The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid H₂X⁺)are K_a1 = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.100 M solution of alanine after 100.00 mL of 0.100 M NaOH has been added? A)9.70 B)10.69 C)11.11 D)12.70

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The correct answer is C because the pH of the solution is 11.11. The pKa of the first dissociation is 2.34 and the pKa of the second dissociation is 9.70. The addition of NaOH will cause the first dissociation to occur, resulting in the formation of H2X+ and OH-. The OH- will then react with H2X+ to form HX and H2O. The HX will then dissociate to form X- and H+. The pH of the solution will be determined by the concentration of H+ ions, which will be equal to the concentration of X- ions. The concentration of X- ions can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([X-]/[HX]). Substituting the values of pKa and [HX] into the equation, we get: pH = 9.70 + log([X-]/[HX]) = 9.70 + log(0.100/0.050) = 11.11.

The Dissociation Equilibrium Constants for the Protonated Form of Alanine