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.The addition of NaOH to alanine will result in the deprotonation of the protonated form of alanine, H2X+, to form the anionic form of alanine, X-. The pH of the solution will increase as a result of the decrease in the concentration of H+ ions.The equilibrium constants for the dissociation of H2X+ are:Ka1 = 4.6 × 10^-3Ka2 = 2.0 × 10^-10The pH of the solution can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the anionic form of the acid, and [HA] is the concentration of the protonated form of the acid.In this case, the pKa of the protonated form of alanine is:pKa1 = -log(4.6 × 10^-3) = 2.34The concentration of the anionic form of alanine can be calculated using the following equation:[A-] = [NaOH] - [H2X+]where [NaOH] is the concentration of NaOH added and [H2X+] is the concentration of the protonated form of alanine.In this case, [NaOH] = 0.100 M and [H2X+] can be calculated using the following equation:[H2X+] = Ka1[A-]/[OH-]where [OH-] is the concentration of hydroxide ions.In this case, [OH-] can be calculated using the following equation:[OH-] = Kw/[H+]where Kw is the ion product of water (1.0 × 10^-14).In this case, [H+] can be calculated using the following equation:[H+] = 10^-pHSubstituting the appropriate values into the above equations, we get:[A-] = 0.100 M - 4.6 × 10^-3 M = 0.0954 M[H2X+] = 4.6 × 10^-3 M * 0.0954 M / 1.0 × 10^-14 M = 4.36 × 10^-12 MpH = 2.34 + log(0.0954 M / 4.36 × 10^-12 M) = 11.11Therefore, the pH of the solution is 11.11.

The Dissociation Equilibrium Constants for the Protonated Form of Alanine