Phyphar Chapter-5
October 3, 2022 | Author: Anonymous | Category: N/A
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Physical Pharmacy: CHAPTER 5
Prediction of solubility General principle: “Like dissolves like”
SOLUTION – a homogeneous dispersion of two or more substances in each other – the dissolved substance SOLUTE – SOLVENT – – the dissolving medium
Solute dissolves best in a solvent with similar chemical properties. 1. Polar solutes dissolve in polar solvents. s olvents. 2. Non-polar solutes dissolve in nonpolar solvents.
3 MOST IMPORTANT TYPES OF SOLUTIONS
METHODS OF EXPRESSING CONCENTRATION 1. % (W/V, W/W, V/V)
•
Solid in liquid
•
Liquid in liquid
2. Molarity (M)
Gas in liquid
3. Molality (m) 4. Normality (N) 5. Osmolarity or Osmolality 6. mole %
•
Dissolution - the transfer of molecules or ions from a solid state into solution
-
rate of solution PERCENTAGE SOLUTION
Solubility- the amount of solid that passes into solution when equilibrium is established between the solution and excess (undissolved) substance. The solution that is obtained under these conditions is said to be saturated. Solubility of the solute in the solvent- referred to as the extent to which the dissolution proceeds under a given set of
experimental conditions Miscibility- when the 2 components forming the solution are either both gases or both liquids
% / / =
wt. of solut solutee (g) × 10 100 0 vol. of solution (mL)
% / / =
wt. of solut solutee (g) × 100 wt. of soluti solution on (g)
vol. of solute solute (mL) 100 0 % / / = vol. of solution solution (mL) × 10 Molarity
= Solubility and Dissolution
- -
are not the same and not n ot necessarily related in practice high drug solubility is usually associated with a high dissolution rate U.S.P. Descriptive Terms of Solubility Description
Parts of solvent that dissolves 1 part of solute (mL)
Molality
=
10,000
2E-PH: Belmonte, Manalo, Sarto, Yan
wt.(g) Equal wt.
molecular weight h
Final Equations: GEW N= L solution
= Slightly soluble
solutee (g) no. of mol moles es (n) wt. of solut = MW x kg solvent kg solvent
Normality gram equivalent wt. of so solute lute (Equivalence) = L solution
N= Very soluble
solutee (g) no. of mole moless (n) wt. of solut = MW x L solution L solution
wt. of solute (g) MW x L so sol′n l′n h
Where: h= total + or – charge for salt = # of replaceable H + for acid = # of replaceable OH + for base
N =
N =
− . − .
N =
N =
..
mEW =
N =
x
−
mEW =
N =
x EW
x
a. Molality b. mf of solute c. mole % solvent d. % by wt.
e. Osmolarity
. ()
a. Molality G solution = 600 mL x 1.082g/mL = 649.2g - 64.92g = 584.28g 584.28g / 1000 = 0.58428kg Mg (24.31) + Cl2(35.45 x 2) = 95.21MW
. ()
64.92g x
. ()
. ()
= ..
N =
Examples: 1. A solution is prepared by dissolving 64.92g of Magnesium chloride in sufficient water to make 600 mL of solution. The density of the solution is 1.082 g/mL. (At. wt.: Mg = 24.31, Cl = 35.45) Calculate:
. ()
=
m=
= 0.681861149mol
95. ()
=
.688649 .5848
. ()
b. Mf of solute N solute = 64.92g x
95.
= 0.6819mol MgCl2
N = M x h
N solvent = 584.28g H2O x
Osmolarity = mOsmol /L solution = (mmol x i) / L solution
Mf solute = =
= (moles x 1000 x i) / L solution =(
.
) / L solution
= M x 1000 x i
32.46mol
+ .689 .689 + 3.46
Mf solvent =
Osmolality = mOsmol /kg solvent = (mmol x i) / kg solvent = (moles x 1000 x i) / kg solvent .
= 8
= 0.02057mol
c. Mole % solvent
..
=(
= 1.1670m
=
+ 3.46
.689 + 3.46
= 0.9794mol
Mole % solvent = mf x 100 = 0.9794mol x 100 = 97.94%
) / kg solvent
..
d. % by weight
= m x 1000 x i
.
= Mole % = mole fraction (mf or X) x 100
=
Mole % solute = mf solute ( ) x 100 Mole % solvent = mf solvent ( ) x 100 Where:
mf solute =
+
mf solvent =
+
mf solute + mf solvent = 1
mole% solute + mole% solvent = 100%
2E-PH: Belmonte, Manalo, Sarto, Yan
. 64.9 649.
x 100
x 100 = 10%w/w
2. A solution of acetic acid is prepared by adding 164.2 16 4.2 g of the acid to sufficient water to make 800 mL of solution at 20°C. The density of the solution at this temperature is 1.026 g/mL. (At. wt.: C = 12, H = 1, O = 16) Calculate: a. Molarity b. mf of solvent c. mole % solute d. % w/v e. Osmolality f. normality
SOLUTIONS OF GASES IN LIQUIDS 1. Effect of pressure on the solubility ● Solubility of a gas over a given solvent increases with an increase in the partial pressure of the gas over the solution ●
Since : 1 = 1 2
1 = 1 ° (1 2 ) If:
1 = 1 ° (1 2 )
Henry’s Law
1 = 1 ° 1 ° 2
- the concentration of dissolved gas in a given solvent is directly proportional to the partial pressure of the gas, temp. remaining constant
1 ° 1 = 1 ° 2
=
Then: vapor pressure lowering ∆ = °
2. Effect of temperature on the solubility ● Solubility of a gas over a given solvent decreases with an increase in temperature therefore increase in kinetic energy ● Higher kinetic energy causes more motion in molecules which break intermolecular bonds and escape from solution.
∆ = ° Solutions of Solids in Liquids Colligative Properties
depend on the concentration (no. of particles) of solute molecules or ions in solution but not on the chemical identity of the solute
•
1. Vapor Pressure lowering 3. Presence of dissolved substance on the solubility ● Presence of dissolved substance(electrolytes) lowers
the solubility of a gas. Solutions of Liquids in Liquids Raoult’s Law
the vapor pressure of a component in a solution is directly proportional to its mole fraction and is equal to the mole fraction multiplied by the vapor pressure in the pure state at that temperature
•
If:
1 = 1 ° (1 2 ) 1 = 1 ° 1 ° 2 1 ° 1 = 1 ° 2
The tendency of the two different sorts of molecules to escape is unchanged.
Then: vapor pressure lowering ∆ = ° ∆ = ° Phase diagram for Pure Water
P is pressure and X is mole fraction P solvent in solution: 1 = 1 ° 1 P solute in solution: 2 = 2 ° 2 P solution: = *if solute is non-volatile*
Therefore:
= = ° = °
2E-PH: Belmonte, Manalo, Sarto, Yan
Adding Solute to Water
❖
OSMOSIS
•
Spontaneous process
•
Diffusion of water across a semi-permeable membrane Semi-permeable membrane allows passage of solvent but not solute
•
Movement of solvent from lower solute concentration to higher solute concentration Movement of solvent from higher solvent concentration to lower solvent concentration
•
•
OSMOTIC PRESSURE
•
Pressure required to stop osmosis PV = nRT
BOILING POINT ELEVATION ∆Tb α m kb H2O = 0.51 ∆Tb = kbm
if P = Π and n/V = M then Π = MRT Π = mRT
Kb Wsolute MWsolute x kgsolvent
CRYOSCOPIC METHOD Π = mRT
∆Tb =
∆Tb = Tb Tb° FREEZING POINT DEPRESSION ∆Tf α m ∆Tf = kfm kf H2O = 1.86
Kf Wsolute
ΔTf
= kfm
∆Tf Kf ∆TfRT Π= Kf m=
*R = 0.0821 L-atm/n-K
∆Tf = MWsolute x kgsolvent ∆Tf = Tf° Tf *Kb = ebullioscopic constant or molal boiling point constant *Kf = cryoscopic constant or molal freezing point constant
Non-electrolytes ΔP1 P1˚X2 ΔTb kbm ΔTf Kfm Π mRT *i = dissociation or Van’t Hoff factor
Electrolytes iP1˚X2 ikbm iKfm imRT
Calculate the Dissociation Value (i) 1. KCl is a 2-ion electrolyte dissociating 80% in a certain concentration. Calculate its dissociation value (i).
80 K+ ion -
80 Cl ion 20 undissociated particles 180 particles represent 1.8 times as many particles as there were before dissociation, i = 1.8
2E-PH: Belmonte, Manalo, Sarto, Yan
2. ZnCl2 is a 3-ion electrolyte dissociating 80% in a certain concentration. Calculate its dissociation value (i). 80 Zn+ ion 80 Cl- ion 80 Cl- ion 20 undissociated particles 260 particles represent 2.6 times as many particles as there were before dissociation, i = 2.6 Non-electrolytes Electrolyes (2-ion) Electrolytes (3-ion) Electrolytes (4-ion) Electrolytes (5-ion)
1.0 1.8 2.6 3.4 4.2
2E-PH: Belmonte, Manalo, Sarto, Yan
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