You make a 1.000 L solution of tetracycline (TTC). It is controlled (buffered) a

Question

You make a 1.000 L solution of tetracycline (TTC). It is controlled (buffered) at pH 7.4, and that solution contains a formal concentration of 0.000100 M TTC_Total and you added 0.00100 M CaCl_2. Solve for the concentrations of all tetracycline species, considering the equilibria below (unlike oxytetracyline, for tetracycline we know all these equilibria; see Werner et al., 2006). H_3 L H_2 L^- + H^+ K_a1 = 10^-8.00 H_2 L^- HL^2- + H^+ K_a2 = 10^-9.82 Ca^2+ + H_2 L^- CaH_2 L^+ K_m1 = 10^3.4 Ca^2+ + HL^2- CaHL K_m2 = 10^5.8 a. Write the charge balance equation for the solution. b. Write the mass balance equations for total L (tetracycline) and total calcium. c. Write out all the equilibrium constant expressions for the reactions in (a): d. How many equations do you have? How many unknowns are there? Is this solvable? e. Solve for all concentrations: [H_3 L], [H_2 L^-], [HL^2-], [CaH_2 L^+], [CaHL], and [Ca^2+ J. You can ignore activity coefficients.

Explanation / Answer

Tetracycline contains ionizable functional groups and it gives several species with charges at different locales and differing net charge; the fractional distribution of each species depends on pH-pKa relationship in the aqueous phase. These species can interact with Ca2+ to form metal-tetracycline complexes in water.

Part.a

To write a charge balance equation,

For positive charges [H+] + 2[Ca2+] + [CaH2L+ ] = Molar Positive Charge in solution

For negative charges [H2L-] + 2[HL2-] = Molar Negative Charge in Solution

To maintain electrical neutrality of solution

Molar positive charge in solution = Molar negative charge in solution

[H+] + 2[Ca2+] + [CaH2L+ ] = [H2L-] + 2[HL2-]

Part.b Mass Balance equations for total Tetracycline

[H3L] = [H2L-] + [HL2-] +[CaH2L+] + [CaHL] = 0.0001M

Mass Balance equation for total Calcium

2[Ca2+] = 2[Cl-] + [CaH2L+] + [CaHL] = 0.001M

[Ca2+] =0.0005 M

4. Equilibrium Constant Expressions

Ka1 = [H2L-] [H+] / [H3L] = 10-8 .................(i)

Ka2 = [HL2-] [H+] / [H2L-] = 10-9.82 ............(ii)

Km1 =[CaH2L+] / [Ca2+] [H2L-] = 103.4 .......(iii)

Km2 =[CaHL] / [Ca2+] [HL2-] = 105.8 ...............(iv)

[H+] = 10-7.4 M, [H3L] = 10-4 , [Ca2+]= 0.0005

Total unknown species are 4 and they are [H2L-], [HL2-] , [CaH2L+] and [CaHL] . They can all be solved by putting the known values one by one in equation (i) to (iv).

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

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