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STAAR Chemistry Review Topic: Solutions

STAAR Chemistry Review Topic: Solutions. TEKS 10 A-J. Student Expectation (SE). 10A – The student describes the unique role of water in chemical and biological systems. INDEX CARD TIME! TITLE: The Importance of Water FRONT : water is a polar molecule

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STAAR Chemistry Review Topic: Solutions

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  1. STAAR Chemistry Review Topic: Solutions TEKS 10 A-J

  2. Student Expectation (SE) 10A – The student describes the unique role of water in chemical and biological systems.

  3. INDEX CARD TIME! TITLE: The Importance of Water FRONT: water is a polar molecule – diagram and labels BACK: water’s unique properties

  4. Mini-Review Water is a polar molecule

  5. Mini-Review Water’s unique properties: • Universal solvent: water’s polarity helps it dissolve ions and polar molecules. • Cohesion: water is attracted to itself. • Surface tension: force of attraction at the surface of the water. It causes the water to bead when placed on wax paper, a non polar surface; enables insects to walk on water. • Adhesion: water is attracted to other polar materials. • Capillary action: water rises inside thin tubes. • High specific heat: water can absorb/release heat with little change of temperature; oceans help moderate atmospheric temperature. • Lower density solid than liquid: water expands as it freezes, causing weathering of rocks; ice floats, reducing risk of entire lakes freezing.

  6. Student Expectation (SE) 10 B– The student develops and uses general rules regarding solubility through investigations with aqueous solutions

  7. INDEX CARD TIME! TITLE: Solubility FRONT: Definition of solubility ,solution, solute, solvent, aqueous, symbol (s), symbol (l) , (g) , (aq). BACK: -solubility rules. - example of balanced net ionic equation

  8. Mini-Review • Solubility: maximum amount of solute that can dissolve in a solvent. • Solution: mixture blended so that properties are the same throughout. • Solute: substance dissolved in another ( solid- liquid- gas). • Solvent: substance doing the dissolving ( solid- liquid-gas). • Aqueous: solution in which the solvent is liquid water. • (s) : the substance is solid • (l): the substance is liquid. • (g): the substance is a gas. • (aq): the material is dissolved in water.

  9. Mini-Review Solubility Rules for ionic compounds Soluble compounds contain CH3COO- , NH4+ , NO3-, CN- , ClO- , ClO2- , ClO3- , ClO4- . They dissociate when dissolved in water ex: NaClO (aq) → Na+ (aq) + ClO- ( aq) For the other soluble or insoluble compounds , refer to solubility table provided . Pay attention to the exceptions. ex: soluble compounds contain Br-, Cl-, I- except compounds of Ag+, Pb2+, and Hg 2+

  10. Mini-Review Solubility Rules for molecular compounds “ Like dissolves like” Non polar molecules (= non metals sharing electrons) dissolve best in non polar solvents, whereas polar and ionic compounds ( metal + non metal) dissolve best in polar solvents. Ex: Oil is non-polar and not soluble in water Balanced net ionic equation Pb (s) + 2AgNO3 (aq) → 2 Ag(s) + Pb(NO3)2 (aq) Pb (s) + 2Ag+ (aq)+ 2NO3 - (aq) → 2 Ag(s) + Pb 2+(aq)+2 NO3- (aq) Spectator ions appear on both sides of an equation but are not directly involved in the reaction ( not chemical change) – they are cancelled . Pb (s) + 2Ag+(aq) → 2 Ag(s) + Pb 2+(aq) Soluble = dissociates

  11. Student Expectation (SE) 10-c The student is expected to calculate the concentration of solutions in units of molarity

  12. INDEX CARD TIME! TITLE: Molarity FRONT: definition and calculation Back: example

  13. Mini-Review Molarity: M = describes concentration of a solute in solution M = moles of solute= mol liter of solution L

  14. Mini-Review Examples: • You dissolve 0.25 moles of sodium sulfate in 1.5L of solution. Calculate the molarity of the solution. Molarity = moles of sodium sulfate = 0.25 = 0.17 M liters of solution 1.5 b) How many liters of a 2.0 M solution of glucose would contain 3 moles of glucose? Liters of solution= moles glucose = 3 = 1.5L Molarity 2

  15. Student Expectation (SE) 10-d The student is expected to use molarity to calculate the dilutions of solutions

  16. INDEX CARD TIME! TITLE: Dilution of a solution FRONT: Definition and equation BACK: example

  17. Mini-Review Dilution of a solution is the reduction of the concentration of a solute by adding more solvent. M initialx V initial= M finalx V final

  18. Mini-Review Example: What volume of 0.5 M NaOH is needed to make 2 L of a 0.075M solution ? M initialx V initial= M finalx V final 0.5 M x V initial= 0.075 M x 2 L V initial = 0.075 M x 2 L =0.3 L 0.5M

  19. Student Expectation (SE) 10-E The student distinguishes types of solutions such as electrolytes and nonelectrolytes and unsaturated, saturated, and supersaturated solutions.

  20. INDEX CARD TIME! TITLE: Types of Solutions FRONT: Define electrolyte, nonelectrolyte, unsaturated, saturated, supersaturated. BACK: • How can you distinguish between electrolytes and nonelectrolytes? ( electrolyte solution circuit) • How can you distinguish between unsaturated,saturated, and supersaturated solutions? ( Solubility graphs)

  21. Mini-Review The particles of an electrolyte ( dissolved ions) allow electrons to flow. note: if the ions are not dissolved ( solid crystal), they will not allow electrons to flow. At 50 °C, we can dissolve a maximum of 30 g of K2Cr2O7 in 100 g of water. The solution is saturated If we add only 20 g, the solution is unsaturated and could dissolve another 10 g of K2Cr2O7. If we add 50 g, we will observe the precipitation of an excess of 20 g of K2Cr2O7 . If we add 30 g of K2Cr2O7 in 100 g of water at 50 °C and gradually lower the temperature of the saturated solution, crystals may not begin to form and the solution is supersaturated ( too much solute remains dissolved).

  22. Student Expectation (SE) 10-F The student investigate factors that influence solubilities and rates of dissolution such as temperature, agitation, and surface area.

  23. INDEX CARD TIME! TITLE: Factors that influence solubility and rate of dissolution. FRONT: - definition of solubility. - definition of rate of dissolution. BACK: - how does temperature influence solubilities and rates of dissolution? - how does agitation influence solubilities and rates of dissolution? - how does surface area influence solubilities and rates of dissolution? - how does pressure influence solubilities and rates of dissolution?

  24. Mini-Review To remember how different factors affect solubility, think about the interaction between the solute particles and the solvent particles. Generally, if the solute and solvent are able to interact more with one another, then solubility increases and dissolution speeds up.

  25. Mini-Review Note: solubility of gases: - increases at higher pressure - decreases at higher temperature; warm water dissolves less oxygen gas than cold water

  26. Student Expectation (SE) 10- G The student defines acids and bases and distinguishes between Arrhenius and Bronsted-Lowry definitions and predicts products in acid-base reactions that form water.

  27. INDEX CARD TIME! TITLE: Acids and Bases. FRONT: - List four properties of acids and bases BACK - What is the Arrhenius definition of an acid and of a base? Give an example. -What is the Bronsted-Lowry definition of an acid and of a base? Give an example. - What is a neutralization reaction? Give an example

  28. Mini-Review

  29. According to Arrhenius: An acid produces H+ in water HCl (aq) → H+(aq) + Cl-(aq) A base produces OH- in water Ca(OH)2(aq) → Ca+2(aq) + 2OH-(aq) According to Bronsted-Lowry An acid donates a proton in a reaction. NH4+ (aq) + OH-(aq) → NH3(aq) + H2O (l) A base accepts a proton in a reaction. NH3(aq) + H2O (l) → NH4+(aq) + OH-( aq) A neutralization reaction: reaction that occurs when an acid and base are mixed to produce a salt and water. Base + Acid → Water + Salt   NaOH   + HCl → H2O  +  NaCl BOH + HA ---> H2O + BA

  30. Student Expectation (SE) 10-H The student understands and differentiates among acid-base reactions, precipitation reactions, and oxidation-reduction reactions.

  31. INDEX CARD TIME! TITLE: Types of Reactions FRONT: - Define and give an example of a Precipitation reaction. - Define and give an example of an acid-base reaction. BACK: -Define and give an example of an oxidation-reduction reaction ( Redox)

  32. Mini-Review In a precipitation reaction a product is a solid precipitate (s) . You need to recall the general rules of solubility . NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq) http://employees.csbsju.edu/hjakubowski/classes/ch111/olsg-ch111/chemrxs/animprecip.gif In an acid-base reaction, protons (H+) are transferred. The product includes a salt and , often but not always , water NaOH (aq)  + HCl (aq) → H2O  +  NaCl (aq) NH3 (aq) + HCl (aq) → NH4Cl (aq) A compound contributed a proton to the other compound and a salt formed In a oxidation-reduction reaction ( REDOX) electrons are transferred. Many types of reactions are oxidation-reduction reactions. ex: All combustion, single displacement, many combination and decompositions. You need to recall the rules for determining Oxidation Numbers. Oxidation number of oxygen : 0 -2 reduction (gain of electrons)

  33. Student Expectation (SE) 10-I The student defines pH and uses the hydrogen or hydroxide ion concentrations to calculate the pH of a solution.

  34. INDEX CARD TIME! TITLE: pH of a solution FRONT: - Define pH - Equation BACK: - Calculate the pH of a solution from the hydrogen ion concentration [H+]. - Calculate the pH of a solution from the hydroxide ion concentration [OH-].

  35. Mini-Review • In water [ H+] = [OH- ] • In an acidic solution [ H+] > [OH-] • In a basic solution [OH-] > [ H+] pH : indicates the concentration of hydrogen ions [ H+] in aqueous solution. In aqueous solutions, pH ranges from 0 (most acidic) to 7 (neutral) to 14 (most basic). EXAMPLES. Find the pH of a solution if [ H+] = 2.9 x 10-3 M ans: pH = -log (2.9 x 10-3 ) = 12.54 →This solution is basic. Find the pH of a solution if [ OH-] = 1.3 x 10-10 M ans: pOH = - log (1.3 x 10-10 ) = 9.9 therefore pH = 14-9.9 = 4.1 → This solution is acidic. pH = -log [ H+] You also need to know pOH = -log [OH-] and pH + pOH = 14

  36. Student Expectation (SE) 10- J The student distinguishes between degrees of dissociation for strong and weak acids and bases

  37. INDEX CARD TIME! TITLE: Degrees of Dissociation for Acids and bases. FRONT:-Define strong acid or base. Give an example. - Define weak acid or base. Give an example. BACK: Dissociation constant: Define and give examples

  38. Mini-Review Acids: Concentration vs. Strength CONCENTRATED DILUTE H+ A- H+ A- H+ A- H+ A-HA A- H+ A- H+ A- H+ A- H+ A - H+ A- HA H+ A- H+ A- H+ A- A- H+ A- H+ A- H+ A- H+ A- H+ H+ A - H + A - H + A -HA H + A - A- H+ A- H+ A- H+ A- H+ A– H+ A- H+ A- H+ A- H+ A- H+ A- H+ A- H+ A- H+ A- H+ A- HA A- H+ A- H+ A- H+ A- H+ H+ A - H+ A - HA A - H+ A - H+ A – H+ A - H+ A - H+ A - HA H+ A - H+ A - H+ A - H+ STRONG ACIDS Dissociate nearly 100% HA H1+ + A- example: HCl WEAK ACIDS Dissociate very little HA H1+ + A- example : CH3COOH STRONG HA HA H+ A- HA HA HA HA HA HA HA HA H+ A- H+ A- HA HA HA HA HA HA HA H+ A- HA HA HA HA HA HA H+A- HA HA H+ A- HA HA HA HA HA HA HA HA H+A- HA HA H+ A- HA HA HA HA HA HA HA H+ A- HA HA HA HA HA HA HA HA HA HA H+ A - HA HA HA HA H + A – HA H + A – HA HA WEAK

  39. Strong vs. Weak Acid Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 508

  40. Mini-Review How can we distinguish between the degrees of dissociation for weak acids and bases ? The degree of dissociation of a weak acid or base in water is represented by their dissociation constant ( Ka or K b ). Stronger acids and bases will have constants (Ka or K b ) that are closer to, or greater than a value of 1 than will weaker acids and bases. The greater the Ka orKb, the more dissociated the acid or base, the stronger they are. Which would be the best conductor of electricity? HNO2 with a Ka = 4.5 x 10-4 or H2SO3 with a Ka= 1.3 x 10-2 ans: H2SO3 = highest Ka = more dissociated = better electrolyte Remember to interpret the scientific notation correctly : 10 -1 ( 0.1 ) is greater than 10-2 (0.01) EXAMPLES. Which would be the weakest base ? NH3 with a Kb= .8 x 10-5 or N2H4 with a Kb = 1.7 x 10-6 ans: N2H4 = lowest Kb = less dissociated

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