Question

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.

Short answer

a) The pH of the 0.500 M alanine hydrochloride is 2.34. b) The pH at the first half-neutralization point is also 2.34. c) The pH at the first equivalence point is somewhere just below 14. The pH at the isoelectric point is 6.02. d) Halfway between the first and second equivalence points, the pH is 9.69. e) At the second equivalence point, the pH is greater than 9.69 but less than 14. f) and g) The specific pH values at given volumes of NaOH and the sketch of the titration curve will be dependant on the initial volume of alanine hydrochloride and the concentrations of alanine hydrochloride and NaOH.

Step-by-step solution

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 }. \]

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, \( pH = pKa_{1} = 2.34 \).

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.

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.

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.

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.

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.

Key concepts

Understanding pKa Values

When discussing amino acid titration, understanding the concept of pKa values is crucial. pKa is a measure of the acidity of a site in a molecule. It indicates how easily a proton (H⁺) is donated. For alanine, a common amino acid, it has two pKa values due to its acidic carboxyl group \(-\mathrm{COOH}\) and its basic amino group \(-\mathrm{NH}_{2}\).
These groups give alanine its diprotic character:

• pKa₁ = 2.34 for the carboxyl group.
• pKa₂ = 9.69 for the amino group.

When pH is lower than pKa, the group tends to be protonated. Conversely, when pH is higher, the group is deprotonated. This helps predict ionization states during titration.
This pKa information helps determine the areas of buffering, where the pH changes minimally with added base or acid. This occurs around the pKa values themselves. It's essential in understanding the progress of a titration curve.

Calculating the Isoelectric Point

The isoelectric point (pI) is a critical concept when analyzing amino acid titration. It is the pH at which the amino acid carries no net charge. For alanine, this is when its positive and negative charges balance.
To calculate the pI, you average the pKa values of the ionizable groups:
\[ pI = \frac{1}{2}(pKa_{1} + pKa_{2}) \]
Substituting alanine's pKa values:
\[ pI = \frac{1}{2}(2.34 + 9.69) = 6.02 \]
At pH 6.02, alanine is in its zwitterionic form, meaning it has both positive and negative charges, but overall, it is neutral. This is significant in practices like electrophoresis, where the isoelectric point determines the position of amino acids in a gel.
Notice how this pI guides the titration curve, indicating a point of minimal electrical mobility and solubility, which is often marked by a flat portion on a titration graph.

The Henderson-Hasselbalch Equation Explained

A fundamental tool in titration is the Henderson-Hasselbalch equation. It relates pH to pKa and the ratio of concentrations of an acid and its conjugate base. The equation is given by:
\[ pH = pKa + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
This equation is particularly useful when calculating the pH at various points in a titration.
For example, at the half-equivalence point of alanine’s titration, the concentration of base equals that of the acid, making the logarithmic term zero. Thus, at this point, \( pH = pKa \).
Throughout titration, the equation helps determine:

• pH before and after equivalence points.
• Buffering regions where pH changes slowly.
• The effects of adding different amounts of titrant.

Using this equation, you can predict how an amino acid will behave in a solution, adjust conditions for experiments, and analyze titration data effectively.