The Composition of Solutions Copyright © Houghton Mifflin Company. All rights reserved.4–24–2 Which of the following solutions contains the greatest.

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Presentation transcript:

The Composition of Solutions

Copyright © Houghton Mifflin Company. All rights reserved.4–24–2 Which of the following solutions contains the greatest number of ions? (Assume complete solubility for all salts.) –One mole of potassium chloride dissolved in 1.0 L of water. –One mole of sodium phosphate dissolved in 1.0 L of water. –One mole of iron(II) nitrate dissolved in 1.0 L of water. –One mole of sodium hydroxide dissolved in 1.0 L of water. –At least two of the above solutions have an equally great number of ions. React 2

Copyright © Houghton Mifflin Company. All rights reserved.4–34–3 Let’s Think About It Draw molecular level pictures showing each solution. Think about relative numbers of ions. How many moles of each ion are in each solution? React 2

Copyright © Houghton Mifflin Company. All rights reserved.4–44–4 Molarity Molarity (M) = moles of solute per volume of solution in liters:

Copyright © Houghton Mifflin Company. All rights reserved.4–54–5 Which of the following solutions contains the greatest number of ions? a) mL of 0.10 M NaCl. b) mL of 0.10 M CaCl 2. c) mL of 0.10 M FeCl 3. d) mL of 0.10 M sucrose. React 3

Copyright © Houghton Mifflin Company. All rights reserved.4–64–6 Let’s Think About It Draw molecular level pictures showing each solution. Think about relative numbers of ions. How many moles of each ion are in each solution? React 3

Copyright © Houghton Mifflin Company. All rights reserved.4–74–7 Notice The solution with the greatest number of ions is not necessarily the one in which: –the volume of the solution is the largest. –the formula unit has the greatest number of ions.

Precipitation Reactions

Copyright © Houghton Mifflin Company. All rights reserved.4–94–9 The Reaction of K 2 CrO 4 (aq) and Ba(NO 3 ) 2 (aq)

Copyright © Houghton Mifflin Company. All rights reserved.4–10 Simple Rules for Solubility 1.Most nitrate (NO 3  ) salts are soluble. 2.Most alkali (group 1A) salts and NH 4 + are soluble. 3.Most Cl , Br , and I  salts are soluble (NOT Ag +, Pb 2 +, Hg 22 + ) 4.Most sulfate salts are soluble (NOT BaSO 4, PbSO 4, HgSO 4, CaSO 4 ) 5.Most OH  salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH) 2, Ca(OH) 2 are marginally soluble) 6.Most S 2 , CO 3 2 , CrO 4 2 , PO 4 3   salts are only slightly soluble.

Copyright © Houghton Mifflin Company. All rights reserved.4–11 Describing Reactions in Solution 1.Molecular equation (reactants and products as compounds) –AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) 2.Complete ionic equation (all strong electrolytes shown as ions) Ag + (aq) + NO 3  (aq) + Na + (aq) + Cl  (aq)  AgCl(s) + Na + (aq) + NO 3  (aq)

Copyright © Houghton Mifflin Company. All rights reserved.4–12 Describing Reactions in Solution (continued) 3.Net ionic equation (show only components that actually react) Ag + (aq) + Cl  (aq)  AgCl(s) Na + and NO 3  are spectator ions.

Copyright © Houghton Mifflin Company. All rights reserved.4–13 You have two separate beakers with aqueous solutions, one with 4 “units” of potassium sulfate and one with 3 “units” of barium nitrate. a)Draw molecular level diagrams of both solutions. React 4

Copyright © Houghton Mifflin Company. All rights reserved.4–14 You have two separate beakers with aqueous solutions, one with 4 “units” of potassium sulfate and one with 3 “units” of barium nitrate. b)Draw a molecular level diagram of the mixture of the two solutions before a reaction has taken place. React 4

Copyright © Houghton Mifflin Company. All rights reserved.4–15 You have two separate beakers with aqueous solutions, one with 4 “units” of potassium sulfate and one with 3 “units” of barium nitrate. c)Draw a molecular level diagram of the product and solution formed after the reaction has taken place. React 4

Stoichiometry of Precipitation Reactions

Copyright © Houghton Mifflin Company. All rights reserved.4–17 a)Draw two pictures, one of solution A and one of solution B. Make sure your pictures show the relative volumes of solutions and the relative numbers of ions. Solution A: 2.00 L 2.00 M AgNO 3 (aq) React 5 Solution B: 3.00 L 1.00 M Na 2 CrO 4 (aq)

Copyright © Houghton Mifflin Company. All rights reserved.4–18 b)Draw a picture of solution C (made by adding solution A to solution B) before any reaction has taken place. Make sure this picture shows the relative volume of solution C and the relative number of ions, as compared with solutions A and B. React 5

Copyright © Houghton Mifflin Company. All rights reserved.4–19 c)Draw a picture of solution C after the reaction has taken place. Make sure this picture shows the relative volume of solution C and the relative number of ions, as compared with solutions A and B. React 5

Copyright © Houghton Mifflin Company. All rights reserved.4–20 d)Calculate the concentrations (in M) of all ions in solution C. React 5

Copyright © Houghton Mifflin Company. All rights reserved.4–21 Let’s Think About It To solve this problem we need to know: –the initial concentration of all ions in solution C before a reaction has taken place. –the balanced equation for the reaction that has taken place. In this case, it makes sense to use the net ionic equation. –which reactant, if either, is limiting. React 5

Copyright © Houghton Mifflin Company. All rights reserved.4–22 Calculating an Answer How many moles of each ion are present in solution? What is the total volume of the solution? What is the definition of molarity?

Copyright © Houghton Mifflin Company. All rights reserved.4–23 You have two mL aqueous solutions. Solution A is a solution of silver nitrate and solution B is a solution of potassium chromate (K 2 CrO 4 ). The masses of the solutes in each of the solutions are the same. When the solutions are added together, a blood red precipitant is formed. After the reaction has gone to completion, you dry the solid and find that it has a mass of grams. a)Calculate the concentration (molarity) of the original potassium chromate solution. React 6

Copyright © Houghton Mifflin Company. All rights reserved.4–24 You have two mL aqueous solutions. Solution A is a solution of silver nitrate and solution B is a solution of potassium chromate (K 2 CrO 4 ). The masses of the solutes in each of the solutions are the same. When the solutions are added together, a blood red precipitant is formed. After the reaction has gone to completion, you dry the solid and find that it has a mass of grams. b)Calculate the concentration (molarity) of the chromate ions in the final solution. React 6

Copyright © Houghton Mifflin Company. All rights reserved.4–25 Let’s Think About It To solve this problem we need to know: –the initial concentrations of all ions in the final solution before a reaction has taken place. –the balanced equation for the reaction that has taken place. In this case, it makes sense to use the net ionic equation. –which reactant, if either, is limiting. React 6

Copyright © Houghton Mifflin Company. All rights reserved.4–26 Calculating an Answer How many moles of each ion is present in solution? What is the total volume of the solution? What is the definition of molarity?

Acid-Base Reactions

Copyright © Houghton Mifflin Company. All rights reserved.4–28 Key Titration Terms Titrant - solution of known concentration used in titration. Analyte - substance being analyzed. Equivalence point - enough titrant added to react exactly with the analyte. Endpoint - the indicator changes color so you can tell the equivalence point has been reached.

Copyright © Houghton Mifflin Company. All rights reserved.4–29 Performing Calculations for Acid-Base Reactions 1.List initial species and predict reaction. 2.Write balanced net ionic reaction. 3.Calculate moles of reactants. 4.Determine limiting reactant. 5.Calculate moles of required reactant or product. 6.Convert to grams or volume, as required.

Oxidation- Reduction Reactions

Copyright © Houghton Mifflin Company. All rights reserved.4–31 Reaction of Sodium and Chlorine

Copyright © Houghton Mifflin Company. All rights reserved.4–32 Rules for Assigning Oxidation States 1.Oxidation state of an atom in an element = 0 2.Oxidation state of monatomic element = charge 3.Oxygen =  2 in covalent compounds (except in peroxides where it =  1) 4.H = +1 in covalent compounds 5.Fluorine =  1 in compounds 6.Sum of oxidation states = 0 in compounds 7.Sum of oxidation states = charge of the ion

Copyright © Houghton Mifflin Company. All rights reserved.4–33 React 7 Find the oxidation states for each of the elements in each of the following compounds: K 2 Cr 2 O 7 CO 3 2- HClO 4 MnO 2 PCl 5 SF 4

Copyright © Houghton Mifflin Company. All rights reserved.4–34 Make a list of nitrogen compounds with as many different oxidation states for nitrogen as you can. React 8

Copyright © Houghton Mifflin Company. All rights reserved.4–35 Which of the following are oxidation- reduction reactions? Identify the oxidizing agent, and the reducing agent. Zn (s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) Cr 2 O 7 2- (aq) + 2OH - (aq)  2CrO 4 2- (aq) + H 2 O(l) O 3 (g) + NO (g)  O 2 (g) + NO 2 (g) 2CuCl (aq)  CuCl 2 (aq) + Cu (s) React 9

Copyright © Houghton Mifflin Company. All rights reserved.4–36 Balancing Oxidation-Reduction Reactions Cr 2 O 7 2- (aq) + SO 3 - (aq)  Cr 3+ (aq) + SO 4 2- (aq) How can we balance this equation?

Copyright © Houghton Mifflin Company. All rights reserved.4–37 Method of Half Reactions Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 - (aq)  + SO 4 2- (aq) How many electrons are involved in each half reaction?

Copyright © Houghton Mifflin Company. All rights reserved.4–38 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 - (aq)  + SO 4 2- (aq) + 2e - How can we balance the oxygen atoms? Method of Half Reactions (continued)

Copyright © Houghton Mifflin Company. All rights reserved.4–39 6e - + Cr 2 O 7 2- (aq)  Cr 3+ (aq) + 7H 2 O H 2 O +SO 3 - (aq)  + SO 4 2- (aq) + 2e - How can we balance the hydrogen atoms? Method of Half Reactions (continued)

Copyright © Houghton Mifflin Company. All rights reserved.4–40 This reaction occurs in an acidic solution. 14H + + 6e - + Cr 2 O 7 2- (aq)  Cr 3+ (aq) + 7H 2 O H 2 O +SO 3 - (aq)  SO 4 2- (aq) + 2e - + 2H + How can we balance the electrons? Method of Half Reactions (continued)

Copyright © Houghton Mifflin Company. All rights reserved.4–41 Method of Half Reactions (continued) 14H + + 6e - + Cr 2 O 7 2- (aq)  Cr 3+ (aq) + 7H 2 O 3[H 2 O +SO 3 - (aq)  SO 4 2- (aq) + 2e - + 2H + ] Cr 2 O 7 2- (aq) + 3SO 3 - (aq) + 8H + (aq)  2Cr 3+ (aq) + 3SO 4 2- (aq) + 4H 2 O(l)

Copyright © Houghton Mifflin Company. All rights reserved.4–42 Half-Reaction Method - Balancing in Base 1.Balance as in acid. 2.Add OH  that equals H + ions (both sides!) 3.Form water by combining H +, OH . 4.Check elements and charges for balance.

Copyright © Houghton Mifflin Company. All rights reserved.4–43 Balance the following oxidation-reduction reactions that occur in acidic solution. ClO - (aq) + I - (aq)  Cl - (aq) + I 3 - (aq) Br - (aq) + MnO 4 - (aq)  Br 2 (l)+ Mn 2+ (aq) CH 3 OH(aq) + Cr 2 O 7 2- (aq)  CH 2 O(aq) + Cr 3+ (aq) React 10