# ch 13 15 pp

Information about ch 13 15 pp

Published on January 3, 2008

Author: bruce

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Slide1:  Complete and Net Ionic Equations Dissociation: When ionic compounds dissolve, water will break up the compounds into ______ (this process is called dissociation. (fig. 1 pg. 435) Sample A pg. 436 Al2(SO4)3 (aq) → 2 Al+3 (aq) + 3 SO4-2 (aq) Ionization: When water can separate elements of a covalent compound (particularly acids) to produce _____. See hydrochloric acid (HCl pg. 441). ions ions HCl (aq) → H+ (aq) + Cl- (aq) Complete and Net Ionic Equations: pg. 439 Single displacement and double displacement neutralization reactions involve ionic compounds dissolved in __________. The driving force of reactions is the formation of a ________, a ________ or ________ (water is produced from a double displacement reaction between an acid and a base called a neutralization reaction). 2. These reactions can be written showing _____ of the ions involved in the reaction (called the complete ionic equation). 3. The complete ionic equation can be observed for ions that do not participate in the reaction called ______________ ions. Elimination of these spectator ions leaves what is called the net ionic equation. 4. We will utilize the solubility rules for ionic compounds given in the reaction unit. water solid gas water all spectator Slide2:  Sample B pg. 440 Zn(NO3)2 (aq) + (NH4)2S (aq)→ ZnS (s) + NH4NO3 (aq) 2 Complete ionic equation Zn+2 (aq) + 2 NO3-1 (aq) + 2 NH4+1 (aq) + S-2 (aq)→ ZnS (s) + 2 NH4+ (aq) + 2 NO3-1 (aq) Net ionic equation Zn+2 (aq) + S-2 (aq)→ ZnS (s) Strong vs. Weak Electrolytes: pg. 442-443 Soluble ionic compounds are _________ electrolytes (compounds which conduct electricity well when dissolved in water). 2. Strong acids are _________ electrolytes (acids which ionize 100%). Strong acids are HCl; HNO3; HBr; HClO4 and H2SO4 into (H+ and HSO4-) 3. To obtain a complete ionic equation only ___________ electrolytes will be broken up into ions. Weak electrolytes will include insoluble ionic compounds and _________ acids (these compounds will not be broken into ions). Weak acids include all acids other than those listed as strong acids. 5. A scientist named Svente Arrhenius classified acids by the ability of equal concentrations to conduct electricity. Strong acids conduct electricity much better than weak acids for _________ concentrations. strong strong strong weak equal The driving force is formation of a solid Slide3:  Neutralization Reactions (a type of double displacement reaction): pg. 487-488 Acids react with bases (hydroxide salts) to form water and a _______ (ionic compound). Fig. 18 pg. 488 Write the complete and net ionic equation for the neutralization reaction of a solution of acetic acid with a solution of sodium hydroxide. HC2H3O2 (aq) + NaOH (aq)→ NaC2H3O2 (aq) + HOH (l) Complete ionic equation HC2H3O2 (aq) + Na+1 (aq) + OH-1 (aq)→ Na+1 (aq) + C2H3O2-1 (aq) + HOH (l) Net ionic equation HC2H3O2 (aq) + OH-1 (aq)→ C2H3O2-1 (aq) + HOH (l) The driving force is formation of water If all ions are spectator ions, then no reaction will occur and there is no net ionic equation. salt Slide4:  Titrations: Titration is a method of determining the _________________ (or amount) of a chemical contained in a solution. A common titration involves the ______________ of an acid and base (called a neutralization reaction). See fig. 8 pg. 516 and 518-519. The long stemmed glassware delivering base into the flask containing the acid is called a ___________. Here base is being added to the acid until exactly _________ of the acid is completely reacted. To determine when all of the acid has completely reacted with the base either an appropriate acid/base indicator (a substance which undergoes a _______ change at a certain pH) or a pH meter is used (plot the pH of the solution as base is added). See Fig. 9 pg. 517. pH is a measurement of the ___________ of the solution. A neutral solution has a pH of 7. As the pH gets lower, the solution becomes more acidic (or as it gets higher it becomes more basic). See fig. 5 pg. 512. The point where exactly all of the acid has reacted is called the __________________ point, if determined by indicator it is called the endpoint (these may not be equal if the indicator is not chosen well, the indicator should change color near the pH at the equivalence point). The substance being analyzed is called the __________, with the substance of known concentration (or amount) is called the titrant (often delivered by buret). The pH of a strong acid/base reaction will be ____ at the equivalence point but when a weak acid is titrated with strong base the pH will be basic (this is due to a base being contained in the salt formed from the reaction-this will be discussed in the next unit). See pg. 513 for selection of the appropriate indicator for different titrations. concentration reaction buret all color acidity equivalence analyte 7 Slide5:  20.0 mL Sample F pg. 520 0.0154 M Ba(OH)2 initial equivalence pt. hydrochloric acid (with indicator) HCl (aq) + Ba(OH)2 (aq) → BaCl2 (aq) + HOH (l) 2 2 27.4 mL Ba(OH)2 . 0.0154 mol Ba(OH)2. 1000 mL Ba(OH)2 ? mol HCl = L HCl 2 mol HCl 1 mol Ba(OH)2 / / 20.0 mL HCl . L HCl 1000 mL HCl / / / / ? mol HCl = L HCl 0.0422 M pH is 7 at the equivalence point for a strong acid titrated with strong base endpoint Slide6:  21.2 mL Practice 1 pg. 521 initial equivalence pt. ? M acetic acid (with indicator) HC2H3O2 (aq) + KOH (aq) → KC2H3O2 (aq) + HOH (l) 15.5 mL KOH . 0.215 mol KOH. 1000 mL KOH ? mol HC2H3O2 = L HC2H3O2 1mol HC2H3O2 1 mol KOH / / 21.2 mL HC2H3O2 . L HC2H3O2 1000 mL HC2H3O2 / / / / ? mol HC2H3O2 = L HC2H3O2 0.157 M 15.5 mL 0.215 M KOH 5.0mL pH is above 7 at the equivalence point for a weak acid titrated with strong base endpoint Slide7:  17.6 mL Practice 2 pg. 521 initial equivalence pt. ? M H2SO4 (with indicator) H2SO4 (aq) + LiOH (aq) → Li2SO4 (aq) + HOH (l) 27.4 mL LiOH . 0.0165 mol LiOH. 1000 mL LiOH ? mol H2SO4 = L H2SO4 1mol H2SO4 2 mol KOH / / 17.6 mL H2SO4 . L H2SO4 1000 mL H H2SO4 / / / / ? mol H2SO4 = L H2SO4 0.0128 M 0.0165 M LiOH 2 2 Endpoint Slide8:  Solution Stoichiometry What volume of 6.0 M H2SO4 is required to neutralize 5.60 g of NaOH? What volume 6.0 M H2SO4 is required to neutralize 55.0 mL of 3.0 M NaOH? ?mL H2SO4 = 5.60 g NaOH . H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 HOH (l) ionizes partially mol NaOH . 40.00 g NaOH / / 1 mol H2SO4 . 2 mol NaOH 1000 mL = 6.0 mol H2SO4 / / / / 12 mL ?mL H2SO4 = 55.0 mL NaOH . 3.0 mol NaOH . 1000 mL NaOH / / 1 mol H2SO4 . 2 mol NaOH / / 1000 mL = 6.0 mol H2SO4 / / 14 mL Slide9:  Acids and Bases: 1. The Latin word acidus means ________ (acids taste sour). Bases taste bitter. 2. Common acids are juices in fruit, vinegar is _____ acetic acid (also called ethanoic acid), phosphoric acid in soft drinks and stomach acid. Common bases are ammonia (_______ used as a household cleaner ), antacid tablets (aluminum hydroxide, magnesium hydroxide or sodium bicarbonate) and lye (sodium hydroxide used in oven cleaner and Draino). See fig. 1 pg. 457. 3. See points on pg. 468 for some acid properties that have been described previously (see points on pg. 471 for base properties that have been described previously). Arrhenius Acids and Bases 1. A Swedish chemist named Svante Arrhenius described acids as substances that increased the ____ concentration in aqueous solutions (HCl (aq) is an acid; HCl (g) is not due to being a covalent compound that produces H+ ions ). 2. He described bases as substances that increased the ______ concentration in solution. 3. Arrhenius described the strength of acids by the ability of equal concentrations of differing acids to _________ electrical current. The greater the current for a given concentration, the more ______ produced in solution (thus the greater the amount of H+). Question: Is there ever a case when a weak electrolyte could conduct more electricity than a strong electrolyte? Answer: yes, if the weak electrolyte is much more ___________________ than the strong electrolyte. 4. A strong acid ionizes _______ (Know these: HCl, HBr, HNO3, HClO4 and the first ionization of H2SO4). Example: HCl (aq) → H+ (aq) + Cl- (aq) Initial number before ionization 100 0 0 Number after ionization by water 0 100 100 sour 5% NH3 H+ OH- conduct ions concentrated 100% Slide10:  % ionization = [H+]final 100 (X is some acid anion) [HX]original % ionization = 100 100 (X is some acid anion) 100 % ionization = 100 % 5. Weak acids ionize ____________ (acetic acid or any acid that is not strong is a weak acid for they only partially ionize in water) Example: HX (aq) ↔ H+ (aq) + X- (aq) (↔ shows reaction goes both ways) Initial number before ionization 100 0 0 Number after ionization by water 95 5 5 % ionization = [H+]final 100 (X is some acid anion) [HX]original % ionization = 5 100 (X is some acid anion) 100 % ionization = 5 % Base Strength depends upon the amount of ______ it produces (NaOH or other soluble hydroxide salts are strong bases, NH3 would be an example of a weak base it only partially ionizes). partially OH- Slide11:  Bronsted-Lowry Acids and Bases: 1. An acid is a proton (H+) __________. In aqueous solution an acids donates an H+ to water to produce a ______________ ion (H3O+). HCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq) acid 2. A base is a proton (H+) _________________. NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq) base 3. The stronger the acid, the __________ its conjugate base (the weaker the acid the stronger its conjugate base and the weaker a base, the stronger its conjugate acid). The conjugate base is the substance formed from the acid donating a proton and the conjugate acid is the substance formed from the base accepting a proton. donor hydronium base acceptor acid weaker Side Note:. The pH at the equivalence point of a weak acid/strong base titration is _____________ than 7 at the equivalence point because the salt that is formed is the conjugate base of the weak acid (making it a base that can accept a proton from water to make OH- and raise the pH). HX (aq) + NaOH (aq) ↔ HOH (l) + NaX (aq) acid greater base Slide12:  Application of conjugate acids and bases: Strong Acid/Strong Base Titration: HCl (aq) + OH- (aq) → H2O (l) + Cl- (aq) conjugate conjugate ________ ________ In a titration of a strong base with a strong acid, the solution will have a pH ___ 7 at the equivalence point (due to Cl- being the conjugate base of a very strong acid it will not want to accept protons). Strong Base/Weak Acid Titration: HC2H3O2 (aq) + OH‑ (aq) → H2O (aq) + C2H3O2- (aq) conjugate conjugate ________ ________ In a titration of a weak acid with a strong base, the solution will have a pH _______ 7 at the equivalence point (due to C2H3O2- being the conjugate base of a weak acid it will accept protons from water and form OH-(aq)) . Strong Acid/Weak Base Titration: HCl (aq) + NH3 (aq) → Cl- (aq) + NH4+ (aq) conjugate conjugate ________ ________ In a titration of a weak base with a strong acid, the solution will have a pH _______ 7 at the equivalence point (due to NH4+ being a weak acid as a conjugate of a weak base but Cl- is too weak of a base to accept any H+ since it is the conjugate of a strong acid). We have seen that water may act as an acid or a base making it an ________________ molecule. acid base acid base = acid acid base base above acid base acid base below amphoteric Slide13:  Polyprotic Acids: 1. Some acids contain more than 1 ______________ hydrogens. Example: H2SO4 has ___, H3PO4 has ___ (HC2H3O2 (or CH3COOH) has ___ ionizable hydrogen). 2. Polyprotic acids donate hydrogens __ at a time. H2SO4 (aq) + H2O (l) → HSO4- (aq) + H3O+ (aq) acid base conj. base conj. acid HSO4- (aq) + H2O (l) ↔ SO4-2 (aq) + H3O+ (aq) acid base conj. base conj. acid H2SO4 is the strongest acid and SO4-2 is the strongest base. HSO4- is _______________. H3PO4 (aq) + H2O (l) ↔ H2PO4- (aq) + H3O+ (aq) acid base conj. base conj. acid H2PO4- (aq) + H2O (l) ↔ HPO4-2 (aq) + H3O+ (aq) acid base conj. base conj. acid HPO4-2 (aq) + H2O (l) ↔ PO4-3 (aq) + H3O+ (aq) acid base conj. base conj. acid __________ is the strongest acid and __________ is the strongest base. __________ and ____________ are amphoteric species. ionizable 2 3 1 1 amphoteric H3PO4 PO4-3 H2PO4- HPO4-2 Titration of a diprotic weak acid with a strong base yields 2 inflection zones. Titration curve for a diprotic weak acid (H2C2O4):  Titration of a diprotic weak acid with a strong base yields 2 inflection zones. Titration curve for a diprotic weak acid (H2C2O4) H2C2O4 (aq) + OH- (aq) → HC2O4- (aq) + H2O (l) HC2O4- (aq) + OH- (aq) → C2O4-2 (aq) + H2O (l) Acid Rain: 1. Non-metal oxides form acids when dissolved in _________. water Example: SO3 (aq) + H2O (l) → H2SO4 (aq) SO3 is the acidic _________________ of H2SO4. anhydride 2. Non-metal oxides are formed from the burning of coal, volcanic activity and in hot engine blocks. 3. What is the acidic anhydride of HNO3? N2O5 Slide15:  pH, pOH and Kw: 1. Water dissociates to a ______ degree. H2O (l) ↔ H+ (aq) + OH- (aq) or H2O (l) + H2O (l) ↔ H3O+ (aq) + OH- (aq) 2. The product of H+ and OH- (or H3O+ and OH-) is called the ionization ____________ of water (Kw). Kw = [H+] [OH-] Kw = 1.0.10-14 at 25oC or Kw = [H3O+] [OH-] [ ] means molarity. 3. For a strong acid, the [H+] or [H3O+] after dissociation ____ the original concentration of the acid (because it will completely ionize). Sample A pg. 502: 4. In pure water: [H+] = [OH-] because of this: 1.0.10-14 = [H+] [H+] 1.0.10-14 = [H+]2 ________ [H+] = √1.0.10-14 [H+] = 1.0.10-7 M pH is a method of describing the [H+] (or [H3O+]) concentration (pH 7 is neutral, below 7 is acidic, above 7 is basic, see fig. 3 and table 3 pg. 503-504). pH = - log [H+] or pH = - log [H3O+] small constant = [HNO3]= 1.0.10-4 M before dissociation [H+] = [H3O+] = 1.0.10-4 M after dissociation [OH-] = Kw / [H+] [OH-] = 1.0.10-14 / [1.0.10-4] [OH-] = 1.0.10-10 M 6. Calculate the pH of pure water below (significant digits are_________ of the decimal in pH values): right pH = - log [1.0.10-7] pH = 7.00 Slide16:  7. pOH is way of describing the [OH-] concentration. pOH = - log [OH-] 8. All equations: pH = -log [H3O+] pH + pOH = 14.00 pOH = -log [OH-] Kw = 1.0.10-14= [H+][OH-] Sample B-E pg. 505-508: pH = ? 2 ways: Determine [H+] [H+] = Kw / [OH-] [H+] = 1.0.10-14 / [1.0.10-3] [H+] = 1.00 .10-11 M pH = -log [H+] pH = -log [1.00.10-11] pH = 11.00 pH = ? 2 ways: 2. Determine pOH pOH = -log [1.0.10-3] pOH = 3.000 pH = 14.00 - pOH pOH = -log [OH-] pH = 14.00 – 3.000 pH = 11.00 C. pH = ? pH = -log [H+] pH = -log [3.4.10-5] pH = 4.47 D. [H3O +] = ? pH = -log [H3O+] [H3O+] = 10-pH [H3O+] = 10-4.0 [H3O+] = 1.10-4 M E. H3O+ = ? E. [H3O+] = ? pH = -log [H3O+] [H3O+] = 10-pH [H3O+] = 10-7.52 [H3O+] = 3.0.10-8 M [OH-] = Kw/ [H3O+] [OH-] = 1.0.10-14 [3.02.10-8] [OH-] = 3.3.10-7 M Slide17:  pH can be utilized to determine the amount of ionization of an acid: If a 1.0 M solution of acetic acid has a pH of 1.30, what is the % ionization of the acid HA (aq) ↔ H+ (aq) + A- (aq) 1.0 M 0 M 0 M before ionization after ionization 1.0 M – 0.05012 M 0.05012 M 0.05012 M [H+] = 10-pH [H+] = 10-1.30 [H+] = 5.012.10-2 M % ionization = [H+] final 100 [HA] initial % ionization = [0.05012 ] 100 [1.0 ] % ionization = 5.0 %

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