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Chapter 5: Solutions. Classifying Solutions. 5.1- The Basics. classification of matter:. All Matter. Mixtures. Pure Substances. Heterogeneous (Mechanical Mixtures and Colloids). Homogeneous (Solutions). Elements. Compounds. is any that has. matter. solid, liquid or gas. mass.
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Chapter 5: Solutions Classifying Solutions 5.1- The Basics • classification of matter: All Matter Mixtures Pure Substances Heterogeneous (Mechanical Mixtures and Colloids) Homogeneous (Solutions) Elements Compounds
is any that has matter solid, liquid or gas mass and volume • are a type of matter that has pure substances definite fixed composition eg) elements and compounds • are combinations of matter that can be mixtures do NOT have definite proportions separated by physical means and
have and the different components are heterogeneousmixtures (mechanical mixtures) varying composition usually visible • are which have solutions homogeneous mixtures uniform composition and the different components are not visible • composed of at least one substance dissolved in another
solvent • is the and is usually the substance present in the dissolver, largest quantity (mass, volume, amount) eg) water • is what is solute dissolved in the solvent eg) salt
Types of Solutions • both solvents and solutes can be solids, liquids or gases • you can have various combinations of solute and solvent phases eg) liquid in liquid – ethylene glycol (antifreeze) solid in liquid – Kool Aid gas in liquid – carbonated beverages solid in solid – alloys liquid in solid – mercury amalgam fillings
an solution is any solution in which is the aqueous water solvent • water is called the therefore we will be concerned mainly with “universal solvent” aqueous solutions • water dissolves a lot of different solutes because of it’s …this is good because …also bad because dissolve in it easily unique properties 75% of the earth (and our bodies) is water toxins
Solutions in our Society • solutions are all around us and affect our lives in many ways • we have developed many to meet the and needs of humans technologies personal industrial eg) hair products, breathalyzer test, tests to monitor drinking water
in using solutions in our lives, care must be taken to ensure responsible use eg) – set up in Canada in 1985 to ensure that companies that deal with chemicals are using them in a safe and appropriate manner Responsible Care Program
using solutions can have intended and unintended consequences for humans and the environment • released into the environment are taken in by organisms toxins • this can have toxic effects immediate eg) H2S(g)…sour gas will cause a loss of consciousness at 700 ppm
sometimes the of the food chain are by the toxins since the levels are not that high…but the toxins are then passed along to the of the food chain lower levels unaffected upper levels • organisms at the of the food chains (like humans!) can end up with of these chemicals top toxic levels eg) mercury, lead, arsenic, PCB’s, DDT etc.
is the increase in the concentration of toxins as you biomagnification (bioaccumulation) move up the food chain
we must assess the andof using technologies that contain certain substances risks benefits eg) heavy metals released from mineral processing, power plants – mercury, arsenic
many chemical reactions (either in our bodies or in the lab) are if the reactants are not dissolvedin very slow water • dissolving in water allows the solute particles to separate, disperse and collide with other solute particles • particle collisions are necessary for to occur reactions
Electrolytes vs Non-Electrolytes • are aqueous solutions that electrolytes (weak and strong) conduct electricity eg) all soluble ionic compounds, very polar molecular compounds (like acids, ammonia) • are aqueous solutions that non-electrolytes do not conductelectricity eg) molecular compounds in solution
5.2: Explaining Solutions • is a dissolving physical change • the of a solid solute are held together by molecules or ions bonds • when dissolving occurs, these bonds and the of the solute become break ions or molecules attracted to the solvent particles H2O NaCl
the 3 processes involved in dissolving are: 1. bonds between molecules or ions of broken endothermic (requires energy) solute – always 2. bonds between molecules of broken solvent – always endothermic 3. bonds between molecules or ions of form solute and solvent – always exothermic
overall energy change in dissolving is equal to the of the three steps sum (law of conservation of energy) • if, the overall dissolving process is more energy is released than is required exothermic • if, the overall dissolving process is less energy is released than is required endothermic
When we dissolve compounds in water, water is not included as a reactant. • Water is involved in the ion interaction but it is not consumed so its not a reactant • We symbolize compounds dissolving in water by write an (aq) as the ‘state’ of the compound
occurs when when they are dissolved in an dissociation ionic compounds break apart into their ions aqueous solution • are used to show what happens to a substance when it is put into dissociation equations water • you can have 4 situations: 1. insoluble ionic or molecular compounds • they to any great extent do not dissolve solubility table • use the for ionic compounds eg) AgCl(s) AgCl(s) C25H52(s) C25H52(s)
2. soluble ionic compounds • dissolve to a great extent to form ions in solution • are broken ionic bonds • use solubility table • includes which turn litmus paper bases blue • the number of ions balance eg) NaCl(s) Na+(aq) + Cl(aq) KOH(s) K+(aq) + OH(aq)
3. soluble molecular compounds • dissolve to form molecules in solution intermolecular forces (LD, DD, HB) • are broken eg) C12H22O11(s) C12H22O11(aq)
Molecular compounds do not have a ‘solubility chart’ but there are a few patterns that can help us. • Non-polar molecules- generally don’t dissolve in water • Polar molecules may be slightly soluble in water • Polar compounds with hydrogen bonding are the most likely to be very soluble in water.
4. acids • they are but are molecular compounds very polar • dissolve to form in solution ions (ionize) • turns litmus paper red • the number of ions balance H2SO4(s) eg) 2H+(aq) + SO42(aq)
Ionization • Ionization: the process by which a neutral atom or molecule is converted to an ion. • Arrhenius proposed that acids ionize by interacting with water solvents • Idea is that the H+ causes the litmus paper to turn color
Examples Write the equations to show what happens to each of the following in water: 1. potassium chloride KCl(s) K+(aq) + Cl-(aq) 2. carbon dioxide CO2(g) CO2(g)
3. solid hydrogen nitrate HNO3(s) + H+(aq) NO3(aq) 4. aluminum sulphate Al2(SO4)3 (s) Al3+ (aq) 2 + 3 SO42- (aq) 5. sodium phosphate decahydrate 3 Na3PO4 10H2O(s) Na+ (aq) + PO43- (aq)
6. gasoline C8H18(l) C8H18(l) 7. barium sulphate BaSO4(s) BaSO4(s)
many chemical reactions (either in our bodies or in the lab) are if the reactants are not dissolvedin very slow water • dissolving in water allows the solute particles to separate, disperse and collide with other solute particles • particle collisions are necessary for to occur reactions
5.3: Solution Concentration the amount of solute relative to the amount of solvent • concentration is dilute = solute,solvent low high concentrated = high solute, solvent low • concentration can be expressed in a variety of ways: ppm, % volume, % mass, mol/L, mmol/L, mg/dL, %
these expressions can be seen on many products that we use in our daily lives and in industry eg) vinegar – toothpaste – blood cholesterol levels – breathalyzer - % (g of alcohol per 100 mL of blood) eg. 5% acetic acid by volume 0.243% by mass NaF 250 mg/dL 0.08
Molar Concentration molarity = number of moles of solute per litre of solvent c = n v n = m M where: c = concentration in mol/L n = number of moles in mol v = volume in L m = mass in g M = molar mass in g/mol
Example 1 A sample of 0.900 mol of NaCl is dissolved to give 0.500 L of solution. What is the molar concentration of the solution? n = 0.900 mol v = 0.500 L c = n v = 0.900 mol 0.500 L = 1.80 mol/L
Example 2 Calculate the concentration of 100 mL of a solution containing 0.300 mol of sulphuric acid. n = 0.300 mol v = 0.100 L c = n v = 0.300 mol 0.100 L = 3.00 mol/L
Example 3 Calculate the molar concentration of a 250 mL solution that has 3.2 g of NaCl dissolved in it. m = 3.2 g v = 0.250 L M = 58.44 g/mol n = m M = 3.2 g 58.44 g/mol = 0.0547…mol c = n v = 0.0547… mol 0.250 L = 0.22 mol/L
Example 4 Calculate the number of moles of Pb(NO3)2 needed to make 500 mL of a 1.25 mol/L solution. c = 1.25 mol/L v= 0.500 L n = cv = (1.25 mol/L)(0.500 L) = 0.625 mol
Example 5 How many litres of 4.22 mol/L solution would contain 3.69 mol of BaCl2? c = 4.22 mol/L n = 3.69 mol v = n c = 3.69 mol 4.22 mol/L = 0.874 L
Example 6 Calculate the mass of the salt required to prepare 1.50 L of a 0.565 mol/L solution of K3PO4. n = cv = (0.565 mol/L)(1.50 L) = 0.8475 mol c = 0.565 mol/L v = 1.50 L M = 212.27 g/mol m = nM = (0.8475 mol)(212.27 g/mol) = 180 g
Assignment • Practice questions
5.3 part 2: Molar Concentration of Ions • you may have to calculate the concentration ofin solution ions • once you have your dissociation equation, you can calculate the concentration of ions in solution using the mole ratio wanted given • are anions negative ions • are cations positive ions
Example 1 Calculate the ion concentrations in a 0.050 mol/L solution of KCl. g w w 1 KCl(s) 1 K+(aq) + 1 Cl-(aq) 1 1 1 1 C = 0.050 mol/L C = 0.050 mol/L C = 0.050 mol/L = 0.050 mol/L = 0.050 mol/L
Example 2 Calculate the ion concentrations in a 0.050 mol/L solution of Al2(SO4)3. g w w 1 Al2(SO4)3(s) 2 Al3+(aq) + 3 SO42-(aq) 2 1 3 1 C = 0.050 mol/L C = 0.050 mol/L C = 0.050 mol/L = 0.10 mol/L = 0.15 mol/L
Example 3 Calculate the concentrations of dissolved Na3PO4(s) 10H2O that gives a 0.30 mol/L concentration of Na+(aq) ions. w g 1 Na3PO4(s) 10H2O 3 Na+(aq) + 1 PO43-(aq) 1 3 C = 0.30 mol/L C = 0.30 mol/L = 0.10 mol/L
Example 4 Calculate the concentration of sodium ions in a NaCl(aq) solution made by dissolving 6.33 g of NaCl(s) in 150 mL of water. g w 1 NaCl(s) 1 Cl-(aq) + 1 Na+(aq) c = .722 mol/L 1 1 m = 6.33 g M = 58.44 g/mol = 0.722 mol/L n = m M = 6.33 g 58.44 g/mol = 0.108…mol c = n v = 0.108…mol 0.150 L = 0.722 mol/L
Assignment • Practice questions
5.4 Part 1: Preparing a Standard Solution • a solution of is called a known concentration standard solution • there are two ways to make a solution: 1. a measured amount ofin a certain volume of dissolve pure solute solvent 2. a standard solution dilute
to prepare a solution of known concentration from a : solid solute Steps mass solute Calculate theof therequired to achieve a specific concentration and volume. n = cv m = nM
1. ______ g (mass) of _______ (solute). Measure
2. the solute in _______ mL of water (half of the volume). Dissolve
3. Transfer solution to a ______ mL volumetric flask.