Question

Consider a solution of 0.0500 M isoleucine. If we call the fully protonated form H2A+, find the pH of a solution of: (a) 0.0500 M H2ACl, (b) 0.0500 M HA, and (c) 0.0500 M NaA. For the intermediate form, HA, calculate the pH using any reasonable approximations. pKa1 = 2.318 pKa2 = 9.758

Answer 1

(a) pH of 0.0500 M H2ACl = 1.88

(b) 0.0500 M HA = 6.04

(c) 0.0500 M NaA = 11.22

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pH of the solution can be calculated by the formula, pH = -log[H3O+]

Henderson–Hasselbalch equation for weak acid/base is given by

pH= pKa +log [A-]/[HA], [HA] is the molarity of undissociated weak acid, [A⁻] is the molarity of this acid's conjugate base.

Similarly,

pOH = pKb + log[BH]/[B]

[B] = Concentration of Base , [BH] = Concentration of salt/acid

pH + pOH = 14

A.

The equilibrium of isoleucine at this pH will be,

H2ACl <==> AH + H+, pKa1 = 2.318

pKa1 = -log(Ka1),

Ka1 = 10^-2.318 = 0.0048

[H2ACl] = 0.0500 M

Consider a fraction x of H2ACl is ionize to HA and H3O+

[H2ACl] = 0.0500 - x M

[H3O+] = x M

[HA] = x M

Ka1 = x^2/(0.0500 - x) = 0.0048

solving, x = 0.01328

[H3O+] = x = 0.01328

pH = -log[H3O+] = -log (0.01328) = 1.8768 = 1.88

B.

Here HA will be in equilibrium with the acid dissociation reaction; HA + H2O <==> H3O+ + A-,

and base dissociation reaction; HA + H2O <==> OH- + HA,

Similarly, dissociation of water; H2O <==> H3O+ + OH-

In such cases, the simplified equation for [H3O+] = sqrt (Ka1 x Ka2)

Ka1 = 0.0048

Ka2 = 10^-9.758 = 1.75 x 10^-10

[H3O+] = Sqrt (0.0048 x 1.75 x 10^-10) = 9.16 x 10^-7

pH = -log ( 9.16 x 10^-7) = 6.04

or it can also be pH = pI = (pKa1 + pKa2)/2 = (2.318 +9.758)/2 = 6.038 = 6.04

C.

The equilibrium of isoleucine at this pH will be,

A- + H2O <==> AH + OH-,

Kb = [HA][OH-]/[A-]

pKa2 = 9.758

pKb = 14-9.758 = 4.242

Kb = 10^-4.242 = 5.727 x 10^-5

Consider a fraction x of A- is ionize to HA and OH-

[A-] = 0.0500 - x M

[OH-] = x M

[HA] = x M

kb = x^2/(0.0500- x) = 5.727 x 10^-5

[OH-]= 0.0016638

pOH = -log 0.0016638 = 0.0016638 = 2.78

pH = 14-2.78 = 11.22