Question

A derivative of the polysaccharide chitosan is being utilized in a spray to treat severe wounds. The chitosan aids in quickly stopping the flow of blood by forming a gel at the wound site. The behavior of the gel is strongly dependent upon the activity of ionic species in the spray. In one process to form the spray, an aqueous solution containing 0.002 M NaCl and 0.006 M CaCl2 is mixed with a solid powder of the chitosan derivative. (a) What is the ionic strength of the precursor solution, before the chitosan is added? (b) Use the Debye-Huckel expression to calculate the activity coefficient of Na+ , Cl- , and Ca2+ in the precursor solution at 25o C.A derivative of the polysaccharide chitosan is being utilized in a spray to treat severe wounds. The chitosan aids in quickly stopping the flow of blood by forming a gel at the wound site. The behavior of the gel is strongly dependent upon the activity of ionic species in the spray. In one process to form the spray, an aqueous solution containing 0.002 M NaCl and 0.006 M CaCl2 is mixed with a solid powder of the chitosan derivative. (a) What is the ionic strength of the precursor solution, before the chitosan is added? (b) Use the Debye-Huckel expression to calculate the activity coefficient of Na+ , Cl- , and Ca2+ in the precursor solution at 25o C.

Answer 1

The ionic strength, I, of a solution is a function of the concentration of all ions present in that solution.

where one half is because we are including both cations and anions, c_i is the molar concentration of ion i (M, mol/L), z_i is the charge number of that ion, and the sum is taken over all ions in the solution. For a 1:1 electrolyte such as sodium chloride, the ionic strength is equal to the concentration, but for MgSO₄ the ionic strength is four times higher. Generally multivalent ions contribute strongly to the ionic strength.

(a) Ionic strength (I) = 1/2(0.002 x 1^2 + 0.002 x 1^2 + 0.006 x 2^2 + 0.012 x 1^2)

                                = 0.02 M

(b) Activity coefficient for,

log(y(Na+)) = -0.51 x 1^1 x sq.rt.(0.02)/1+3.3 x 0.45 x sq.rt.(0.02)

y(Na+] = 0.872    

log(y(Cl-)) = -0.51 x 1^1 x sq.rt.(0.02)/1+3.3 x 0.3 x sq.rt.(0.02)

y(Cl-) = 0.865   

log(y(Ca2+)) = -0.51 x 2^1 x sq.rt.(0.02)/1+3.3 x 0.6 x sq.rt.(0.02)

y(Ca2+) = 0.595