Use the systematic treatment of equilibrium to determine the pH and concentratio

Question

Use the systematic treatment of equilibrium to determine the pH and concentrations of species in 1.00 L of solution containing 0.070 mol lysine (HL), 0.055 mol aspartic acid (H2A), and 0.0080 mol NaOH. Consider just acid-base chemistry. Ignore ion pairing and activity coefficients. Lysine (HL) is derived from the triprotic acid H3L2 with pKH3L2 = 1.77, pKH2L = 9.07, and pKHL = 10.82. Aspartic acid (H2A) is derived from the triprotic acid H3A with pKH3A = 1.990, pKH2A = 3.900, and pKHA– = 10.002.

Solve for
pH

H3L^2+

H2L^+

HL

L^-

H3A^+

H2A

HA^-

A^2-

H^+

OH^-

Na^+

Explanation / Answer

Lysine concentration = 0.07 mol

NaOH = 0.008 mol

reacts to form 0.008 mol of L-

HL + OH- <==> L- + H2O

remaining HL = 0.07 - 0.008 = 0.062 mol

pH = pKa + log(L-/HL)

     = 10.82 + log(0.008/0.062) = 9.93

[L-] = 0.008 M

[HL] = 0.062 M

[H+] = 1.17 x 10^-10 M

pH = 9.93

[OH-] = 8.51 x 10^-5 M

HL + H2O <==> H2L+ + OH-

Kb1 x 10^-14/8.51 x 10^-10 = x^2/0.062

[H2L+] = x = 8.53 x 10^-4 M

H2L + H2O <==> H3L2+ + OH-

Kb3 = 1 x 10^-14/0.017 = x^2/8.53 x 10^-4

[H3L2+] = x = 2.24 x 10^-8 M

For aspartic acid H2A

initial = 0.055 mol

react with 0.008 mol NaOH

[HA-] = 0.008 mol

[H2A] = 0.055 - 0.008 = 0.047 mol

H2A + H2O <==> H3A+ + OH-

Kb3 = 1 x 10^-14/0.010 = x^2/0.047

[H3A+] = x = 2.14 x 10^-7 M

Na+ = 0.008 M

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.