A volume of of solution of the protonated form of an amino acid, methionine () is titrated with . Th — Ionic Equilibrium Chemistry Question
Question
A volume of $30\text{ ml}$ of $0.06\text{ M}$ solution of the protonated form of an amino acid, methionine ($\text{H}_2\text{A}^+$) is titrated with $0.09\text{ M} – \text{NaOH}$. The pH of the resulting solution after addition of $20\text{ ml}$ of base, is ($pK_{a1} = 2.28$, $pK_{a2} = 9.72$)
💡 Solution & Explanation
Initial moles of $\text{H}_2\text{A}^+ = 30\text{ ml} \times 0.06\text{ M} = 1.8\text{ mmol}$. Moles of NaOH added = $20\text{ ml} \times 0.09\text{ M} = 1.8\text{ mmol}$. The strong base exactly neutralizes the first acidic proton, converting all of the $\text{H}_2\text{A}^+$ entirely into the amphiprotic zwitterion $\text{HA}$. For an amphiprotic species, the pH is given by the average of the two $pK_a$ values: $\text{pH} = (pK_{a1} + pK_{a2})/2 = (2.28 + 9.72)/2 = 12.00 / 2 = 6.0$. Therefore, correct answer is 6.