[{\text{HIn}}^{\text{-}}\right]\text{=}\frac{{\text{F}}_{\text{In}}}{{\text{R}}_{\text{In}}\text{+1}}\text{=16}\text{.9μm}\\ \\ \left[{\text{In}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{In}}{\text{F}}_{\text{In}}}{\left[{\text{H}}^{\text{+}}\right]\left({\text{R}}_{\text{In}}\text{+1}\right)}\text{=33}\text{.1μm}\\ \\ \left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\text{=2}\text{.94}\left[{\text{CO}}_{\text{2(aq)}}\right]\text{=8}\text{.9μM}\\ \\ \left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[{\text{CO}}_{\text{2(aq)}}\right]}{{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}\text{=0}\text{.0019}\left[{\text{CO}}_{\text{2(aq)}}\right]\text{=0}\text{.003μM}\end{array}$

Ionic strength =

$\begin{array}{l}\frac{\text{1}}{\text{2}}{\displaystyle \sum {\text{c}}_{\text{i}}{\text{z}}_{\text{i}}{}^{\text{2}}\text{=}}\left\{\left[{\text{Na}}^{\text{+}}\right]{\text{.1}}^{\text{2}}\text{+}\left[{\text{H}}^{\text{+}}\right]{\text{.1}}^{\text{2}}\text{+}\left[{\text{OH}}^{\text{-}}\right]{\text{.1}}^{\text{2}}\text{+}\left[{\text{In}}^{\text{2-}}\right]{\text{.2}}^{\text{2}}\text{+}\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]{\text{.1}}^{\text{2}}\text{+}\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]{\text{.2}}^{\text{2}}\text{+}\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]{\text{.2}}^{\text{2}}\right\}\\ \\ \text{=125μM}\end{array}$