How many millilitres of 1.00 M KOH should be added to 100 mL of solution containing 10.0 g of histidine hydrochloride

[His? HCI (HisH1) (CI2), FM 191.62 ] to get a pH of 9.30 ?

The derivation of fractional composition equations for a diprotic system follows the same pattern used for the monoprotic system.

A material that behaves like an acid and has at least two hydrogens is referred to as diprotic acid in general. An acid, according to Bronsted-theory, Lowry's is a chemical species capable of giving away hydrogen ions to other atoms it is reacting with.

Given Formula mass of ${\mathrm{H}}_2{\mathrm{His}}^{+}{\mathrm{Cl}}^{-}\mathrm{is}191.62$

The weight of ${\mathrm{H}}_2{\mathrm{His}}^{+}{\mathrm{Cl}}^{-}10.00\mathrm{g}$

The required amount of 1.00M KOH solution

The required $0.{0521}_9$mol of KOH + x amount

$\mathrm{HHis}+{\mathrm{OH}}^{-}\to {\mathrm{His}}^{-}$

Initial mol:$0.{0521}_9\mathrm{x}-$

Final mol:$0.{0521}_{9-x}-\mathrm{x}$

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_3+\log \frac{\left[{\mathrm{His}}^2\right]}{\left|{\mathrm{HHis}}^2\right|} \\ =9.30 \\ =9.28+\log \frac{\mathrm{x}}{0.{0521}_{9_{\mathrm{x}}}} \\ \mathrm{x}=0.{0267}_0 \end{gathered}$$

$$\begin{gathered} \mathrm{molKOH}=0.{0521}_9+0.{0267}_0 \\ =78.9\mathrm{mL} \\ 1.00\mathrm{MKOH} \end{gathered}$$

Thus, the required amount of Potassium hydroxide solution 78.9mL