— Analytical Chemistry Chemistry Question
Buffer from Biological Acid: Lysine
Acid Dissociation Constants
Name: Lysine | Carboxylic acid pKa: 2.16 | Ammonium pKa: 9.06 | Substituent pKa: 10.54
One of essential amino acids, lysine, is normally depicted using the molecular structure below: Note that the amine group on the left is part of the substituent group. Is lysine likely to exist in this form when dissolved in neutral aqueous solution? If not, write the correct form.
Model Answer
No. The correct form is as shown:
Draw the molecular structures for dominant forms of lysine that are present in aqueous solution and arrange them in order from the most acidic form to the most basic form. Use Na + or Cl - to balance the charge. Label each structure with the name of the compound.
To prepare buffer solution, you start with the most acidic form of lysine that has the concentration of 0.100 mol dm -3 with the volume of 100 cm 3 . Calculate the volume of KOH solution with a concentration of 0.500 mol dm -3 you should add to obtain a pH of 9.5?
Model Answer
Let us mark:
H3L 2+ - the most acidic form
H2L + - the first intermediate form
HL - the second intermediate form
L – - the most basic form
c(H3L 2+ ) = [H3L 2+ ] + [H2L + ] + [HL] + [L – ] = 0.1 mol dm -3
total n(H3L 2+ ) = (0.100 mol dm -3 )(0.1 dm 3 ) = 0.01 mol
total n(H3L 2+ ) = n(H3L 2+ ) + n(H2L + ) + n(HL) + n(L – ) = 0.01 mol
At pH = 9.5, [H + ] = 10 -9.5
At pH = 9.5, the equilibrium amount of substances of all forms can be calculated as follows:
n(H3L 2+ ) = 1.14 × 10^-3 mol (negligible)
n(H2L + ) = 2.497 × 10^-3 mol
n(HL) = 6.88 × 10^-3 mol
n(L - ) = 6.27 × 10^-4 mol
Titration reaction to convert H3L 2+ to the desired forms;
H3L 2+ + OH – → H2O + H2L +
H2L + + OH – → H2O + HL
HL + OH – → H2O + L –
1 mol of KOH required = (2.497×10 -3 ) + 2(6.88×10 -3 ) + 3(6.27×10 -4 ) = 1.814×10 -2 mol
1 cm 3 of KOH solution required = (1000 cm 3 / 0.5 mol) 1.813×10 -2 mol = 36.28 cm 3
Dissolve 5.00 g of the neutral zwitterion form of lysine in 100.0 cm 3 of pure water. Determine pH of the solution once equilibrium is reached.
Model Answer
In this case, Ka1 is Ka(carboxylic acid), Ka2 is Ka(ammonium) and Ka3 is Ka (substituent).
HL is the second intermediate form
Since Ka2 and Ka3 are small,
c(HL) = (5.00 g * (1 mol / 146.19 g)) / 0.100 dm^3 = 0.342 mol dm^-3
[HL] = 0.342
[H + ] = 1.59×10 -10
pH = 9.80
The alternative calculation is pH = (pKa2 + pKa3) / 2 = (9.06 + 10.54) / 2 = 9.80.
Determine the equilibrium concentrations of all other forms of lysine present in the solution prepared in question 10.4.
Model Answer
HL L – + H + Ka3 = 10 –10.54
HL + H2O H2L + + OH – Kb2 = Kw / Ka2 = 1×10 –14 / 1×10 –9.06 = 1.15×10 –5
From question 10.4: [H + ] = 1.59×10 –10
Then [OH -] = 6.29×10 –5
At equilibrium:
[L - ] = (10^-10.54 × 0.250) / (1.59×10^-10) = 0.0453
[H2L + ] = (1.15×10^-5 × 0.25) / (6.29×10^-5) = 0.0457
Where: cr(HL) = 0.342
[HL] = 0.250
H2L + + H2O H3L 2+ + OH –
Kb3 = Kw / Ka1 = 10^-14 / 10^-2.16 = 1.45 × 10^-12
[H3L 2+ ] = (1.45×10^-12 × 0.0457) / (6.29×10^-5) = 1.05 × 10^-9