Spectrophotometric properties of a particular indicator are given below:

A solution with a volume of 20.0 mL containing 1.40 ×10^-5 M indicator plus 0.050 0 M benzene-1,2,3-tricarboxylic acid was treated with 20.0 mL of aqueous KOH. The resulting solution had an absorbance at 604 nm of 0.118 in a 1.00-cm cell. Calculate the molarity of the KOH solution.

Volume of aqueous KOH=20mL

Volume of the solution containing indicator= 20mL

Concentration of the indicator=1.40 ×10^-5 M

Concentration of benzene-1,2,3-tricarboxylic acid= 0.050 M

Solution absorbance =0.118

Wavelength =604 nm

Let the absorbance can be denoted by the symbol A

$\begin{array}{c}{A}_{\lambda }={\epsilon }_{I{n}^{-}}{C}_{I{n}^{-}}L\\ C_{I{n}^{-}}=concentrationofI{n}^{-}\\ L=Lengthofthecell\end{array}$

Therefore, we can write

$\begin{array}{c}{A}_{604}={\epsilon }_{I{n}^{-}}{C}_{I{n}^{-}}\left(1.00\right)\\ 0.118=4.97\times {10}^4\times C_{I{n}^{-}}\times \left(1.00\right)\\ C_{I{n}^{-}}=2.37\times {10}^{-6}M\end{array}$

A 20 mL solution containing indicator was diluted with a 20 mL of aqueous KOH solution.

The formal concentration of the indicator will be

$\begin{array}{l}=\frac{1.40\times {10}^{-5}\times 20}{20+20}M\\ =0.7\times {10}^{-5}M\\ =7\times {10}^{-6}M\end{array}$

$\begin{array}{c}\left[HIn\right]=\left(7\times {10}^{-6}-2.37\times {10}^{-6}\right)M\\ =4.63\times {10}^{-6}M\end{array}$

Therefore, the pH will be

$\begin{array}{c}pH=p{K}_{In}+\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}\\ =7.95+\log \frac{2.37\times {10}^{-6}}{4.63\times {10}^{-6}}\\ =7.66\end{array}$

In the solution there is a triprotic acid named benzene-1,2,3-tricarboxylic acid. It can be denoted as H₃A.

Following are the dissociation constants (pK_a) of this triprotic acid.

pK₁=2.88, pK₂=4.75, pK₃=7.13

As the pH obtained is 7.66 therefore, main species is A^3- and second main species is HA^2-

The number of moles of OH^- can be calculated as follows

$\begin{array}{c}pH=p{K}_3+\log \frac{\left[A^{3-}\right]}{\left[H{A}^{2-}\right]}\\ 7.66=7.13+\log \frac{x}{1.00-x}\\ =0.772mmol\end{array}$

The total KOH added =2.772 mmol

Molarity of KOH

$\begin{array}{l}=\frac{2.772}{20}M\\ =0.1386M\end{array}$