High School Chemistry Rapid Learning Series - 19

Rapid Learning Center
Chemistry :: Biology :: Physics :: Math

Rapid Lea...
High School Chemistry Rapid Learning Series - 19

Learning Objectives
By completing this tutorial you will learn…
How solu...
High School Chemistry Rapid Learning Series - 19

Forming Solutions

5/45

Definition: Solution, Solute & Solvent
Solution...
High School Chemistry Rapid Learning Series - 19

Steps for Solution Formation
In order for a solution to form:

1

Interm...
High School Chemistry Rapid Learning Series - 19

Why Don’t Oil and Water Mix?
Let’s look at the IMF’s broken and formed.
...
High School Chemistry Rapid Learning Series - 19

Factor’s Affecting
Solubility

11/45

Pressure - 1
How does pressure aff...
High School Chemistry Rapid Learning Series - 19

Pressure - 2
How does pressure affect solubility of a gas in a
liquid?
I...
High School Chemistry Rapid Learning Series - 19

Temperature-Dissolving Gases - 2
When dissolving gases:

If gas particle...
High School Chemistry Rapid Learning Series - 19

Temperature—Dissolving Solids
For most solids dissolving, increasing the...
High School Chemistry Rapid Learning Series - 19

Saturation of
Solutions

19/45

Definitions: Saturation
Unsaturated Solu...
High School Chemistry Rapid Learning Series - 19

How are Supersaturated Solutions Formed?
How can a solution hold more so...
High School Chemistry Rapid Learning Series - 19

Definition: Concentration
Concentration – Measure of the
qua t ty o so u...
High School Chemistry Rapid Learning Series - 19

Percent by Mass
To determine concentration in % by mass:
% mass =
Exampl...
High School Chemistry Rapid Learning Series - 19

Percent Mass/Volume
To determine concentration in % mass/volume:
% mass ...
High School Chemistry Rapid Learning Series - 19

Molality
To determine molality (m):
n
m=
kg solvent
Example:

m = molali...
High School Chemistry Rapid Learning Series - 19

Dilutions - 1
Often, you are supplied with a more concentrated
solution ...
High School Chemistry Rapid Learning Series - 19

Dilution Example
A dilution calculations example:
M 1 × V1 = M 2 × V2

E...
High School Chemistry Rapid Learning Series - 19

Solution Stoichiometry
Example: If you need 15.7 g Ba(OH)2 to precipitat...
High School Chemistry Rapid Learning Series - 19

Definition: Electrolyte

Electrolyte – Ionic compound
which dissolves in...
High School Chemistry Rapid Learning Series - 19

Breaking Compounds into Electrolytes
How do you break up a compound when...
High School Chemistry Rapid Learning Series - 19

Definition: Colloid

Colloid Solution with
C ll id – S l ti
ith
solute p...
High School Chemistry Rapid Learning Series - 19

Learning Summary

Solutions are composed
p
of solute and solvents.

The ...
High School Chemistry Rapid Learning Series - 19

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  1. 1. High School Chemistry Rapid Learning Series - 19 Rapid Learning Center Chemistry :: Biology :: Physics :: Math Rapid Learning Center Presents … p g Teach Yourself High School Chemistry in 24 Hours 1/45 http://www.RapidLearningCenter.com Solutions HS Ch i t R id Learning Series Chemistry Rapid L i S i Wayne Huang, PhD Kelly Deters, PhD Russell Dahl, PhD Elizabeth James, PhD Rapid Learning Center www.RapidLearningCenter.com/ © Rapid Learning Inc. All rights reserved. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 1
  2. 2. High School Chemistry Rapid Learning Series - 19 Learning Objectives By completing this tutorial you will learn… How solutions form. Factors affecting solubility. F t ff ti l bilit Concentrations of solutions. Calculations with solution concentrations. Electrolyte solutions. Colloids. 3/45 Concept Map Previous content Chemistry New content Studies Matter Solution Formation Solute & Solvent properties govern One type of Solutions Solute & Solvent ratios described by Various factors affect Concentration Solubility Used in calculations Dilutions Stoichiometry 4/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 2
  3. 3. High School Chemistry Rapid Learning Series - 19 Forming Solutions 5/45 Definition: Solution, Solute & Solvent Solution – Homogeneous mixture, solution = solute + solvent. Solute Substance being dissolved, present in a S l t –S b t b i di l d ti smaller quantity (solute dissolves into solvent). Solvent – Substance (in a greater quantity) doing the dissolving. Water is called the “Universal Solvent Solvent” because it s used for so it’s many solutions. Solutions can be any combination of solids, liquids and gases! 6/45 Solute vs Solvent Mnemonic: Dissolve solute into solvent = “The police came, the thief hide the lute in the vent.” © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 3
  4. 4. High School Chemistry Rapid Learning Series - 19 Steps for Solution Formation In order for a solution to form: 1 Intermolecular forces in the solvent are broken ( (“expanding the solvent”): Solvent - - X- - Solvent p g ) 2 Intermolecular forces in the solute are broken (“expanding the solute”): Solute - - X - - Solute 3 Solute and solvent particles form new intermolecular forces together: Solute –– Solvent 7/45 Energy of Solution Formation 1 Intermolecular forces in the solvent are broken. 2 Intermolecular forces in the solute are broken. 3 Solute and solvent particles form new intermolecular forces together. IMF – Intermolecular Force Breaking IMF’s requires energy Forming IMF’s releases energy If energy required >> energy released… gy q gy the solution will not form. If energy required similar or < energy released… The solution will form. 8/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 4
  5. 5. High School Chemistry Rapid Learning Series - 19 Why Don’t Oil and Water Mix? Let’s look at the IMF’s broken and formed. IMF’s Broken IMF’s Formed Water—London Dispersion Oil & Water together— W t t th London Dispersion Water—Dipole-Dipole Water—Hydrogen bonding The two are more stable separate than together. Oil—London Dispersion 4 IMF’s are broken. Only 1 IMF can be formed (and it’s the weakest kind). Much more energy would be required than released. Therefore, the solution doesn’t form. 9/45 “Like Dissolves Like” This general rule that can be followed for solution formation stems from: Compounds that have similar properties and bond characteristics form similar intermolecular forces. If similar intermolecular forces can be formed between 2 compounds as within each compound, a solution is likely to form. Things with “Like” bond types will dissolve other things with “Like” bond types. Things dissolves other things with “Like” bond properties (“Like dissolves Like”): 10/45 Examples: Polar compounds (HCl) dissolve polar compounds (H2O). Non-polar (C5H12) dissolves non-polar (C6H6), etc. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 5
  6. 6. High School Chemistry Rapid Learning Series - 19 Factor’s Affecting Solubility 11/45 Pressure - 1 How does pressure affect solubility of a gas in a liquid? Gas molecules above the solvent cause pressure. When a gas molecules comes into contact with the surface of the solvent, it can be “dissolved”. 12/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 6
  7. 7. High School Chemistry Rapid Learning Series - 19 Pressure - 2 How does pressure affect solubility of a gas in a liquid? If there are more gas th particles above the solution (a higher pressure)… More will come in contact with the t t ith th solvent and “dissolve”. Higher pressure 13/45 Higher solubility for gases (Pressure has no affect on solids or liquids dissolving!) Temperature-Dissolving Gases - 1 When dissolving gases: As A gas particles move around, when ti l d h they reach the surface of the solution, they can “escape” if they have enough energy. 14/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 7
  8. 8. High School Chemistry Rapid Learning Series - 19 Temperature-Dissolving Gases - 2 When dissolving gases: If gas particles have ti l h more energy (higher temperature)… They will move faster, reach the surface more often, and have more energy to “escape” when they do reach the surface. Higher temperature Lower solubility for gases 15/45 Dissolving Gases Gases dissolve in a liquid best at: High Pressure Low Temperature 16/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 8
  9. 9. High School Chemistry Rapid Learning Series - 19 Temperature—Dissolving Solids For most solids dissolving, increasing the temperature of the solvent: Increase in solvent temperature increases energy of solvent particles. Increased energy allows for more IMF’s of the solvent to be broken (the solvent is more “expanded”). There is more “room” for solute particles in between the solvent particles. For most solids, as temperature increases, solubility increases. 17/45 Solubility Curves The solubility of a solid at different temperatures is shown in a solubility curve: Solubility changes a lot as temperature is increased. Solubility is barely affected by change in temperature. 18/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 9
  10. 10. High School Chemistry Rapid Learning Series - 19 Saturation of Solutions 19/45 Definitions: Saturation Unsaturated Solution – The solvent can still hold more solute particles. Saturated Solution – The solvent is holding as many solute particles as it can possibly hold. Supersaturated Solution – The solvent is holding more solute particles (crystals) than it should be able to at that temperature. More solute can dissolve. No more solute can dissolve. More solute can crystallize. 20/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 10
  11. 11. High School Chemistry Rapid Learning Series - 19 How are Supersaturated Solutions Formed? How can a solution hold more solute particles than it usually can at that temperature? Solution is saturated. Temperature of T t f solution is raised. Solution is no longer saturated at the new temperature. More solute particles are added. Solution is slowly and carefully cooled back down. Solute particles remain dissolved. Solution is now supersaturated to form crystals. 21/45 Concentrations of Solutions 22/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 11
  12. 12. High School Chemistry Rapid Learning Series - 19 Definition: Concentration Concentration – Measure of the qua t ty o so ute per quantity quantity of solute pe qua t ty of solvent or total solution. Common Units: Percent Composition by Mass (%) Molarity M (Moles of solute per volume solution L) Molality m (Moles of solute in 1kg solvent) y ( g ) Mole Fraction (ratio of moles in one component over moles of all components) 23/45 Definition: Concentrated & Dilute Concentrated – Large ratio of solute : solvent. Dilute Small ti f l t Dil t – S ll ratio of solute : solvent. l t Do not use the terms “weak” or “strong” to describe concentrations -those words have very specific meanings in chemistry. 24/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 12
  13. 13. High School Chemistry Rapid Learning Series - 19 Percent by Mass To determine concentration in % by mass: % mass = Example: mass solute ×100 mass solution The mass units must match! A sample contains 1.25 g NaCl in 100 mL of water. (1 mL water = 1 g water). Determine the % by mass of the solution. Mass solute = 1.25 g NaCl Mass solution = 1.25 g NaCl + 100 g water = 101.25 g % by mass = ? % mass = 1.25 g NaCl ×100 = 1.23% NaCl 101.25 g solution 25/45 Percent by Volume To determine concentration in % by volume: % volume = Example: volume solute ×100 volume solution The volume units must match! What volume of water is needed to make a 15% by volume solution of alcohol if you have 5 mL of alcohol? Volume solute = 5 mL Volume solution = 5 mL solute + x mL solvent = (5+x) mL % by volume = 15% 15% = 5mL × 100 (5 +x)mL x mL = (5 + x)mL = 5mL ×100 15% 5mL ×100 − 5mL 15% x = 28 mL solvent 26/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 13
  14. 14. High School Chemistry Rapid Learning Series - 19 Percent Mass/Volume To determine concentration in % mass/volume: % mass / volume = Example: mass solute ×100 volume solution Use mass in grams. Use volume in milliliters. Vinegar is a 5% solution of acetic acid in water (by mass/volume). What mass of acetic acid (CH3COOH) is in 45 mL of vinegar? Mass solute = ? g Volume solution = 45 mL % mass/volume = 5% 5% = mass acetic acid ×100 45mL (45mL) × 5% = mass acetic acid 100 2.25 g CH3COOH 27/45 Molarity One of the most commonly used concentration units is Molarity (M): M = Molarity n n = moles solute M= V V = volume solution (in Liters) Example: Find the molarity if 12.5 g NaCl is dissolved in 500 mL of water. Solute = 12.5 g NaCl = 0.214 mole NaCl Solution = 500 mL = 0.500 L 0.214mole NaCl Molarity = Molarity = ? M 0.500 L 12.5 g NaCl 1 mole NaCl = 0.214 mole NaCl 58.44 g NaCl 500 mL 28/45 0.428 M NaCl 0.001 L 1 mL = 0.500 L © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 14
  15. 15. High School Chemistry Rapid Learning Series - 19 Molality To determine molality (m): n m= kg solvent Example: m = molality n = moles solute How many moles NaNO3 are needed to make a 0.24 m solution with 1.5 L (1.5 kg) of water? Solute = ? mole NaNO3 Solvent = 1.5 kg molality = 0.24 m 0 24 0.24m = mole NaNO3 1.5 kg 1.5kg × 0.24m = mole NaNO3 0.36 mole NaNO3 29/45 Molarity vs Molality Mnemonic: MolaRity (moles to Liter solution) and MolaLity (moles to kg solvent) = “Rose to Lover and Loyal to King!” Calculations with Concentrations 30/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 15
  16. 16. High School Chemistry Rapid Learning Series - 19 Dilutions - 1 Often, you are supplied with a more concentrated solution than you need. In order to dilute a solution to a lower concentration, more solvent is added. dd d The moles of solute did not change: moles solute before = moles solute after Solvent Solute 31/45 Dilutions - 2 Often, you are supplied with a more concentrated solution than you need. In order to dilute a solution to a lower concentration, more solvent is added. added The moles of solute did not change: moles solute before = moles solute after And if M= n V Then n1 = n2 n = M ×V M 1 × V1 = M 2 × V2 The original molarity times the volume of the original solution = the new molarity times the new volume of the solution. 32/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 16
  17. 17. High School Chemistry Rapid Learning Series - 19 Dilution Example A dilution calculations example: M 1 × V1 = M 2 × V2 Example: 2 volume units must match! You need 55 mL of 0.10 M HCl solution. You currently have 12 M HCl solution. What volume of the concentrated solution will you dilute to 55 mL? M1 = 12 M HCl V1 = ? mL M2 = 0 10 M HCl 0.10 V2 = 55 mL 12 M × V1 = 0.10 M × 55mL V1 = 0.10 M × 55mL 12 M V1 = 0.46 mL 33/45 Using Molarity in Conversions Molarity is used to convert between moles and liters. Example: If 0.85 moles NaOH are needed and you have a 1.5 M solution, how many liters of the solution do you need? From concentration: 1.5 mole NaOH = 1 L 0.85 mole NaOH 1 L 0.57 = ________ L 1.5 mole NaOH 34/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 17
  18. 18. High School Chemistry Rapid Learning Series - 19 Solution Stoichiometry Example: If you need 15.7 g Ba(OH)2 to precipitate, how many liters of 2.5 M NaOH solution is needed? 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl From balanced equation: q 2 mole NaOH 1 mole Ba(OH)2 Concentration of NaOH: 2.5 mole NaOH = 1 L Molar Mass of Ba(OH)2: 1 mole Ba(OH)2 = 171.35 g 15.7 g Ba(OH)2 ( ) 1 mole Ba(OH)2 171.35 g Ba(OH)2 2 mole NaOH 1 mole Ba(OH)2 1 L NaOH 2.5 mole NaOH 0.0733 = ________ L NaOH 35/45 Electrolyte Solutions 36/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 18
  19. 19. High School Chemistry Rapid Learning Series - 19 Definition: Electrolyte Electrolyte – Ionic compound which dissolves in water, producing free-floating ions. Free-floating ions can conduct electricity (hence “electro”). e.g. When dissolved in water: NaCl Na+ + ClCa(NO3)2 Ca2+ + 2 NO3- 37/45 Definition: Strong, Weak & Non-Electrolytes Strong Electrolyte – Most of the ions dissociate and are free floating. g Weak Electrolyte – Only some of the ions dissociate and are free floating (weakly conducts electricity). Non-Electrolyte – Dissolved substance does not produce ions at all (does not conduct electricity). 38/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 19
  20. 20. High School Chemistry Rapid Learning Series - 19 Breaking Compounds into Electrolytes How do you break up a compound when forming electrolytes? 1 Do not break up polyatomic ions. 2 Use subscripts that are not a part of a polyatomic ion as coefficients. e.g. CaCl2 doesn’t have “Cl2” ions, it has 2 “Cl” ions. Example: Break up the following strong electrolytes: Na3PO4 (NH4)2CO3 3 Na+ + PO432 NH4+ + CO32- 39/45 Colloids 40/45 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 20
  21. 21. High School Chemistry Rapid Learning Series - 19 Definition: Colloid Colloid Solution with C ll id – S l ti ith solute particles large enough to deflect light as it travels through the solution. 41/45 Definition: Tyndall Effect Tyndall Effect – Property exhibited by colloids. The scattering light is visible through the solution. Light coming in Light going out The light is not scattered, and is not seen traveling through the solution. Solution The light is scattered, and is seen traveling through the colloid. Light coming in 42/45 Light going out Colloid © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 21
  22. 22. High School Chemistry Rapid Learning Series - 19 Learning Summary Solutions are composed p of solute and solvents. The l ti Th solution process is governed by energetics of solution formation and factors affecting solubility. Colloids are mixtures which have large enough g g solute particles to scatter light (exhibit the Tyndall Effect). Concentration is an expression of the ratio of i f th ti f solute to solvent particles. Concentrations are used in dilution and stoichiometry calculations. 43/45 Congratulations You have successfully completed the core tutorial Solutions Rapid Learning Center © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 22
  23. 23. High School Chemistry Rapid Learning Series - 19 Rapid Learning Center Chemistry :: Biology :: Physics :: Math What’s N t Wh t’ Next … Step 1: Concepts – Core Tutorial (Just Completed) Step 2: Practice – Interactive Problem Drill Step 3: Recap – Super Review Cheat Sheet Go for it! 45/45 http://www.RapidLearningCenter.com © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 23

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