Electrolyte Solution PDF

Electrolyte solution By Pharmacist Muhanad S. AL Ani  Van’t Hoff Factor(i) Van’t Hoff corrects for the fact that the number of particles you thrown into solution is not always the number of particles that determine the magnitude of the property. The relationship between t...

Electrolyte solution By Pharmacist Muhanad S. AL Ani  Van’t Hoff Factor(i) Van’t Hoff corrects for the fact that the number of particles you thrown into solution is not always the number of particles that determine the magnitude of the property. The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: actual number of particles in solution after dissociation Van’t Hoff factor (i) = number of formula units initially dissolved in solution  Ion Pairing At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle.  Examples The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). – NaCl i=2 – KNO3 i=2 – Na3PO4 i=4 Ion Pairing Ion pairing is most important in concentrated solutions. As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Colligative Properties of Electrolyte Solutions π= M R T it was found that the π for electrolytes is approximately 2,3, and more times larger than expected from this equation. So for electrolyte solution the equation will be π=i M R T i= correction factor (Van’ts Hoff factor)  so for the other colligative properties ΔTb = i Kb m ΔTf = i Kf m ΔP = i P1 ̊ m At what temperature will a 5.4 molal solution of NaCl freeze? Kf= 1.86 ∆Tf = i Kf m ∆Tf = 2 x 1.86 x 5.4 ∆Tf = 20.1 Co FP = 0 – 20.1 = -20.1 Co Degree of dissociation : To differentiate between the strong and weak electrolytes by the fraction of the moles ionized which is the degree of dissociation Ac = A0 Ac = equivalent conductance at conc. (c) Ao = equivalent conductance at infinite dilution. The correlation between i and i = 1+ (v-1) i-1 = v : is the no. of ions v-1 Example: The freezing point of a 0.10 m solution of acetic acid is -0.188 C̊. Calculate the degree of ionization of acetic acid at this conc. Acetic acid dissociates into two ions. Kf= 1.86 Ionic strength : Relate the interionic attraction for solution of strong electrolytes and for solutions of weak electrolytes together with salts and other electrolytes, such as exist in buffer systems. 1 ² ² ² μ = 2 ( c1 z 1 + c2 z2 + ………ci zi ) 1 i ² μ = 2 Σ ci zi C is the molar conc. of ions 1 Z is the valence we divided by 2 because +v and –v ion pairs contribute to the total electrostatic interaction, whereas we are interested in the contribution of each ion separately So μ represents the contribution to the electrostatic forces of the ions _ of all types. + ex. KCl K + Cl 0.01 M 0.01 M 0.01 M 1 ² ² μ= 2 [ ( 0.01 x 1 ) +( 0.01 x 1 )] =0.01 +1 -2 Na2So4 Na 2 + So 4 0.01 M 2 x 0.01 M 0.01 M 1 ² ² μ= 2 [ ( 0.02 x 1 ) + ( 0.01 x 2 )] = 0.03 μ total = μ KCl + μ Na2So4 ( for solution of them) Example: What is the ionic strength of (a) 0.01 M KCl, (b) 0.01M BaSO4,( c ) 0.01 M Na2SO4 (d) what is the ionic strength of a solution containing all three electrolytes. Example: A buffer contains 0.3 mole of K2HPO4 and 0.1 mole of KH2PO4 per liter of solution. Calculate the ionic strength of the solution.  The L value: ΔTf = i Kf m used for the colligative properties of nonelectrolyte and electrolyte solutions. in dilute solution it can be modified slightly by ΔTf = L c L = i Kf c = M molar conc. At a conc.of a drug that is isotonic with body fluids L = i Kf = L iso = 1.9 for non electrolytes = 2.0 for weak electrolytes = 3.4 for univalent electrolytes