Amino acids contain both an acidic carboxylic acid group \((-\mathrm{COOH})\) and a basic amino group \(\left(-\mathrm{NH}_{2}\right)\) The amino group can be protonated (that is, it has an extra proton attached) in a strongly acidic solution. This produces a diprotic acid of the form \(\mathrm{H}_{2} \mathrm{A}^{+}\), as exemplified by the protonated amino acid alanine. The protonated amino acid has two ionizable protons that can be titrated with \(\mathrm{OH}^{-}\) For the \(-\mathrm{COOH}\) group, \(\mathrm{pK}_{\mathrm{a}_{1}}=2.34 ;\) for the \(-\mathrm{NH}_{3}^{+}\) group, \(\mathrm{p} K_{\mathrm{a}_{2}}=9.69 .\) Consider the titration of a 0.500 M solution of alanine hydrochloride with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution. What is the \(\mathrm{pH}\) of \((\mathrm{a})\) the \(0.500 \mathrm{M}\) alanine hydrochloride; (b) the solution at the first half- neutralization point; (c) the solution at the first equivalence point? The dominant form of alanine present at the first equivalence point is electrically neutral despite the positive charge and negative charge it possesses. The point at which the neutral form is produced is called the isoelectric point. Confirm that the \(\mathrm{pH}\) at the isoelectric point is \(\mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}_{1}}+\mathrm{p} \mathrm{K}_{\mathrm{a}_{2}}\right)\) What is the \(\mathrm{pH}\) of the solution (d) halfway between the first and second equivalence points? (e) at the second equivalence point? (f) Calculate the pH values of the solutions when the following volumes of the \(0.500 \mathrm{M} \mathrm{NaOH}\) have been added to \(50 \mathrm{mL}\) of the \(0.500 \mathrm{M}\) alanine hydrochloride solution: \(10.0 \mathrm{mL}, 20.0 \mathrm{mL}, 30.0 \mathrm{mL}, 40.0 \mathrm{mL}, 50.0 \mathrm{mL}\) \(60.0 \mathrm{mL}, 70.0 \mathrm{mL}, 80.0 \mathrm{mL}, 90.0 \mathrm{mL}, 100.0 \mathrm{mL},\) and \(110.0 \mathrm{mL}\) (g) Sketch the titration curve for the 0.500 M solution of alanine hydrochloride, and label significant points on the curve.

Step by step solution

01

Determining pH of alanine hydrochloride solution

Initially, before any base is added, the pH is determined by the acidic \( -COOH \) group. We use the information provided that the \( pKa \) of this \( -COOH \) group is 2.34 to calculate the pH of the solution using the following formula for a weak acid in water: \[ pH = pKa - \log {[Base]/[Acid]} \] Because alanine hydrochloride fully dissociates into its ionic form in water, the concentration of acidic protons is 0.5 M and there is no base present yet. Hence, the pH of the solution is \( pH = 2.34 - \log {[0]/[0.5]} = 2.34 }. \]

02

Determining pH at the first half-neutralization

At the first half-neutralization point, half of the acidic protons have been neutralized by the added base. At this point, the ratio of [Base]/[Acid] is 1, and the pH equals the \( pKa \) of the protonated group being neutralized. Here, it is the \( -COOH \) group. So, at the first half-neutralization point, \( pH = pKa_{1} = 2.34 \).

03

Determining pH at the first equivalence point

At the first equivalence point, all of the acidic protons of the \( -COOH \) group have been neutralized. Now, the pH is determined by the \( -NH3^{+} \) group. We use the Henderson-Hasselbalch equation, \( pH = pKa_{2} + \log {[Base]/[Acid]} \), to determine the pH. Since all alanine molecules are fully deprotonated, the [Base]/[Acid] ratio is infinity, and the pH at the first equivalence point is \( 14 + \log {[1/0]} = \infty \). However, in practicality, the pH at the first equivalence point will be slightly less than 14.

04

Calculating pH at the isoelectric point

The isoelectric point is where the dominant form of alanine will be electrically neutral. The pH at the isoelectric point can be calculated by the formula: \( pH = \frac{1}{2} \times (pKa_{1} + pKa_{2}) = \frac{1}{2} \times (2.34 + 9.69) = 6.02 \). So pH at the isoelectric point is 6.02.

05

Calculating pH between first and second equivalence points, and at the second equivalence point

For these calculations, the Henderson-Hasselbalch equation is used again. Halfway between the first and second equivalence points, the ratio of [Base]/[Acid] is 1 and hence the pH is \( pKa_{2} = 9.69 \). At the second equivalence point, all the \( -NH3^{+} \) groups are deprotonated, so the pH is greater than 9.69 but less than 14.

06

Calculating pH values at given volumes of NaOH

This is similar to the previous steps, but you need to calculate the concentrations of the acid and base at the specific added volumes of NaOH and use the Henderson-Hasselbalch equation to find the pH at each volume. You must also consider that the total volume changes with the addition of NaOH solution.

07

Sketch the titration curve

The titration curve starts at a low pH (2.34 in this case), then increases slowly until it nears the first half-equivalence volume, where it rises sharply (vertical section of the curve) until it reaches the first equivalence point. After this, it increases slowly again until it nears the second half-equivalence volume, where it rises sharply again until it reaches the second equivalence point. After this, it slowly rises and finally levels off as the pH reaches a value below 14. The titration curve will have significant inflection points at the mid-point of the sharp rises, these points correspond to the half-equivalence points.

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