17012018 Flashcards
(21 cards)
Solution
- Homogenous mixture of 2/2+ substances
- e.g. see table 13.1
- Classify by the amount of solute dissolved
Unsaturated solution
- Contains less solute than the solvent has the capacity to dissolve at a specific temp
Saturated solution
- Contain max amount of solute that will dissolve in a solvent at a specific temp
Supersaturated solutions
- Contain more dissolved solute than is present in a saturated solution
- Generally unstable
Solvation
- Occur when solute molecules are separated frm one another & surrounded by solvent molecules
- Depend on 3 types of interactions:
1) Solute-solute interactions
2) Solvent-solvent interactions
3) Solute-solvent interactions
Intermolecular forces (IMFs)
- Ion-dipole (e.g. NaCl / KI in H2O)
- Dipole-induced dipole (e.g. He / CO2 in H2O)
- Ion-induced dipole (e.g. Fe2+ & O2)
Entropy of a system
- Measure of how dispersed / spread out its energy is
- Natural tendency for the energy of a system to become dispersed (entropy increase)
Like dissolves like
- 2 substances w/ similar type & magnitude of IMFs are likely to be soluble in each other
- e.g.: CCl4 & C6H6 > nonpolar / nonpolar
CH3OH & CH3CH2OH > polar / polar
Miscible
- 2 liquids are said to be miscible: completely soluble in each other in all proportions
Concentration
- Amount of solute relative to volume of a solution or to the amount of solvent in a solution
- Concentration choice is based on the experiment
Molarity (M)
= moles of solute / liters of solution
Mole fraction of component A (X_A)
= moles of A / sum of moles of all components
Molality (m) & % by mass
- No. of moles of solute dissolved in 1kg (1000g) solvent
= moles of solute / mass of solvent (in kg)
% by mass
- mean parts per hundred & the multiplier is 100
= (mass of solute / (mass of solute + mass of solvent) )*100%
Example: Dissolve 170.1g of glucose (C6H12O6) in enough water to make a liter of solution. Density of the solution = 1.062 g/mL
A) Concentration in molality:
170.1g / (180.2g/mol) = 0.9440 mol glucose (L of solution)
1 L of solution * 1062 g/L > 1062 g
1062 g - 170.1 g = 0.892 kg H2O
m = 0.9440 mol glucose / 0.892 kg H2O = 1.06 m
B) Concentration in percent by mass
(170.1 g glucose / 1062 g solution) * 100% = 16.02% glucose by mass
C) Concentration in parts per million
(170.1 g / 1062 g solution) 1,000,000 = 1.60210^5 ppm glucose
Molarity (use when…)
- Titration
- Gravimetric analysis
Mole Fraction (use when…)
- Gases
- Vapor pressure
Molality & percent by mass (use when…)
- Independent of temp
Example: “Rubbing alcohol” is a mixture of isopropyl alcohol (C3H7OH) and water that is 70% isopropyl alcohol by mass (density = 0.79 g/mL at 20 C). Express the concentration of rubbing alcohol:
A) molarity
(790 g solution / L of solution) * (70 g C3H7OH / 100 g solution)
= 553 g C3H7OH (L solution)
(553 g C3H7OH / L of solution) * (1 mol / 60.09 g C3H7OH)
= 9.20 mol C3H7OH / L solution = 9.20 m
B) molality (rmb to mk it in kg!!!)
790 g solution - 553 g C3H7OH = 237 g H2O = 0.237 kg H2O
9.20 mol C3H7OH / 0.237 kg H2O = 39 m
Factors that affect solubility
1) Temperature: affect most of substances
2) Pressure: influence solubility of a gas
> Henry’s law (c = kP): solubility of a gas in a liquid is proportional to the pressure of the gas over the solution
- where c = molar concentration (mol/L), P = pressure (atm), k = proportionality constant called Henry’s law constant
Example: Calculate the concentration of carbo dioxide in a soft drink that was bottled under a partial pressure of 5.0 atm CO2 at 25C
A) before the bottle is opened
B) after the soda has gone flat at 25C
Henery’s law constant for CO2 in water at this temp
= 3.110^-2 mol/Latm
Assume partial pressure of CO2 in air is 0.0003 atm & Henry’s law constant for the soft drink is the same as that for water
A) c = kP
c = (3.110^-2 mol/Latm)(5.0 atm) = 1.610^-1 mol/L
B) c = (3.110^-2 mol/Latm)(0.0003 atm) = 910^-6 mol/L
