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Stoichiometry & Reactions

Stoichiometry & Reactions. {4.1-Properties of Solutions. Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes. Solutions. Dissociation.

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Stoichiometry & Reactions

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  1. Stoichiometry & Reactions {4.1-Properties of Solutions

  2. Solutionsare defined as homogeneous mixtures of two or more pure substances. • The solvent is present in greatest abundance. • All other substances are solutes. Solutions

  3. Dissociation • When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them. • This process is called dissociation.

  4. Dissociation • An electrolyte is a substances that dissociates into ions when dissolved in water.

  5. Solutions • An electrolyte is a substance that dissociates into ions when dissolved in water. • Anonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

  6. Electrolytes & Nonelectrolytes Soluble ionic compounds tend to be electrolytes.

  7. Electrolytes & Nonelectrolytes Molecular compounds tend to be nonelectrolytes, except for acids and bases.

  8. Electrolytes • A strong electrolyte dissociates completely when dissolved in water. • A weak electrolyte only dissociates partially when dissolved in water.

  9. Electrolytes • A strong electrolyte dissociates completely when dissolved in water. • A weak electrolyte only dissociates partially when dissolved in water.

  10. Strong electrolytes are... • Strong acids • Strong bases

  11. Strong electrolytes are... • Strong acids • Strong bases • Soluble ionic salts

  12. Concept Check 4.1 Relating relative numbers of anions and cations to chemical formulas PROBLEM 4.1a: The accompanying diagram represents an aqueous solution of either MgCl2, KCl, or K2SO4. Which solution does the drawing best represent? Analyze & Plan:We are asked to associate the charged spheres in the diagram with ions present in a solution of an ionic substance. We examine each ionic substance given to determine the relative numbers and charges of its ions .We then correlate these ionic species with the ones shown in the diagram. Solve The diagram shows twice as many cations as anions, consistent with the formulation K2SO4.

  13. Concept Check 4.1 Relating relative numbers of anions and cations to chemical formulas PROBLEM 4.1b: If you were to draw diagrams representing aqueous solutions of (a) NiSO4, (b)Ca(NO3)2, (c) Na3PO4, (d) Al2(SO4)3, how many anions would you show if each diagram contained six cations? Answer:(a) 6, (b) 12, (c) 2, (d) 9

  14. Concept Check 4.2 Identifying strong, weak, and nonelectrolytes PROBLEM 4.2a: Classify these dissolved substances as strong, weak, or nonelectrolyte: CaCl2, HNO3, C2H5OH (ethanol), HCOOH (formic acid), KOH. Analyze & Plan: We are given several chemical formulas and asked to classify each substance as a strong electrolyte, weak electrolyte, or nonelectrolyte. We can predict whether a substance is ionic or molecular based on its composition. Most ionic compounds we encounter in this text are composed of a metal and a nonmetal, whereas most molecular compounds are composed only of nonmetals.

  15. Concept Check 4.2 Identifying strong, weak, and nonelectrolytes PROBLEM 4.2a: Classify these dissolved substances as strong, weak, or nonelectrolyte: CaCl2, HNO3, C2H5OH (ethanol), HCOOH (formic acid), KOH. Solve Two compounds fit the criteria for ionic compounds: CaCl2 and KOH. Because Table 4.3 tells us that all ionic compounds are strong electrolytes, that is how we classify these two substances. The three remaining compounds are molecular. Two, HNO3 and HCOOH, are acids. Nitric acid, HNO3, is a common strong acid, as shown in Table 4.2, and therefore is a strong electrolyte. Because most acids are weak acids, our best guess would be that HCOOH is a weak acid (weak electrolyte). This is correct. The remaining molecular compound, C2H5OH, is neither an acid nor a base, so it is a nonelectrolyte.

  16. Concept Check 4.2 Identifying strong, weak, and nonelectrolytes PROBLEM 4.2b: Consider solutions in which 0.1 mol of each of the following compounds is dissolved in 1 L of water: Ca(NO3)2 (calcium nitrate), C6H12O6 (glucose), NaCH3COO (sodium acetate), and CH3COOH (acetic acid). Rank the solutions in order of increasing electrical conductivity, based on the fact that the greater the number of ions in solution, the greater the conductivity. Answer: C6H12O6 (nonelectrolyte) < CH3COOH (weak electrolyte, existing mainly in the form of molecules with few ions) < NaCH3COO (strong electrolyte that provides two ions, Na+ and CH3COO–) < Ca(NO3)2(strong electrolyte that provides three ions, Ca2+ and 2 NO3–)

  17. Stoichiometry & Reactions {4.5-Concentrations of Solutions

  18. moles of solute Molarity (M) = volume of solution in liters Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution:

  19. Mixing a Solution • To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute. • The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

  20. Concept Check 4.3 Calculating Molarity PROBLEM 4.3a: Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution. Answer:

  21. Concept Check 4.3 Calculating Molarity PROBLEM 4.3b: Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL of solution. Answer: 0.278 M

  22. Concept Check 4.4 Calculating Molar Concentrations of Ions PROBLEM 4.4a: What is the molar concentration of each ion present in a 0.025 M aqueous solution of calcium nitrate? Answer:

  23. Concept Check 4.4 Calculating Molar Concentrations of Ions PROBLEM 4.4b: What is the molar concentration of K+ ions in a 0.015 M solution of potassium carbonate? Answer: 0.030 M

  24. Concept Check 4.5a Using molarity to calculate grams of solute PROBLEM 4.5a: How many grams of Na2SO4 are required to make 0.350 L of 0.500 M Na2SO4? Answer:

  25. Concept Check 4.5 Using molarity to calculate grams of solute PROBLEM 4.5b: (a) How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4? (b) How many milliliters of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt? Answer:(a) 1.1 g, (b) 76 mL

  26. Dilution • One can also dilute a more concentrated solution by • Using a pipet to deliver a volume of the solution to a new volumetric flask, and • Adding solvent to the line on the neck of the new flask.

  27. Concept Check 4.6 Preparing a solution by dilution PROBLEM 4.6a: How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4? Answer:

  28. Concept Check 4.6 Preparing a solution by dilution PROBLEM 4.6b: (a) What volume of 2.50 M lead(II) nitrate solution contains 0.0500 mol of Pb2+? (b) How many milliliters of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution? (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting stock solution? Answer: (a) 0.0200 L = 20.0 mL, (b) 5.0 mL, (c) 0.40 M

  29. Stoichiometry & Reactions {4.2-Precipitation Reactions

  30. Precipitation Reactions • Reactions that result in the formation of an insoluble product; occur when pairs of oppositely charged ions attract so strongly they form an insoluble ionic solid precipitate-insoluble solid formed by a reaction in solution

  31. Solubility Guidelines • Solubility-amount of substance that can be dissolved in a given quantity of solvent at a given temperature insoluble-any substance with solubility < 0.01 mol/L (attraction of ions in solid too strong for water molecules to separate significantly and thus remain undissolved)

  32. Solubility Guidelines • Predicting precipitate formation: • Identify ions present in the reactants • Consider product formations (possible cation/anion combinations) • Use solubility rules to determine if any of the combinations are insoluble

  33. Concept Check 4.7 Using solubility rules PROBLEM 4.7a: Classify these ionic compounds as soluble or insoluble in water: (a) sodium carbonate, Na2CO3, (b) lead sulfate, PbSO4. Answer: (a) According to Table 4.1, most carbonates are insoluble. But carbonates of the alkali metal cations (such as sodium ion) are an exception to this rule and are soluble. Thus, Na2CO3 is soluble in water. (b) Table 4.1 indicates that although most sulfates are water soluble, the sulfate of is an exception. Thus, PbSO4is insoluble in water.

  34. Concept Check 4.7 Using solubility rules PROBLEM 4.7b: Classify the following compounds as soluble or insoluble in water: (a) cobalt(II) hydroxide, (b) barium nitrate, (c) ammonium phosphate. Answer: (a)insoluble, (b) soluble, (c)soluble

  35. Metathesis (Exchange) Reactions AgNO3(aq) + KCl(aq) AgCl(s) + KNO3(aq) AX + BY  AY + BX Metathesis comes from a Greek word that means “to transpose.” Also called “double displacement/replacement” reactions It appears as though the ions in the reactant compounds exchange, or transpose, ions:

  36. Concept Check 4.8 Predicting a metathesis reaction PROBLEM 4.8a: (a) Predict the identity of the precipitate that forms when aqueous solutions of BaCl2 and K2SO4 are mixed. (b) Write the balanced chemical equation for the reaction. Answer: (a) The reactants contain Ba2+, Cl–, K+, and SO42– ions. Exchanging the anions gives us BaSO4 and KCl. According to Table 4.1, most compounds of SO42– are soluble but those of Ba2+ are not. Thus, BaSO4 is insoluble and will precipitate from solution. KCl is soluble. (b) From part (a) we know the chemical formulas of the products, BaSO4 and KCl. The balanced equation is BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2 KCl(aq)

  37. Concept Check 4.8 Predicting a metathesis reaction PROBLEM 4.8b: (a) What compound precipitates when aqueous solutions of Fe2(SO4)3 and LiOH are mixed? (b) Write a balanced equation for the reaction. (c) Will a precipitate form when solutions of Ba(NO3)2 and KOH are mixed? Answer: (a) Fe(OH)3 (b) Fe2(SO4)3(aq) + 6 LiOH(aq)  2 Fe(OH)3(s) + 3 Li2SO4(aq) (c)NR (both possible products, Ba(OH)2 and KNO3, are water soluble)

  38. Solution Chemistry • It is helpful to pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution). • If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction.

  39. Molecular Equation The molecular equation lists the reactants and products in their molecular form, even though they may actually exist in solution as ions: • PRO-explicit about what reactant solutions are and what products you obtain • PRO-closely describes what you actually do in the laboratory or in an industrial process • CON- does not tell you what is happening at the level of ions (i.e., it does not give you an ionic-theory interpretation of the reaction) Ca(OH)2(aq)+ Na2CO3(aq)CaCO3(s)+ 2NaOH(aq)

  40. Ionic Equation In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. • represent other reactants and products by the formulas of the compounds, indicating any soluble substance by (aq) after its formula and any insoluble solid substance by (s) after its formula Ca2+(aq) + 2OH-(aq) + 2Na+(aq) + CO32-(aq) CaCO3(s) + 2Na+(aq)+ 2OH-(aq)

  41. Ionic Equation Ca2+(aq) + 2OH-(aq) + 2Na+(aq) + CO32-(aq) CaCO3(s) + 2Na+(aq)+ 2OH-(aq) • PRO-represents each substance by its predominant form in the reaction mixture (indicates which species exist as ions in solution-strong electrolytes, which species exist primarily as molecules in solution-weak electrolytes, and which are insoluble species-precipitates represented by the formula of the compound) • CON-contains ions that do not take part in the reaction (appear on both reactant and product sides of the equation)

  42. Net Ionic Equation The net ionic equation includes only ions and molecules directly involved in a reaction (spectator ions are canceled) • spectator ions- an ion in an ionic equation that does not take part in the reaction; appear on both the reactant and product sides of the equation and can be canceled from both sides to express the essential reaction that occurs Ca2+(aq) + 2OH-(aq) + 2Na+(aq) + CO32-(aq) CaCO3(s) + 2Na+(s) + 2OH-(aq) Ca2+(aq) + CO32-(aq) CaCO3(s)

  43. Net Ionic Equation • PRO-generality of the net ionic equation; from the net ionic equation, you can see that mixing any solution of calcium ion with any solution of carbonate ion will give this same reaction

  44. Writing Net Ionic Equation • Write a balanced molecular equation. • Dissociate all strong electrolytes. • Cross out anything that remains unchanged from the left side to the right side of the equation. • Write the net ionic equation with the species that remain. • Comment If none of the ions in an ionic equation is removed from solution or changed in some way, all ions are spectator ions and a reaction does not occur (NR).

  45. Net Ionic Equation e.g., Calcium nitrate reacts with potassium carbonate Ca(NO3)2(aq) + K2CO3(aq) CaCO3(s) + 2KNO3(aq) Ca2+(aq) + 2NO3-(aq) + 2K+(aq) + CO32-(aq) CaCO3(s) + 2K+(aq) + 2NO3-(aq) Ca2+(aq) + CO32-(aq) CaCO3(s)

  46. Concept Check 4.9 Writing a net ionic equation PROBLEM 4.9a: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of calcium chloride and sodium carbonate are mixed. Answer: CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2 NaCl(aq) Ca2+(aq) + 2 Cl–(aq) + 2 Na+(aq) + CO32–(aq)  CaCO3(s) + 2 Na+(aq) + 2 Cl–(aq) Ca2+(aq) + CO32–(aq)  CaCO3(s)

  47. Concept Check 4.9 Writing a net ionic equation PROBLEM 4.9b: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of silver nitrate and potassium phosphate are mixed. Answer: 3 Ag+(aq) + PO43–(aq)  Ag3PO4(s)

  48. Stoichiometry & Reactions {4.3-Acids, Bases, and Neutralization Reactions

  49. Properties of Acids & Bases • Acids and bases are some of the most important electrolytes. • Common property of acid is a sour taste; of a base is a bitter taste and “soapy” feel • Both have the ability to cause color changes in certain dyes

  50. Properties of Acids & Bases • acid-base indicator-a dye used to distinguish between acidic and basic solutions by means of the color changes it undergoes in these solutions. • e.g., tea-lightens its amber color when lemon juice (citric acid) is added; red cabbage juice-changes to green then yellow when a base is added and back to red when an acid is added; litmus-turns red in acidic solutions and blue in basic solution; and phenolphthalein-colorless in acidic solution and pink in basic solution

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