CH. 13 SOLUTIONS 13.1-.5 Process at molecular level Solubility- solute/solvent/solution Factors affect solubility Concentration Expressions Colligative Properties Equations Equations Henry’s Law S gas = K H * P gas Various concentration expressions van’t Hoff factor freezing pt depression T = K f *mi boiling pt elevation T = K b *mi Raoult’s Law P soln = (X*P o ) solvent
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CH. 13 SOLUTIONS 13.1-.5 Process at molecular level Solubility- solute/solvent/solution Factors affect solubility Concentration Expressions Colligative.
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CH. 13 SOLUTIONS 13.1-.5
Process at molecular levelSolubility- solute/solvent/solutionFactors affect solubilityConcentration ExpressionsColligative Properties
Solution Colloid particles: individual atoms, lrg molecules or sm molecules ions, sm molecules not separate out
Solute: subst being dissolvedSolvent: subst doing the dissolveH2O: universal solvent
“LIKE DISSOLVES LIKE”subst w/ similar inter- forces will dissolve in each otherinteraction bet solute & solvent
Miscible: subst dissolve in each other
Electrolyte: subst dissoc into ion, conducts electrical current all ionic cmpds, strong acidsNonelectroylte: no dissoc or very little %, not conduct current weak acids, covalent molecules
COLLIGATIVE PROPERTIES OF SOLUTIONS
Depends on number of solute particles in solution.Properties of dilute solns of nonvolatile solute in volatile solvent
1. Lowering vp2. Elevate bp 3. Lower fp4. Osmotic P
Calculating: 1) VP of sovent 2) FP depression & BP elevation of solvent 3) molar mass, M, of solute from FP data 4) value of mole number, i, of solute from FP data
Boiling Point Elevation, Freezing Point Depression,Vapor Pressure Reduction (Raoult’s)
Depends on [solute]Examine [solute] affects property of pure solvent
Ex 2: a soda bottle at 25oC contains CO2 gas at 5.0 atm over the liquid. Assume partial PCO2 is 4.0*10-4 atm. Calculate [CO2] before & after bottle opened.KH,CO2 = 0.032 mol/L-atm (from table)
Pressure Effect relation gas P & concen dissolved gas Sgas = KH* Pgas
Practice ProblemIf CO2 partial pressure is 3.0*10-4 atm, what is the [CO2] at 250C?
Temp vs Solubility
solidsolid more soluble @ higher T more soluble @ higher T heat absorbed to form soln solute + solvent + heat <---> sat soln DHsoln > 0 incr T, incr rate dir -----> gas DHsolute = 0 since gas particles already sep DHhydra < 0 heat released for gases in H2o solute + H2O <----> sat soln + heat DHsoln < 0 incr T, decr gas solubility rate can lead to thermal pollution
TEMPERATURE EFFECT ON SOLUBILITY OF VARIOUS SUBSTANCES
total soln = solute + solvent
Mass % = [mass subst in soln/total mass soln]*100 (%) ppm: * 106 ppb: * 109
Mole Fraction (X) = mols solute/mols soln (use in Raoult’s Law) (mol % = X * 100)
Molarity (M) = mols solute/L soln (mols/L) vol affected by T
molality (m) = mols solute/Kg solvent (mols/Kg) mass not affected by T
CONCENTRATION EXPRESSIONS
Conversionsconvert mol to mass: use molar massconvert mass to vol: use density
23.6 % HF by mass, states: 23.6 g HF/100 g soln
0.050 ppm states: 0.050 g solute in million (106) g soln, or 0.050 mg/Kg also ≈ 0.050 mg/L
50 ppb states: g solute in billion (109) g soln 50 g/Kg => 50 g/L
PP1) Patient is given 30.0 g glucose in 150 mL solution, what is the mass %? ppm? ppb?
PP2) Patient is given 30.0% glucose solution, what is the mass glucose in 250.0 g H2O?
MASS % PRACTICE PROBLEM
MOLARITY -- MOLALITYPP3) Find M of a soln w/ 8.98 g lithium nitrate in 505 mL
PP4) Find m of soln w/ 164 g HCl in 753 mL H2O
MASS %
30.0% is 30.0 g solute in 100.0 g solution.So, 30 g glucose for 70.0 g H2O
answer unit egiven valu PP2) facorunit
glucose g 107 OH g 70.0
glucose g 30.0 OH g 250.0
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100(g)soln mass
(g) solute of mass % mass PP1) 20.0% 100
g 150
g 30.0 % mass
ppm 10*2 10g 150
g 30.0 ppm 56 ppb 10*2 10
g 150
g 30.0 ppb 89
MOLARITY -- MOLALITY
mols = (8.98 g)/(68.9 g/mol) = 0.130 mols M = 0.130 mol/.505 L = 0.258 M
mols = (164 g)/(36.5 g/mol) = 4.49 mol m = (4.49 mol)/(0.753 Kg) = 5.96 m
M (L) vol mols g PP3) wt.form.
m (Kg)solvent mols g PP4) wt.form.
PHASE DIAGRAM PURE H2O
Pre
ssur
e -
atm
Temperature - oC
1 atm
VP pure solvent
VP solution
Tb
bp pure H2O bp solution
fp solution fp pure H2OTf
VAPOR PRESSURE OF SOLNS
Raoult’s Law Psoln = (X*Po)solvent
Note relation:soln: 50/50% solute/solvet molecules, Xsolv = 0.5, then Psoln is 0.5Po
solv
soln: 3/4 soln is solvent particles, Xsolv = 0.75, then Psoln is 0.75Po
solv
Idea behind this: nonvolatile solute just dilutes the solvent
Remember!!! Ideal Gas obeys ideal-gas eqn and ideal soln obeys Raoult’s LawReal soln approx ideal when……. low [ ], similar molecular sizes, similar inter- attractions
VP of soln containing nonvolatile solutes given by: RAOULT’S LAW
Example:1 mol glucose, result => lower VP same as 0.5 mol NaCl
Reasoning behind this ???> both nonvolatile> form 1 mol of particles
PRACTICE PROBLEMA nonelectrolyte solution is prepared by dissolving 0.250 g in 40.0 g of CCl4.The normal soln BP is increased 0.357oC. Find the molecular wt. of the solute.
Know:i = 1, nonelectrolyteTb = 0.357oC
Find:Kb CCl4: 5.02oC/mKg solvent = 0.040 Kg CCl4
Calculation:Tb = Kb*m*i m = Tb/(Kb*i )
m = (0.357oC)/(5.02oC/m) = 0.0711 m
So, soln contains: 0.0711 mol solute/0.040 Kg solvent = 6.25 g/Kgmeans:0.0711 mol = 6.25 g
then 1 mol: (6.25 g)/(0.0711 mol) = 87.9 g or M = 87.9 amu
Ion-Dipole Hydration shell: ion surrounded by H2O molecules; attraction of H2O
H-Bonding imprt in aq solns; prime reason for solubility in H2O; factor of solubility for many biological & organic subst
Dipole-Dipole factor for solubility of polar-polar molecules
Dispersion factor in NP-NP subst
SOLUTION TYPES
Forces in Solution
Li+1+ Cl-1 + H2OIons of solid attracted to dipole of H2O;attraction as strong as ion-attraction;H2O “substitutes” bet ions
CH4 + H2O NP + Polar NP attraction too weak for H2O sub H-bonding bet H2O molecules too strong