Unit 4 Solutions - Gr.11 chem PDF

Preview

Summary

Download Unit 4 Solutions - Gr.11 chem PDF

Description

Solutions & Solubility Solutions: • a homogenous mixture of two or more substances • homogenous = same properties throughout • composed of a solute and a solvent • solutions can consist of the same substance but have different concentrations (unlike compounds that always have the same proportions). • generally we think of solutions in the liquid phase, however solutions can exist as combinations of a variety go phases. Liquid Solvent • solid dissolved in liquid • the ocean • liquid dissolved in liquid • anti-freeze • gas dissolved in liquid • pop Gas Solvent • solid dissolved in gas • naphelene • liquid dissolved in gas • humidity • gas dissolved in gas • natural gas Solid Solvent • solid dissolved in solid • structural steel • liquid dissolved in solid • liquid mineral • gas dissolved in solid • ice contains air

Unit 4 - Solutions Review

Classify Solutions by Solute Type: 1. Presence of electrolytes • (substances that allow an electric current to pass through) • dissolved ions are electrolytes • dissolved molecular compounds are not examples; ! ! ! MgCl2 (aq) → conducts electricity ! ! ! → contains electrolytes ! ! CO2 (aq) → does not conduct electricity ! ! ! → does not contain electrolytes 2. Presence of acids and bases • (substances that produce H+ or OH- ions in solutions; both produce ions, so therefore they will conduct electricity) • test pH (litmus turns red in acid, and litmus turns blue in base) examples; ! ! H2SO4 (aq) → sulphuric acid turns litmus red ! ! Mg(OH)2 (aq) → magnesium hydroxide turns litmus blue

Solubility & Saturation: • solubility is the amount of subtle that will dissolve in a given amount of solvent at a specific temperature • 25℃ • increase of temperature → increase of solubility (not including gases) • unsaturated solution is where more subtle can be dissolved in the solution at that temperature • saturated solution is where the maximum amount of solute is dissolved in the solution at a specifi c temperature • supersaturated solution is where more than the maximum amount of solute is dissolved in the solution • unstable

Unit 4 - Solutions Review

Solution Concentrations: • solution concentrations give you an indication of the amount of solute per volume of solution • there are a number of ways to calculate or express solution concentrations 1. Percentage by Volume ( % V/V) ! Concentration = Volume of Solute ! ! ! ——————— X 100% ! ! ! Volume of Solution • usually used for liquid solute and liquid solvent • examples; vinegar, hydrogen peroxide 2. Percentage Mass by Volume ( % m/V) ! Concentration = Mass of Solute ! ! ! ——————— X 100% ! ! ! Volume of Solution • usually used for solid solute and liquid solvent 3. Percentage by Mass ( % m/m) ! Concentration = Mass of Solute ! ! ! ——————— X 100% ! ! ! Mass of Solution • usually used for solid solute and solid solvent

Unit 4 - Solutions Review

Very Small Concentrations: • parts per million (ppm) and parts per billion (ppb) • a concentration of 1ppm means that there is 1 part solute per million parts of solution 1. Parts Per Million ! ! ppm = Mass of Solute ! ! ! ——————— X 106 % ! ! ! Mass of Solution 2. Parts Per Billion ! ! ppb = Mass of Solute ! ! ——————— X 109 % ! ! Mass of Solution • units for the mass of solute and solution need to be the same example; ! ! ppm = 0.005mg ! ! ! ——————— X 106 % ! ! ! 1kg ! ! ! ! ! !

! ! ! ! ! !

ppm = 5.00 X 10-6 g ! ——————— X 106 % ! 1000g ppm = 0.005ppm ! ↓ this is equivalent to 5ppb

Unit 4 - Solutions Review

Molar Concentrations: • concentration in mols of solute per litre of solution → c = n ÷ v • this gives us another quantity that we can relate to mols

!

→→→→→→→→→→! !

mass ! !

! ! ! !

!

!

!

→→→→→→→→→→

mols! !

!

!

# of particles

←←←←←←←←←←! ↕ ←←←←←←←←←← ! ! ! ! ↕ ! ! ! ! ↕ ! ! ! ! ↕ concentration

example; 14.5g of calcium chloride is dissolved in water to produce a 225mL solution. What is the molar concentration of the solution? step 1: chemical formula ! CaCl2 step 2: molar mass ! 110.98 g/mol step 3: moles of solute ! n = m÷M ! n = 14.5 ÷ 110.98 ! n = 0.13065 mols step 4: concentration ! c = n÷V ! c = 0.13065 ÷ 0.225L ! c = 0.581 mol/L

Unit 4 - Solutions Review

Ionic Compounds Dissolving: • ionic compounds break apart into all the atoms that made up the molecule and will float around in the substance • the ions will be surrounded by water molecular as they dissolve (this process is called hydration) • aqueous solutions of ionic compounds will conduct electricity

Predicting the Solubility of Ionic Compounds: • the greater the ionic charge, the greater the force of attraction between ions, so the less the solubility • the greater the atomic radius, the weaker the attraction between ions, so the greater the solubility

Molecular Compounds Dissolving: • molecular compounds do not break apart into the atoms that make up the molecule but stay as a whole molecule and float around the substance • “like dissolves like” • polar solutes molecules are attracted to the partial charges of water molecule int the solute molecules dissolving • non-polar molecules are not attracted to the water molecules so they will not dissolve in the water • aqueous solutions of molecular compounds do not conduct electricity

Temperature and Solubility: • as the temperature increases, so does the solubility • a solubility chart is a graph showing the relationship between solubility and temperature • temperature has the opposite effect on the solubility of gases • temperature has no effect on solutes

Unit 4 - Solutions Review

Preparing Lab Solutions: stock solutions - a concentrated solution that is used to prepare dilute versions of solution for lab use example; you have 1.25 mol/L solution, what mass of cobalt (II) chloride dehydrate was used? (c = 1.25 mol/L of CoCl2 · 2H2O) (V = 500mL → 0.500L) step 1: fi nd moles from c and V n=cXV n = (1.25mol/L) (0.500L) n = 0.625 mols step 2: fi nd mass from moles m=nXM m = (0.625 mols) (165.87g/mol) m = 104g of CoCl2 · 2H2O

M = 58.93 +35.45(2) + 18.02(2) M = 165.87g/Mol

dilute solutions - prepared by diluting a stock solution with water example; 500mL of 0.10 mol/L solution of HCl is required for a lab. we could prepare this from our 6.0mol/L HCl stock solution that is required step 1: start with equation nstock = ndilute or

c1V1 = c2V2 step 2: plug number into equation (6.0mol/L) V1 = (0.10 mol/L) (500.00mL) 6V1 = 50 — — V1 = 8.3 mL of stock solution Unit 4 - Solutions Review

Chemical Reactions in Solutions: • when ionic compounds dissolve they split into ions • ionic compounds (s) → ions (aq) examples; ! ! ! !

NaCl dissolves in water NaCl(s) → Na+(aq) + Cl

Mg3(PO4)2 dissolves in water Mg3(PO4)2 → 3Mg2+ + 2PO43-

• for a chemical reaction in an aqueous environment we can treat compounds as dissolved ions Balanced chemical equation: !

Na2SO4(aq) + CaCl2(aq) → 2NaCl(aq) + CaSO4(s)

Total ionic equation: • all aqueous ions compounds are split into ions ! 2Na+(aq) + SO3- 4(aq) + Ca3+(aq) + 2Cl+(aq) → 2Na+ + 2Cl(aq) + CaSO4(s) Net ionic equation: • remove any identical ions that are on both sides of the chemical equation - these are called spectator ions !

SO2- 4(aq) + Ca2+(aq) → CaSO4(s)

example; Balanced chemical equation: !

CuSO4(aq) + 2NaOH(aq) → Na2SO4(aq) + Cu(OH)2(s)

Total ionic equation: !

Cu2+(aq) + SO2- 4(aq) + 2Na+(aq) + 2OH-(aq) → 2Na+ + SO4 2-(aq) + Cu(OH)2(s)

Net ionic equation: !

Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)

Unit 4 - Solutions Review

Qualitative Analysis Solutions: 1. solution colour 2. flame colours 3. precipitation of specific compounds - create a series of reactions that will precipitate one ion (or compound) at a time

Solutions in Stoichiometry: • we can determine quantities in chemical reactions when we are dealing with solids (last unit) and solutions (this unit) m = n X M! ! !

!

mole rations !

→→→→→→→→→→! !

mass ! !

! ! !

!

!

!

!

!

!

n=cXv

→→→→→→→→→→

mols! !

!

!

# of particles

←←←←←←←←←←! ↕ ←←←←←←←←←← ! ! ! ! ↕ ! ! ! ! ↕ concentration

example; 225mL of 0.45mol/L of silver nitrate solution reacts with 475mL of 0.12mol/L of potassium chloride in solution a) determine the limiting reactant b) calculate the mass of precipitate produced c) calculate the remains concentration of the excess reactant a) step 1: balanced equation ! ! AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)

! ! !

step 2: moles of each reactant ! nAgNO3 = c X v ! ! = 0.45mol/L X 0.255L ! ! = 0.101mol

! !

step 3: limiting reactant ! ∴ KCl is the limiting reactant because is it less than AgNO3, and I know ! ! ! that because its a 1:1 ratio

nKCl = c X v = 0.12mol/L X 0.475L = 0.0570mo

Unit 4 - Solutions Review

b) step 1: moles of precipitate produced ! ! nAgNO3 = 0.0570mol KCl X 1 mol AgCl ! ! ! ! ! ! ————— ! ! ! ! ! ! 1 mol KCl ! ! ! = 0.0570 mol step 2: mass of precipitate produced ! ! m = 0.0570 X 143.32 ! ! m = 8.2g c) step 1: moles of excess reactant remaining ! ! nAgNO3 = ntotal - nused ! ! ! = 0.101mol - 0.057 mol ! ! ! = 0.0440mol

! ! ! !

step 2: concentration of ! c=n ! ! —!! ! V!! ! ! !

excess reactant solution ! c = 0.0440 mol ! ————— ! 0.0700 L ! c = 0.063 mol/L

→→ ( total volume)

Unit 4 - Solutions Review

Acids & Bases: Acids

Bases

pH

less than 7

greater than 7

Chemical Formula

starts with H

ends with OH

Produced By

non-metal oxide + water

metal oxide + water

Neutralization Reaction

add base! HCl + NaOH

add acid! NaCl + H2O

Taste

sour

bitter

Other Properties

corrosive! reacts with metals to produce hydrogen gas

corrosive

Reaction with Litmus

turns red

blue

Electrical Conductivity

yes

yes

Theories: • one of the earliest theories is the Arrhenius definition • Arrhenius acid = a compound that ionizes in the presence of water and release H+ ions • HCl(g) → H+(aq) + Cl-(aq) • Arrhenius base = a compound that dissociates in the presence of water and release OHions • NaOH(s) → Na+(aq) + OH-(aq) examples; - H2SO4(g) → 2H+(aq) + SO42-(aq) - Mg(OH)2 → Mg2+(aq) + 2OH-(aq)

Unit 4 - Solutions Review

Bronsted - Lowry Acids & Bases bronsted - lowry acid - a hydrogen ion (H+) donor - a hydronium ions is produced because H+ ions will react with water very quickly bronsted - lowry base - hydrogen ion (H-) donor - ammonia is an example of a substance that acts as a bronsted lowry base but not as an Arrhenius base

What is pH: • it is a measure of hoe acidic a solution is (which means it is related to the concentration of H+ ions in a solution) • pH is based on a logarithmic scale, which means that it is based on powers of 10 • a change of one pH unit means a 10x change in concentration of H+ • a substance that has a pH = 10x as acidic as a substance that is pH = 3; and 100x as acidic as a substance that is pH = 4

How can we Calculate pH: pH = -log[H+]!

→→

[H+] = 10-pH! !

→→

example; if the [H+] = 0.0010mol/L then pH = -log[0.0010 mol/L] pH = 3 example; the pH of a solution is 9.7, what is the [H+]? [H+4] -=Solutions 10-9.7 Unit Review [H+] = 1.995 x 10-10 mol/L

Strong vs. Weak Acids & Bases: • the terms strong and weak refers to the extent of ionization (acids) or dissociation (bases) • a strong acid will ionize completely (all reactant split to become products) • H2SO4 is an example of a diuretic acid (can release 2 H+ ions). • ammonia, NH3 is an example of a weak base

Strong & Weak vs. Concentrated & Dilute: • strong vs. weak refers to the extent of ionization • concentrated/ dilute refers to the concentration of solute example; ! ! ! ! ! !

0.10 mol/L HCl ! ! ↓! ! pH = 1! !

vs. ! ! !

0.10 mol/L HCH3OO ! ↓ ! pH = -0.3

Unit 4 - Solutions Review...