91Ó°ÊÓ

Volume of phosphate buffer=90mL

Volume of the solution containing indicator= 10mL

Concentration of the indicator=5.00 ×10^-4 M

Concentration of phosphate buffer= 0.1 M

pH=7.5

Wavelength =435 nm

At pH 7.5 , presence of H₂In in solution is negligible. As it can be seen from the given reaction that pK₁=1.00, therefore we can write

$\begin{array}{c}\left[HI{n}^{−}\right]+\left[I{n}^{2−}\right]=5.00×{10}^{−5}\\ pH=p{K}_2+\log \frac{\left[I{n}^{2−}\right]}{\left[HI{n}^{−}\right]}\\ 7.5=7.95+\log \frac{\left[I{n}^{2−}\right]}{5.00×{10}^{−5}−\left[I{n}^{2−}\right]}\\ \left[I{n}^{2−}\right]=1.31×{10}^{−5}M\end{array}$

$\begin{array}{c}\left[HI{n}^{−}\right]=5.00×{10}^{−5}−\left[I{n}^{2−}\right]\\ =3.69×{10}^{−5}M\end{array}$

Let the absorbance can be denoted by the symbol A

$\begin{array}{c}{A}_{\mathrm{λ}}=\mathrm{Î\mu }CL\\ C_I=\text{ }concentration\text{ }of\text{ }the\text{ â¶Ä‰}subs\tan ce\\ L=\text{ }Length\text{ }of\text{ }the\text{ }cell\end{array}$

Therefore, we can write

$\begin{array}{c}{A}_{435}={\mathrm{Î\mu }}_{435}\left[C_{HI{n}^{−}}\right]\left(1.00\right)+{\mathrm{Î\mu }}_{435}\left[C_{I{n}^{2−}}\right]\left(1.00\right)\\ ={\mathrm{Î\mu }}_{435}\left[HI{n}^{−}\right]\left(1.00\right)+{\mathrm{Î\mu }}_{435}\left[I{n}^{2−}\right]\left(1.00\right)\\ =\left[\left(1.80×{10}^4\right)×\left(3.69×{10}^{−5}\right)×\left(1.00\right)\right]+\left[\left(1.15×{10}^4\right)×\left(1.31×{10}^{−5}\right)×\left(1.00\right)\right]\\ =0.815\end{array}$