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Chapter 4: Types of Chemical Reactions and Solution Stoichiometry

Chapter 4: Types of Chemical Reactions and Solution Stoichiometry. Mr. Redmond. Aqueous Solutions. dissolved in water Properties: Electrical Conductivity Strong Electrolytes- conducts well Weak Electrolytes- conducts little Nonelectrolytes- does not conduct. Strong Electrolyte.

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Chapter 4: Types of Chemical Reactions and Solution Stoichiometry

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  1. Chapter 4: Types of Chemical Reactions and Solution Stoichiometry • Mr. Redmond

  2. Aqueous Solutions • dissolved in water • Properties: Electrical Conductivity • Strong Electrolytes- conducts well • Weak Electrolytes- conducts little • Nonelectrolytes- does not conduct

  3. Strong Electrolyte • Substances that completely ionize when dissolved in water • Types: • Soluble Salts • Strong Acids • Strong Bases

  4. Soluble Salts • NaCl dissolved in water • Practically all of the sample becomes Na+ and Cl- ions

  5. Acids (Sour) • Arrhenius states: an acid is a substance that produces H+ ions (protons) when it is dissolved in water • HCl, HNO3, and H2SO4 all virtually ionize when placed in water, therefore are considered strong acids.

  6. Bases (Bitter) • When dissolved in water it produces OH-, hydroxide. • Strong bases: NaOH, KOH

  7. Weak Electrolytes • When placed in water they produce few ions. • Weak Acid • Weak Base

  8. Weak Acid • Acetic Acid, dissociates only to a slight extent in an aqueous solution. Only about 1% of its molecules ionize. • HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

  9. Weak Base • Ammonia solution is basic because it produces some OH- ions • NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

  10. Nonelectrolytes • Substances that dissolve but do not dissolve into ions that can conduct electricity • Ethanol, C2H5OH and sucrose, C12H22O11

  11. Molarity • Molarity= moles of solute per volume of solution in liters • M = Molarity = (moles of solute)/(liters of solution) • Solute- a substance dissolved in a liquid • Solution- a homogeneous mixture

  12. Sample Ex. 4.2 • You dissolve 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution. Find the molarity.

  13. Sample Ex. 4.2 cont. • First we calculate the number of moles of HCl...... • Change volume from mL to L....... • Now divide and find Molarity.......

  14. Sample Ex. 4.4 • Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2 • Formula: ZnCl2(s) Zn2+(aq) + 2Cl-

  15. Sample Ex. 4.4 cont. • You have 1.0 x 10-3 M ZnCl2 • After it dissociates you have 1.0 x 10-3 M Zn2+ and 2.0 x 10-3 M Cl- • Use the known information to calculate moles of Cl-

  16. Sample Ex. 4.5 • Typical blood serum is about 0.14 M NaCl. What volume of blood serum contains 1.0 mg NaCl?

  17. Sample Ex. 4.5 cont. • How many moles of NaCl are in 1 mg NaCl? • What volume of 0.14 M NaCl solution contains the number of moles of NaCl in 1 mg NaCl?

  18. Sample Ex. 4.6 • To analyze the alcohol content of a certain wine, a chemist needs 1.0 L of 0.200 M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be weighed out to make this solution?

  19. Sample Ex. 4.6 cont. • Determine the moles of potassium dichromate required. • Convert moles into grams

  20. Dilutions • Most times when you purchase chemicals for a lab they come in a concentrated form call stock solutions. • Water is added to these to give the desired molarity of the particular solution

  21. Dilution • Important to understand that: moles of solute after dilution = moles of solute before dilution • For example: if we need 500 mL of 1.00 M acetic acid from a 17.4 M stock solution of acetic acid, we would need to dilute it.

  22. Dilution • First, determine the number of moles of acetic acid in the final solution...... • We then use the volume of 17.4 M acetic acid that contains the calculated amount of moles • We then use this information to solve for the volume

  23. Sample Ex. 4.7 • What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution?

  24. Sample Ex. 4.7 • Determine moles of H2SO4 in 1.5 L of a 0.10 M H2SO4 • Find the volume of 16 M H2SO4 that contains 0.15 mol H2SO4

  25. Dilution • The equation: M1V1 = M2V2 , may help you remember that the moles before dilution is equal to the moles after dilution • Repeat the last example using this equation

  26. Dilution • M1V1 = M2V2 • M1=16M, M2= 0.10M, V2=1.5 L • Solve for V1

  27. Types of Reactions • Precipitation reactions • Acid-base reaction • Oxidation-reduction reaction

  28. Precipitation Reactions • Sometimes when two solutions are mixed an insoluble substance is formed, this is called a precipitation reaction • The solid substance that separates from the solution is called the precipitate

  29. Precipitation Reactions • Say we perform a precipitation reaction with an aqueous solution of potassium chromate K2CrO4(aq), which is yellow, and mix it with a colorless aqueous solution containing barium nitrate Ba2(NO3)2(aq) • When these are mixed a yellow precipitate forms

  30. Precipitation Reactions What is the yellow solid forming?

  31. Precipitation Reactions • Remember in virtually every case, when a solid containing ions dissolves in water, the ions separate • From this we know that we have K+, CrO42-, Ba2+, and NO3-

  32. Precipitation Reactions • The reaction is: K2CrO4(aq) + Ba(NO3)2(aq) product • It may be easier to see what is happening if we write the equation a little different 2K+(aq)+CrO42-(aq)+Ba2+(aq)+2NO3-(aq) product • What are the different ways these ions could combined?

  33. Precipitation Reactions • Here are our choices: K2CrO4, KNO3, BaCrO4, or Ba(NO3)2 • We can quickly exclude K2CrO4 and Ba(NO3)2 since they are the reactants • It takes a little background knowledge to figure the rest out

  34. Precipitation Reactions • K+ ion and the NO3- ion are both colorless, thus KNO3 would precipitate white. • CrO42- ion is yellow as noted earlier thus the yellow solid must be BaCrO4 • But what happened to K+ and NO3-?

  35. Precipitation Reactions • The other ions are left dissolved in the solution. If we removed our precipitate and then dissolved the water we would find KNO3(s) left in the beaker

  36. Precipitation Reactions

  37. Sample Ex. 4.8 • Use the table just given to solve the problems. Predict what will happen when you mix the following pairs. • KNO3(aq) and BaCl2(aq) • Na2SO4(aq) and Pb(NO3)2(aq) • KOH(aq) and Fe(NO3)3(aq)

  38. Sample Ex. 4.8 • Split up the ions and pair them. The table show that KCl and Ba(NO3)2 are soluble in water, thus no precipitate forms. • NaNO3 is soluble but PbSO4 is not. • K+ and NO3- salts are soluble but Fe(OH)3 is only slightly soluble, thus it will precipitate

  39. Describing Reactions in Solutions • formula equation • complete ionic equation • net ionic equation

  40. Formula Equation • The typical equation we normally write • K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq)

  41. Complete Ionic Equation • In a complete ionic equation, all the substances that are strong electrolytes are represented as ions • 2K+(aq)+CrO42-(aq)+Ba2+(aq)+2NO3-(aq)BaCrO4(s) + 2K+ + 2NO3-

  42. Net Ionic Equation • A net ionic equation only includes those solution components directly involved in the reaction. We do not write the spectator ions (K+ and NO3-) • Ba2+(aq) + CrO42-(aq) BaCrO4(s)

  43. Sample Ex. 4.9 • For the following reaction, write the formula equation, the complete ionic equation, and the net ionic equation. • Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate

  44. Stoichiometry of Precipitation Reaction • Just like in Chapter 3, we first convert all quantities into moles • Second, we form the proper molar ratio using the coefficients of the balanced equation • Use the limiting reactant when necessary

  45. Stoichiometry of Precipitation Reaction • Special attention needs to be given to what reaction actually occurs. Use the complete ionic formula and your background knowledge of the reactants to determine the reaction • We must use the molarity and volume to determine the moles of reactant

  46. Sample Ex. 4.10 • Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.

  47. Sample Ex. 4.10 • Write the complete ionic equation • Refer to the solubility table • Determine what the spectator ions are and what compound is the precipitate • Convert molarity and volume to moles • Use the proper molar ratio then convert to grams

  48. Acid-Base Reactions • Arehenius’ concept of acids and bases • Acids produce H+ ions when dissolved in water • Bases produce OH- ions when dissolved in water • Not all acids or bases have H+ or OH- so a more general definition is needed

  49. Acid-Base Reactions • Bronsted-Lowery Model • Acids are proton donors • Bases are proton acceptors • How do we predict acid-base reactions?

  50. Acid-Base Reactions • Take the example of mixing solutions of HCl and NaOH. We know that HCl is a strong acid and NaOH is a strong base. • Write the ions that are present and look at the solubility table, will NaCl precipitate? • Because water is a nonelectrolyte, large quantities of H+ and OH- ions cannot coexist in the solution. What happens?

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