Question

The both ${}^{\left[H_{2}A\right]\mathbf{=}\left[H{A}^{‐}\right]}$are equal as found to be ${}^{\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{54}}$

Short answer

The both ${}^{\mathbf{[}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{A}\mathbf{]}\mathbf{=}\mathbf{[}\mathbf{H}\mathbf{}{\mathbf{A}}^{\mathbf{‐}}\mathbf{]}}$ are found to be equal${}^{\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{54}}$

Step-by-step solution

Definition of Henderson-Hassel Balch equation

• The Henderson Hasselbalch equation approximates the link between a solution's pH or pOH, its pKa or pKb, and the ratio of dissociated chemical species concentrations.

Determine the both  are equal as found to be 

The both ${}^{\mathbf{[}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{A}\mathbf{]}\mathbf{=}\mathbf{[}\mathbf{H}\mathbf{}{\mathbf{A}}^{\mathbf{‐}}\mathbf{]}}$are equal, and the pH must be determined.

Henderson-Hassel Balch equation

$$\begin{gathered} \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{1}}\mathbf{+}\mathbf{log}\frac{\left[H{A}^{-}\right]}{\left[H_{2}A\right]} \\ \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{2}}\mathbf{+}\mathbf{log}\frac{\left[A^2\right]}{\left[H{A}^{-}\right]} \end{gathered}$$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

It are aware of this.

The purity of the intermediate species' pH,${}^{{\mathbf{H}}_{\mathbf{2}\mathbf{}}{\mathbf{Arg}}^{\mathbf{+}}}$

Arginine is a kind of amino acid ${}^{\left[{\mathrm{Arg}}^{-}\right]}$at ${}^{{\mathbf{pK}}_{\mathbf{3}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{1}}$is 4.00

${\mathbf{H}}_{\mathbf{2}}{\mathbf{Arg}}^{\mathbf{+}}\mathbf{\approx }\frac{\mathbf{1}}{\mathbf{2}}\left({\mathrm{pK}}_1+{\mathrm{pK}}_2\right)\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{40}$

${}^{\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{54}}$(and a pH exceeding ${}^{\mathbf{=}{\mathbf{pK}}_{\mathbf{2}}}$ )

Therefore ${}^{\left[H_2{\mathrm{Arg}}^{+}\right]=\left[{\mathbf{Arg}}^{-}\right]}$at ${}^{\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{54}}$

Both of them ${}^{\mathbf{[}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{A}\mathbf{]}\mathbf{=}\mathbf{[}\mathbf{H}\mathbf{}{\mathbf{A}}^{\mathbf{‐}}\mathbf{]}}$are found to be equal ${}^{\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{54}}$.