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• The Henderson鈥揌asselbalch equation, ${}^{\mathrm{pH}=\mathrm{pKa}+\log \left[\mathrm{A}\right]/\left[\mathrm{HA}\right]}$ , is used to calculate the ${}^{\mathrm{pH}}$of a buffer.
• The equilibrium concentrations of the conjugate acid鈥揵ase pair utilized to form the buffer solution are${}^{\left[\mathrm{HA}\right]}$and ${}^{\left[\mathrm{A}\right]}$ in this equation.
• The pH of the solution is equal to the acid's pKa when${}^{\left[\mathrm{HA}\right]=\left[\mathrm{A}\right]}$. Jay came up with the idea.

A solution which contains acetic acid, oxalic acid, ammonia and pyridine with ${}^{\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00}}$Calculate the fraction of ammonia that is unprotonated.

Consider the reaction

${\mathbf{NH}}_{\mathbf{4}}^{\mathbf{+}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{+}{\mathbf{NH}}_{\mathbf{3}}\mathbf{+}{\mathbf{H}}^{\mathbf{+}}$

The value of localid="1654777119771" ${}^{{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{245}}$${}^{{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{245}}$

By using the Henderson-Hasselbach equation

$$\begin{gathered} \mathbf{pH}\mathbf{}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\left(\left[{\mathrm{NH}}_3\right]/\left[{\mathrm{NH}}_4^{+}\right]\right) \\ \mathbf{9}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{245}\mathbf{+}\mathbf{log}\left(\left[{\mathrm{NH}}_3\right]/\left[{\mathrm{NH}}_4^{+}\right]\right) \\ \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{245}\mathbf{=}\mathbf{log}\left(\left[{\mathrm{NH}}_3\right]/\left[{\mathrm{NH}}_4^{+}\right]\right) \\ {\mathbf{10}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{245}}\mathbf{=}\left[{\mathrm{NH}}_3\right]\mathbf{/}\left[{\mathrm{NH}}_4^{+}\right] \\ \mathbf{0}\mathbf{.}\mathbf{569}\mathbf{=}\left[{\mathrm{NH}}_3\right]\mathbf{/}\left[{\mathrm{NH}}_4^{+}\right] \end{gathered}$$

Hence, this is the fraction of protonated ammonia

In order to get the fraction of ammonia which is unprotonated we will do the following:

$$\begin{gathered} \mathbf{鈭�}\left(\mathrm{unprotonated}\right)\mathbf{}\mathbf{=}\mathbf{}\frac{\left[{\mathrm{NH}}_3\right]}{\left[{\mathrm{NH}}_3\right]\mathbf{+}\left[{\mathrm{NH}}_4^{+}\right]} \\ \mathbf{鈭�}\left(\mathrm{unprotonated}\right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{5969}}{\mathbf{0}\mathbf{.}\mathbf{569}\mathbf{+}\mathbf{1}} \\ \mathbf{鈭�}\left(\mathrm{unprotonated}\right)\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{36} \end{gathered}$$

Therefore, the fraction of unprotonated ammonia is data-custom-editor="chemistry" ${}^{\mathbf{0}\mathbf{.}\mathbf{36}}$.