91影视

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

Consider the reaction, $\mathrm{B}+\mathrm{HA}鈫�{\mathrm{BH}}^{+}+{\mathrm{A}}^{鈭�}$

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

Calculate theof aniline,

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

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

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

Hence, for the reactionis,

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

Evaluate the value of pH,

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

Thus, the absorbance is,

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

Hence, the initial [Hln] is,

$[H鈭�\mathrm{n}]=\left[\mathrm{HIn}\right]鈰�\frac{\mathrm{V}}{{\mathrm{V}}_{\text{dilute}}}$

Thus,

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

Substitute the value of [Hln] in the equation,

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

Therefore, the concentration of Hln is,

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

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

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

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

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