Challenging your acid-base prowess. A solution was prepared by mixing 25.00mL of 0.800Maniline, 25.00mLsulfanilic acid, andand then diluting to 100.0mL. (stands for protonated indicator.)

The absorbance measured at550nmin 5.00 - cmwas 0.110.Find the concentrations of$\text{HIn and In and}{p}_{\mathrm{a}}\text{for HIn}$

In the given amount of solution, the quantity of solute present is called concentration.

Consider the reaction, $\mathrm{B}+\mathrm{HA}\to {\mathrm{BH}}^{+}+{\mathrm{A}}^{-}$

$\mathrm{B}\text{is aniline, and}\mathrm{HA}\text{is sulfanilic acid}$

Calculate theof aniline,

$\begin{array}{r}\mathrm{n}\left(\text{aniline}\right)=\left[\text{aniline}\right]\cdot \mathrm{V}\\ \mathrm{n}\left(\text{aniline}\right)=0.080\mathrm{M}\cdot 25\cdot {10}^{-3}\mathrm{L}\\ \mathrm{n}\left(\text{aniline}\right)=2\cdot {10}^{-3}\mathrm{mol}\end{array}$

Find of sulfanilic acid,

$$\begin{gathered} \mathrm{n}\left(\mathrm{sulfanilc}\mathrm{acid}\right)=[\mathrm{sulfanilic}\mathrm{acid}].\mathrm{V} \\ \mathrm{n}\left(\mathrm{sulfanilc}\mathrm{acid}\right)=0.060\mathrm{M}.25.{10}^{-3}\mathrm{L} \\ \mathrm{n}\left(\mathrm{sulfanilc}\mathrm{acid}\right)=1.5.{10}^{-3}\mathrm{mol} \end{gathered}$$

Calculate the value of K

$\begin{array}{r}\mathrm{K}=\frac{{\mathrm{K}}_{\mathrm{a}}\cdot {\mathrm{K}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{w}}}\\ \mathrm{K}=\frac{{10}^{-3.232}\cdot \frac{{\mathrm{K}}_{\mathrm{w}}}{{10}^{-4.601}}}{{\mathrm{K}}_{\mathrm{w}}}\\ \mathrm{K}=23.39\end{array}$

Hence, for the reactionis,

$\begin{array}{r}\mathrm{K}=\frac{\left[{\mathrm{BH}}^{+}\right]}{\left[\mathrm{B}\right]\left[\mathrm{HA}\right]}\\ \frac{{\mathrm{x}}^2}{\left(2\cdot {10}^{-3}-\mathrm{x}\right)\cdot \left(1.5\cdot {10}^{-3}-\mathrm{x}\right)}=23.39\\ \frac{{\mathrm{x}}^2}{3\cdot {10}^{-6}-2\cdot {10}^{-3}\mathrm{x}-1.5\cdot {10}^{-3}\mathrm{x}+{\mathrm{x}}^2}=23.39\\ {\mathrm{x}}^2=7.017\cdot {10}^{-5}-0.04678\mathrm{x}-0.0351\mathrm{x}+23.39{\mathrm{x}}^2\\ 22.39{\mathrm{x}}^2-0.08188\mathrm{x}+7.017\cdot {10}^{-5}=0\\ \mathrm{x}=1.371\cdot {10}^{-3}\mathrm{mol}\end{array}$

Evaluate the value of pH,

$\begin{array}{r}\mathrm{pH}={\mathrm{pK}}_{{\mathrm{BH}}^{+}}+\log \left(\frac{\left[\mathrm{B}\right]}{{\mathrm{BH}}^{+}}\right)\\ \mathrm{pH}=4.601+\log \left(\frac{2\cdot {10}^{-3}-1.371\cdot {10}^{-3}}{1.371\cdot {10}^{-3}}\right)\\ \mathrm{pH}=4.26\end{array}$

Thus, the absorbance is,

$\begin{array}{r}\mathrm{A}={\dot{o}}_1\cdot \mathrm{b}\cdot \left[\mathrm{HIn}\right]+{\dot{o}}_2\cdot \mathrm{b}\cdot \left[\mathrm{In}\right]\\ 0.110=2.26\cdot {10}^4{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\cdot 5\mathrm{cm}\cdot \left[\mathrm{HIn}\right]+1.53\cdot {10}^4{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\cdot 5\mathrm{cm}\cdot \left[\mathrm{In}\right]\\ 0.110=113000\cdot \left[\mathrm{HIn}\right]+76500\cdot \left[\mathrm{In}\right]\end{array}$

Hence, the initial [Hln] is,

$[H\mid \mathrm{n}]=\left[\mathrm{HIn}\right]\cdot \frac{\mathrm{V}}{{\mathrm{V}}_{\text{dilute}}}$

Thus,

$\begin{array}{r}\left[\mathrm{HIn}\right]=1.23\cdot {10}^{-4}\mathrm{M}\cdot \frac{1\cdot {10}^{-3}\mathrm{L}}{100\cdot {10}^{-3}\mathrm{L}}\\ \left[\mathrm{HIn}\right]=1.23\cdot {10}^{-6}\mathrm{M}\\ \text{Thus,}\\ \left[\mathrm{HIn}\right]=1.23\cdot {10}^{-6}-\left[{\mathrm{In}}^{-}\right]\end{array}$

Substitute the value of [Hln] in the equation,

$\begin{array}{r}0.110=113000\cdot \left(1.23\cdot {10}^{-6}-\left[{\mathrm{In}}^{-}\right]\right)+76500\cdot \left[{\mathrm{In}}^{-}\right]\\ 0.110=0.13899-113000\left[{\ln }^{-}\right]+76500\cdot \left[{\ln }^{-}\right]\\ 36500\cdot \left[{\mathrm{In}}^{-}\right]=0.02899\\ \left[{\ln }^{-}\right]=7.94\cdot {10}^{-7}\mathrm{M}\end{array}$

Therefore, the concentration of Hln is,

$\begin{array}{c}\left[HI\mathrm{In}\right]=1.23\cdot {10}^{-6}-7.94\cdot {10}^{-7}\\ \left[H\mathrm{HI}\right]=4.36\cdot {10}^{-7}\mathrm{M}\end{array}$

Obtain the concentration of${\mathrm{pK}}_{\mathrm{a}}$ ,

$\begin{array}{r}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{\left[{\mathrm{n}}^{-}\right]}{\left[\mathrm{HIn}\right]}\right)\\ {\mathrm{pK}}_{\mathrm{a}}=\mathrm{pH}-\log \left(\frac{\left[{\mathrm{n}}^{-}\right]}{\left[\mathrm{HIn}\right]}\right)\\ {\mathrm{pK}}_{\mathrm{a}}=4.26-\log \left(\frac{7.94\cdot {10}^{-7}}{4.36\cdot {10}^{-7}}\right)\\ {\mathrm{pK}}_{\mathrm{a}}=4.00\end{array}$

Thus, the concentration of Hln,In and ${\mathrm{pK}}_{\mathrm{a}}$are,

$\begin{array}{r}\left[\mathrm{HIn}\right]=4.36\cdot {10}^{-7}\mathrm{M}\\ \left[{\mathrm{In}}^{-}\right]=7.94\cdot {10}^{-7}\mathrm{M}\\ {\mathrm{pK}}_{\mathrm{a}}=4.00\end{array}$