Slide 1 / 208 Slide 2 / 208 cids and ases Slide 3 / 208 Slide 4 / 208 Table of ontents: cids and ases lick on the topic to go to that section Properties of cids and ases onjugate cid and ase Pairs mphoteric Substances Strong cids and ases utoionization of Water ph Weak cids and ases Polyprotic acids The Relationship etween K a and K b cidase Properties of Salt Solutions Factors ffecting cid Strength Properties of cids and ases Return to the Table of contents Slide 5 / 208 Properties of cids What is an cid? cids release hydrogen ions into solutions cids neutralize bases in a neutralization reaction. cids corrode active metals. cids turn blue litmus to red. Slide 6 / 208 Properties of ases ases release hydroxide ions into a water solution. ases neutralize acids in a neutralization reaction. ases denature protein. ases turn red litmus to blue. ases taste bitter. cids taste sour.
Slide 7 / 208 rrhenius cids and ases rrhenius's definition of acids and bases dates back to the 1800's. It is now considered obsolete since it only relates to reactions in water aqueous solutions. rrhenius defined acids and bases this way: n acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. Hl + H 2O H 3O + + l H 3O + is called a hydrated proton or a hydronium ion. base is a substance that, when dissolved in water, increases the concentration of hydroxide ions. Slide 8 / 208 rønstedlowry cids and ases The rønstedlowry definition dates back to the early 1900's and is considered the modern definition of acids and bases. This definition is more general and it works for all reactions; not just in those in water n acid is a proton, H +, donor. base is a proton, H +, acceptor. NH 3 + Hl NH 4 + + l NH 3 + H 2O NH 4 + + OH Slide 9 / 208 rønstedlowry cids and ases rønstedlowry acid: must have a removable (acidic) proton or must transfer a proton to another substance + NH 3 + Hl NH 4 + l Slide 10 / 208 rønstedlowry cids and ases Hl +H 2O l + H 3O + Hl donates the proton and acts as a rønstedlowry acid. H 2O accepts the proton and acts as a rønstedlowry base. rønstedlowry base: must have a pair of nonbonding electrons or must accept a proton H N + Hl NH 4 + + l H H Slide 11 / 208 Slide 12 / 208 Lewis cids Lewis cids rønstedlowry acids replaced rrhenius acids because the former were more general: rrhenius acids could only be defined in aqueous (water) solutions. rønstedlowry acids don't have that limitation. rrehenius acids only substances dissolved in H 2O Similarly, rønstedlowry acids are limited to substances that gain or lose hydrogen. rønstedlowry acids only substances that gain or lose H + The most general approach is that of Lewis acids; which do not require an aqueous environment or an exchange of hydrogen. Lewis acids are defined as electronpair acceptors. toms with an empty valence orbital can be Lewis acids. H 3 H 3 H H 3 + + OH 2 H 3 O H 3 H 3 H Lewis cid
Slide 13 / 208 Lewis ases Lewis bases are defined as electronpair donors. Slide 14 / 208 1 rønstedlowry base is defined as a substance that. nything that could be a rønstedlowry base is a Lewis base. Lewis bases can interact with things other than protons, however. Therefore, this definition is the broadest of the three. H 3 H 3 + + OH 2 H 3 O H 3 Lewis ase H 3 H 3 H H increases [H+] when placed in H 2O decreases [H+] when placed in H 2O increases [OH] when placed in H 2O acts aa s proton acceptor acts as a proton donor Slide 14 () / 208 Slide 15 / 208 1 rønstedlowry base is defined as a substance that. 2 rønstedlowry acid is defined as a substance that. increases [H+] when placed in H 2O decreases [H+] when placed in H 2O increases [OH] when placed in H 2O acts aa s proton acceptor The ronstedlowry definition of a base is a substance that is a proton acceptor. increases K a when placed in H 2O decreases [H + ] when placed in H 2O increases [OH ] when placed in H 2O acts as a proton acceptor acts as a proton donor acts as a proton donor Slide 15 () / 208 Slide 16 / 208 2 rønstedlowry acid is defined as a substance that. 3 Which of the following compounds could never act as an acid? increases K a when placed in H 2O The ronstedlowry definition of an acid is a substance that is a proton donor. decreases [H + ] when placed in H 2O increases [OH ] when placed in H 2O acts as a proton acceptor acts as a proton donor 2 SO 4 HSO 4 H 2 SO 4 NH 3 H 3 OOH
Slide 16 () / 208 3 Which of the following compounds could never act as an acid? SO 4 2 HSO 4 H 2 SO 4 NH 3 In order to be an acid you must be either be a substance that increases the H + concentration or a substance that donates a proton or an electron pair acceptor. SO4 2 fits none of these definitions. H 3 OOH Slide 17 / 208 4 ccording to the following reaction model, reactant is acting like an acid? H 2SO 4 H 2 O H 3 O + HSO 4 H 2O + H 2SO 4 H 3O + + HSO 4 None of the above Slide 17 () / 208 4 ccording to the following reaction model, reactant is acting like an acid? H 2O + H 2SO 4 H 3O + + HSO 4 Slide 18 / 208 5 ccording to the following reaction, which reactant is acting like a base? H 3O + + HSO 4 H 2O + H 2SO 4 H 2SO 4 H 2 O H 3 O + HSO 4 None of the above H2SO4 is donating a proton so it is the reactant that is acting like an acid. H 2 SO 4 H 2 O H 3 O + HSO 4 None of the above Slide 18 () / 208 5 ccording to the following reaction, which reactant is acting like a base? H 3O + + HSO 4 H 2O + H 2SO 4 Slide 19 / 208 6 For the following reaction, identify whether the compound in bold is behaving as an acid or a base. H 3PO 4 + H 2O H 2PO 4 + H 3O + H 2 SO 4 H 2 O H 3 O + HSO 4 None of the above HSO4 is accepting a proton so it is the reactant that is acting like an base. cid ase Neither oth None of the above
Slide 19 () / 208 6 For the following reaction, identify whether the compound in bold is behaving as an acid or a base. H 3PO 4 + H 2O H 2PO 4 + H 3O + cid ase Neither oth None of the above H3PO4 is donating a proton so it is acting like an acid. Slide 20 / 208 7 For the following reaction, identify whether the compound in bold is behaving as an acid or a base. cid ase oth H 3PO 4 + H 2O H 2PO 4 + H 3O + Neither None of the above Slide 20 () / 208 7 For the following reaction, identify whether the compound in bold is behaving as an acid or a base. H 3PO 4 + H 2O H 2PO 4 + H 3O + Slide 21 / 208 8 Which of the following cannot act as a Lewis base? l NH 3 cid ase oth Neither None of the above H2PO4 is accepting a proton so it is acting like an base. F 3 N H 2 O Slide 21 () / 208 8 Which of the following cannot act as a Lewis base? l Slide 22 / 208 9 In the reaction F 3 + F F 4 F 3 acts as a/an acid. NH 3 F 3 N Lewis ase is an electron pair donor and any substance that is a ronsted Lowry base would be a Lewis base. F3 is not a ronsted Lowry base and can not act as an electron pair donor. H 2 O rrhenius ronstedlowry Lewis rrhenius, ronstedlowry, and Lewis rrhenius and ronstedlowry
Slide 22 () / 208 Slide 23 / 208 9 In the reaction F 3 + F F 4 F 3 acts as a/an acid. rrhenius ronstedlowry Lewis rrhenius, ronstedlowry, and Lewis rrhenius and ronstedlowry F3 is accepting an electron pair from F and is acting as a Lewis cid. onjugate cid ase Pairs Return to the Table of contents Slide 24 / 208 onjugate cids and ases The term conjugate comes from the Latin word conjugare, meaning to join together. Reactions between acids and bases always yield their conjugate bases and acids. remove H + HNO 2(aq) + H 2O(l) NO 2 (aq) + H 3O + (aq) cid ase onjugate onjugate base acid add H+ Slide 25 / 208 onjugate cids and ases remove H + HNO 2(aq) + H 2O(l) NO 2 (aq) + H 3O + (aq) cid ase onjugate onjugate base acid add H+ fter the acid donates a proton, the result is called its conjugate base. fter the base accepts a proton, the result is called its conjugate acid. Slide 26 / 208 What Happens When an cid issolves in Water? Which is the acid? Which is the base? cids in Water Water acts as a rønstedlowry base and takes a proton (H+) from the acid. s a result, the conjugate base of the acid and a hydronium ion are formed. H 2SO 4 + H 2O H 3O + + HSO 4 Slide 27 / 208 10 In liquid ammonia, the reaction represented below occurs. In the reaction NH 4 + acts as a catalyst both an acid and a base the conjugate acid of NH 3 the reducing agent the oxidizing agent 2NH 3 NH 4 + + NH 2 Question from the ollege oard
Slide 27 () / 208 10 In liquid ammonia, the reaction represented below occurs. In the reaction NH 4 + acts as Slide 28 / 208 11 What are the conjugate bases of HlO 4, H 2S, PH + 4, HO 3? 2NH 3 NH 4 + + NH 2 a catalyst both an acid and a base the conjugate acid of NH 3 the reducing agent the oxidizing agent NH4 + acts as the conjugate acid of NH3. lo 4+, HS, PH 3, O 3 lo 4, HS, PH 3, O 2 3 2 lo 4, HS 2 3 2, PH 3, O 3 lo 4, H 2S, PH 3, O 3 2 Question from the ollege oard Slide 28 () / 208 11 What are the conjugate bases of HlO 4, H 2S, PH + 4, HO 3? lo 4+, HS, PH 3, O 3 To create a conjugate base for an lo 4, HS, PH 3, O 2 acid 3 you must remove an H and decrease the charge by 1. 2 lo 4, HS 2 3 2, PH HlO 3, O 4's conjugate base is lo 4, 3 H 2S's is HS, and PH + 4 is PH 3 and HO 3 is O 2 3. lo 4, H 2S, PH 3, 2 O 3 100% ionized in H2O cid strength increases cid Slide 29 / 208 cid and ase Strength ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 SO4 2 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 + NH3 HO3 2 O3 HPO4 PO4 2 3 H2O OH OH O 2 Strong Weak Negligible Strong Weak H2 H ase strength increases H4 H3 100% protonated in H2O Strong acids completely dissociate in water. Their conjugate bases are quite weak. cid Proton Negative ion 100% ionized in H2O cid strength increases cid Slide 30 / 208 cid and ase Strength ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 2 SO4 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 + NH3 HO3 O3 2 2 HPO4 3 PO4 H2O OH OH O 2 Strong Weak Negligible Strong Weak H2 H H4 ase strength increases 100% H3 protonated in H2O Weak acids only partially dissociate in water. Their conjugate bases are weak bases. cid Proton Negative ion 100% ionized in H2O cid strength increases cid Strong Weak N Slide 31 / 208 cid and ase Strength ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 2 SO4 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 + NH3 HO3 O3 2 Strong Weak Negligible HPO4 2 PO4 3 H2O OH OH O 2 H2 H H4 H3 ase strength increases 100% protonated in H2O Substances with negligible acidity do not dissociate in water. They will not readily give up protons. Their conjugate bases are exceedingly strong.
Slide 32 / 208 cid and ase Strength In any acidbase reaction, equilibrium will favor the reaction in which the proton moves toward the stronger base. In other words, a stronger base will "hold onto" its proton whereas a strong acid easily releases its proton(s). Hl (aq) + H 2O (l) H 3O + (aq) + l (aq) acid base conj. acid conj. base n alternative way to consider equilibrium is that it will favor the reaction WY from the stronger acid. In this example, H 2O is a much stronger base than l, so the proton moves from Hl to H 2O Slide 33 / 208 cid and ase Strength onversely, Hl is a much stronger acid than the hydronium ion, so equilibrium lies very far to the right K >>1 Hl (aq) + H 2O (l) H 3O + (aq) + l (aq) acid base conj. acid conj. base 100% ionized in H2O cid strength increases cid ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 SO4 2 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 NH3 + HO3 O3 2 Strong Weak Negligible g Weak Negligible HPO4 2 PO4 3 H2O OH OH O 2 H2 H ase strength increases H4 H3 100% protonated in H2O Slide 34 / 208 cid and ase Strength Slide 35 / 208 cid and ase Strength onsider this equilibrium between acetic acid and acetate ion: H 3OOH (aq) + H 2O (l) H 3O + (aq) + H 3OO (aq) H 3OOH (aq) + H 2O (l) If you look for the stronger acid: quilib lies away from the stronger acid. H 3O + (aq) + H 3OO (aq) oes equilibrium lie to the left (K<1) or to the right (K>1)? If you look for the stronger base: quilib favors this base accepting a proton. 100% ionized in H2O cid strength increases cid ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 SO4 2 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 NH3 + HO3 O3 2 HPO4 2 PO4 3 H2O OH OH O 2 Strong Weak Neg Strong Weak Negligible H2 H H4 H3 ase strength increases 100% protonated in H2O Since the hydronium ion is a stronger acid than acetic acid, equilibrium lies to the left (K<1). Slide 36 / 208 cid and ase Strength cetic acid is a weak acid. This means that only a small percent of the acid will dissociate. Slide 37 / 208 12 Strong acids have conjugate bases. The double headed arrow is used only in weak acid or weak base dissociation equations. H 3OOH (aq) + H 2O (l) H 3O + (aq) + H 3OO (aq) single arrow is used for strong acid or strong bases which dissociate completely. strong weak neutral negative NaOH Na + (aq) + OH (aq)
Slide 37 () / 208 12 Strong acids have conjugate bases. Slide 38 / 208 13 Hr, hydrobromic acid is a strong acid. This means that it. strong weak neutral negative aqueous solutions of Hr contain equal concentrations of H + and OH does not dissociate at all when it is dissolved in water cannot be neutralized by a base dissociates completely to H + and r when it dissolves in water Slide 38 () / 208 13 Hr, hydrobromic acid is a strong acid. This means that it. aqueous solutions of Hr contain equal concentrations of H + and OH does not dissociate at all when it is dissolved in water cannot be neutralized by a base Strong acids and bases dissociate completely. dissociates completely to H + and r when it dissolves in water Slide 39 / 208 14 For the following reaction, determine which side of the equilibrium is favored. H 3PO 4 + H 2O H 2PO 4 + H 3O + the right side the left side Neither side is favored 100% ionized in H2O cid strength increases cid ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 SO4 2 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS Strong Weak Negligible ng Weak Negligible H2PO4 HPO4 2 NH4 + NH3 2 HO3 O3 2 3 HPO4 PO4 H2O OH OH O 2 H2 H H4 H3 ase strength increases 100% protonated in H2O Slide 39 () / 208 14 For the following reaction, determine which side of the equilibrium is favored. H 3PO 4 + H 2O H 2PO 4 + H 3O + the right side the left side Neither side is favored The left side is favored because H2PO4 is cid a stronger base asethan H2O and the side of the 100% equilibrium Hl that is favored l is the ionized side where H2SO4 the stronger HSO4 base in H2O has accepted HNO3 the proton. NO3 H3O + H2O 2 HSO4 SO4 H3PO4 H2PO4 [This HF object is a pull F H2H3O2 2H3O2 H2O3 HO3 H2S HS cid strength increases Strong Weak Negligible ng Weak Negligible H2PO4 HPO4 2 NH4 + NH3 HO3 O3 2 2 HPO4 3 PO4 H2O OH OH O 2 H2 H H4 H3 ase strength increases 100% protonated in H2O Slide 40 / 208 15 For the following reaction, determine which side of the equilibrium is favored. H 3O + + HSO 4 H 2O + H 2SO 4 the right side the left side Neither side is favored 100% ionized in H2O cid strength increases cid ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 SO4 2 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 + NH3 HO3 O3 2 Weak Negligible Strong Weak Negligible HPO4 2 PO4 3 H2O OH OH O 2 H2 H H4 H3 ase strength increases 100% protonated in H2O
Slide 40 () / 208 15 For the following reaction, determine which side of the equilibrium is favored. H 3O + + HSO 4 H 2O + H 2SO 4 Slide 41 / 208 16 For the following reaction, determine which side of the equilibrium is favored. HNO 3 + H 2O H 3O + + NO 3 the right side the left side Neither side is favored cid ase The 100% right side Hl is favored because l ionized H2O is a stronger H2SO4 base than HSO4 in H2O HNO3 NO3 HSO4 and the side of the equilibrium that H3O is + favored H2O is the 2 HSO4 SO4 side where the stronger base H3PO4 H2PO4 has accepted the proton. HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 + NH3 HO3 O3 2 cid strength increases Weak Negligible Strong Weak Negligible HPO4 2 PO4 3 H2O OH OH O 2 H2 H H4 H3 ase strength increases 100% protonated in H2O the right side the left side Neither side is favored 100% ionized in H2O cid strength increases cid ase Hl l H2SO4 HSO4 HNO3 NO3 H3O + H2O HSO4 SO4 2 H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS H2PO4 HPO4 2 NH4 NH3 + HO3 O3 2 Strong Weak Negligible g Weak Negligible HPO4 2 PO4 3 H2O OH OH O 2 H2 H H4 H3 ase strength increases 100% protonated in H2O Slide 41 () / 208 Slide 42 / 208 16 For the following reaction, determine which side of the equilibrium is favored. HNO 3 + H 2O H 3O + + NO 3 the right side the left side Neither side is favored cid ase The right side is favored because 100% H2O is a stronger Hl base than l NO3 ionized H2SO4 HSO4 in and H2O the side of the equilibrium HNO3 NO3 that is favored H3O + is the side where H2O the stronger base HSO4 has accepted SO4 2 the proton. H3PO4 H2PO4 HF F H2H3O2 2H3O2 H2O3 HO3 H2S HS 2 H2PO4 HPO4 + NH4 NH3 HO3 O3 2 cid strength increases Strong Weak Negligible g Weak Negligible HPO4 2 PO4 3 H2O OH OH O 2 H2 H H4 H3 ase strength increases 100% protonated in H2O mphoteric Substances Return to the Table of ontents Slide 43 / 208 mphoteric Substances Slide 44 / 208 mphoteric Substances If a substance can act both as an acid and base, it is known as amphoteric. For example, water can act as a base or acid depending on the situation. Hl + H 2O l + H 3O + bove, water accepts a proton, thus acting as a base. NH 3 +H 2O NH 4 + + OH bove, water donates a proton, thus acting as an acid. HO 3 HSO 4 nother term for amphoteric is amphiprotic. For each of the following substances, write two equations, one showing it as a ronstedlowry acid and another showing it as a ronstedlowry base. H 2O
Slide 45 / 208 17 substance that is capable of acting as both an acid and as a base is. Slide 45 () / 208 17 substance that is capable of acting as both an acid and as a base is. autosomal conjugated amphoteric saturated miscible autosomal conjugated amphoteric saturated miscible Slide 46 / 208 Slide 47 / 208 18 Write the equations and equilibrium expressions for HS when it is acting like a rønstedlowry acid and when it Students type their answers here is acting like a rønstedlowry base. Strong cids and ases Return to the Table of ontents The seven strong acids are: Slide 48 / 208 Strong cids Recall, strong acids completely ionize in solution. Hl hydrochloric acid Hr hydrobromic acid HI hydroiodic acid HNO 3 nitric acid H 2SO 4 sulfuric acid HlO 3 chloric acid HlO 4 perchloric acid Slide 49 / 208 Strong cids The seven strong acids are strong electrolytes because they are 100% ionized. In other words, these compounds exist totally as ions in aqueous solution. For the monoprotic strong acids (acids that donates only one proton per molecule of the acid), the hydronium ion concentration equals the acid concentration. [cid] = [H 3O + ] So, if you have a solution of 0.5 M Hl, then [H 3O + ] = 0.5 M Memorize this list.
Slide 50 / 208 Strong ases ll strong bases are group of compounds called "metal hydroxides." ll alkali metals in Group I form hydroxides that are strong bases: LiOH, NaOH, KOH, etc. Only the heavier alkaline earth metals in Group II form strong bases: a(oh) 2, Sr(OH) 2, and a(oh) 2. gain, these substances dissociate completely in aqueous solution. In other words, NaOH exists entirely as Na + ions and OH ions in water. Slide 51 / 208 19 What is the hydroxide ion concentration of a 0.22 M calcium hydroxide solution? 0.11 0.22 0.44 0.88 Not enough information. Slide 51 () / 208 19 What is the hydroxide ion concentration of a 0.22 M calcium hydroxide solution? Slide 52 / 208 20 What is the concentration of H + in a 25ml solution of 0.05M Hl when diluted to final volume of 100ml? 0.11 0.22 alcium hydroxide's formula is 0.44 a(oh) 2. For every 1 mole of a(oh) 2 there are 2 moles of OH. 0.88 If you have a 0.22M solution of a(oh) 2 then the concentration of Not enough information. OH is 2x as much or 0.44M. Slide 52 () / 208 20 What is the concentration of H + in a 25ml solution of 0.05M Hl when diluted to final volume of 100ml? To determine the new concentration you can use the formula M 1V 1=M 2V 2 M 1 =.05M, V 1= 25 ml and V 2 = 100 ml M 2 = M 1V 1/V 2 M 2=.05M x 25 ml /100 ml M 2 = 0.0125M Hl The concentration of H + equals the concentration of the Hl solution and therefore equals 0.0125M. Slide 53 / 208 21 What is the [H + ] ion concentration of a 50 ml solution of 0.025M H 2SO 4, when diluted with 150 ml of water?
Slide 53 () / 208 21 What is the [H + ] ion concentration of a 50 ml solution of 0.025M H 2SO 4, when diluted with 150 ml of water? Using M1V1= M2V2. You are given: M1 = 0.025M, V1= 50 ml and V2=100 ml. M2 = M1V1/V2 M2 = 0.025M x 50 ml/100 ml M2 = 0.0125M For every 1 mole of H2SO4 there are 2 moles of H +. If you have a 0.0125M solution of H2SO4 then the concentration of H + is 2x as much or 0.0250M Slide 54 / 208 22 solution of 25 ml of 0.1M Hl and 50 ml of 0.5M HNO 3 are mixed together. What is the [H+] ion concentration of the resulting solution? Slide 55 / 208 utoionization of Water Slide 56 / 208 utoionization of Water s we have seen, water is amphoteric, meaning that it can act as either an acid or a base. In pure water, a few molecules act as bases and a few act as acids, in a process referred to as autoionization. H O + H O H O H + O H Return to the Table of contents H H The double headed arrow indicates that both the forward and reverse reactions occur simultaneously. H H 2O (l) + H 2O (l) H 3O + (aq) + OH (aq) Slide 57 / 208 utoionization of Water H 2O (l) + H 2O (l) H 3O + (aq) + OH (aq) When there is an equilibrium state, the ratio of products to reactants yields a constant. This value is known as the equilibrium constant, K and will be discussed in more depth later in this unit. Slide 58 / 208 IonProduct onstant In most dilute acid and base solutions, the concentration of undissociated water, remains more or less a constant. We can thus disregard the denominator in the equilibrium expression. K = [H3O+ ] x [OH ] [H 2O] x [H 2O] becomes K w = [H 3O + ] x [OH ] ll concentrations are in M, molarity, as designated by brackets, [ ]. K = H 2O (l) + H 2O (l) [H 3O + ] x [OH ] K = [H 2O] x [H 2O] H 3O + (aq) + OH (aq)
Slide 59 / 208 IonProduct onstant K w = [H 3O + ] x [OH ] Slide 60 / 208 23 The magnitude of K w indicates that. This special equilibrium constant, K w is referred to as the ionproduct constant for water. t 25, K w = 1.0 x 10 14. Since this is such a small number, we conclude that pure water contains relatively very few ions. water ionizes to a very small extent the autoionization of water is exothermic water ionizes very quickly water ionizes very slowly Slide 60 () / 208 Slide 61 / 208 23 The magnitude of K w indicates that. 24 The ionproduct constant for water, K w is represented by water ionizes to a very small extent the autoionization of water is exothermic water ionizes very quickly water ionizes very slowly Kw is a very small number that means the amount of reactants is much greater than the amount of products. Water ionizes to very small extent. [H 2O] 2 [H 3O + ] x [OH ] [H 3O + ] + [OH ] [H 3O + ] [OH ] Slide 61 () / 208 Slide 62 / 208 24 The ionproduct constant for water, K w is represented by [H 2O] 2 [H 3O + ] x [OH ] [H 3O + ] + [OH ] Kw = [H 3O + ] x [OH ] [H 3O + ] [OH ] 25 What is the [H + ] ion concentration of a solution with an [OH ] ion concentration of 1 x 10 9? Try to solve without a calculator! 1 x 10 14 1 x 10 5 1 x 10 5 1.23 x 10 5 1 x 10 7
Slide 62 () / 208 25 What is the [H + ] ion concentration of a solution with an [OH ] ion concentration of 1 x 10 9? Try to solve without a calculator! 1 x 10 14 K w = [H + ][OH ] [H + ] = Kw /[OH ] 1 x 10 5 [H + ] = 1 x 10 14 / 1 x 10 9 [H + ] = 1x 10 5 1 x 10 5 1.23 x 10 5 1 x 10 7 Slide 63 / 208 26 What is the [OH ] ion concentration of a solution with an [H + ] ion concentration of 1 x 10 4? Try to solve without a calculator! 1 x 10 14 1 x 10 10 1 x 10 5 1.23 x 10 9 1 x 10 10 Slide 63 () / 208 26 What is the [OH ] ion concentration of a solution with an [H + ] ion concentration of 1 x 10 4? Try to solve without a calculator! 1 x 10 14 1 x 10 10 1 x 10 5 K w = [H + ][OH ] [H + ] = Kw /[OH ] [H + ] = 1 x 10 14 / 1 x 10 4 [H + ] = 1x 10 10 Slide 64 / 208 27 What is the [OH ] ion concentration of a solution with an [H + ] ion concentration of 1.23 x 10 9? 1 x 10 14 4.34 x 10 6 1.23 x 10 5 1.23 x 10 9 1 x 10 10 1.23 x 10 5 8. 13 x 10 6 Slide 64 () / 208 Slide 65 / 208 27 What is the [OH ] ion concentration of a solution with an [H + ] ion concentration of 1.23 x 10 9? 1 x 10 14 4.34 x 10 6 1.23 x 10 5 1.23 x 10 5 K w = [H + ][OH ] [H + ] = Kw /[OH ] [H + ] = 1 x 10 14 / 1.23 x 10 9 [H + ] = 8.13 x 10 6 ph 8. 13 x 10 6 Return to the Table of contents
Slide 66 / 208 ph It is a measure of hydrogen ion concentration, [H + ] in a solution, where the concentration is measured in moles H + per liter, or molarity. The ph scale ranges from 014. ph is defined as the negative base10 logarithm of the concentration of hydronium ion. ph = log [H 3O + ] Slide 67 / 208 ph ph is defined as the negative base10 logarithm of the concentration of hydronium ion. Hydrogen ion concentration, [H+] in moles/liter ph = log [H 3O + ] ph 1.0 x 10 1 1 1.0 x 10 2 2 1.0 x 10 10 10 Is the relationship between [H + ] and ph a direct or an inverse one? Slide 68 / 208 Slide 69 / 208 ph ecause of the base10 logarithm, each 1.0point value on the ph scale differs by a value of ten. solution with ph = 9 has a hydrogen ion concentration, [H + ], that is ten times more than a ph = 10 solution. solution with ph = 8 has a hydrogen ion concentration, [H + ], that is 10 2 or 100 times more than a ph = 10 solution. solution with ph = 7 has a hydrogen ion concentration, [H + ], that is 10 3 or 1000 times more than a ph = 10 solution. Slide 70 / 208 Slide 70 () / 208 28 The molar concentration of hydronium ion, [H 3O + ], in pure water at 25 is. 28 The molar concentration of hydronium ion, [H 3O + ], in pure water at 25 is. 0 1 7 10 7 10 14 0 1 7 10 7 If the solution is neutral the concentration of [H3O + ] = [OH ] and each has a concentration of 1x10 7. 14 10
Slide 71 / 208 29 solution with ph = 3 has a hydrogen ion concentration that is than a solution with ph = 5. Slide 71 () / 208 29 solution with ph = 3 has a hydrogen ion concentration that is than a solution with ph = 5. 2x more 2x less 100x more 100x less 2x more 2x less 100x more 100x less The solution with a ph of 3 has a hydrogen ion concentration is 100 x more than a solution with a ph of 5, because ph is a log function and the difference is 1 x 10 3 vs 1 x 10 5 which is 100 x more. Slide 72 / 208 30 solution with ph = 14 has a hydrogen ion concentration that is than a solution with ph = 11. Slide 72 () / 208 30 solution with ph = 14 has a hydrogen ion concentration that is than a solution with ph = 11. 3x more 3x less 1000x more 1000x less 3x more 3x less 1000x more The solution with a ph of 14 has a hydrogen ion concentration is 1000 x less because ph is a log function and the difference is 1 x 10 14 vs 1 x 10 11 which is 1000 x less. 1000x less Slide 73 / 208 Slide 74 / 208 ph ph Therefore, in pure water I S ph = log [H 3O + ] = 7.00 ph = log (1.0 # 10 7 ) = 7.00 [H+] > [OH] There are excess hydrogen ions in solution. [H+] < [OH] There are excess hydroxide ions in solution. n acid has a higher [H 3O + ] than pure water, so its ph is <7. base has a lower [H 3O + ] than pure water, so its ph is >7. Solution type [H +](M) [OH] (M) ph value cidic >1.0x10 7 <1.0x10 7 <7.00 Neutral =1.0x10 7 =1.0x10 7 =7.00 asic <1.0x10 7 > 1.0x10 7 >7.00 Solution type [H +](M) [OH] (M) ph value cidic >1.0x10 7 <1.0x10 7 <7.00 Neutral =1.0x10 7 =1.0x10 7 =7.00 asic <1.0x10 7 > 1.0x10 7 >7.00
These are the ph values for several common substances. Slide 75 / 208 More basic More acidic ph 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 leach Soapy water mmonia Milk of Magnesia aking soda Sea water istilled water Urine lack coffee Tomato juice Orange juice Lemon juice Gastric acid Slide 76 () / 208 31 For a basic solution, the hydrogen ion concentration is than the hydroxide ion concentration. Slide 76 / 208 31 For a basic solution, the hydrogen ion concentration is than the hydroxide ion concentration. greater than less than equal to Not enough information. Slide 77 / 208 32 For an acidic solution, the hydroxide ion concentration is than the hydrogen ion concentration. greater than less than equal to Not enough information. In a basic solution the concentration of hydrogen ion is less than the concentration of hydroxide ion. greater than less than equal to Not enough information. Slide 77 () / 208 Slide 78 / 208 32 For an acidic solution, the hydroxide ion concentration is than the hydrogen ion concentration. 33 Which solution below has the highest concentration of hydroxide ions? greater than less than equal to Not enough information. In an acidic solution, the concentration of hydroxide ion is less than the concentration of hydrogen ion. ph = 3.21 ph = 7.00 ph = 8.93 ph = 12.6
Slide 78 () / 208 33 Which solution below has the highest concentration of hydroxide ions? Slide 79 / 208 34 Which solution below has the lowest concentration of hydrogen ions? ph = 3.21 ph = 7.00 ph = 8.93 The smaller the ph the more acidic the solution. The larger the ph the more basic the solution. The solution with the highest concentration of hydroxide ion, the most basic solution, has the largest ph in this case 12.6. ph = 12.6 ph = 1.98 ph = 8.53 ph = 5.91 ph =11.4 Slide 79 () / 208 34 Which solution below has the lowest concentration of hydrogen ions? Slide 80 / 208 35 For a 1.0M solution of a weak base, a reasonable ph would be. ph = 1.98 ph = 8.53 ph = 5.91 The smaller the ph the more acidic the solution. The larger the ph the more basic the solution. The solution with the highest concentration of hydrogen ion, the most acidic solution, has the smallest ph in this case 1.98. ph =11.4 2 6 7 9 13 Slide 80 () / 208 35 For a 1.0M solution of a weak base, a reasonable ph would be. Slide 81 / 208 36 For a 1.0M solution of a strong acid, a reasonable ph would be. 2 6 7 9 13 weak base would have a relatively high ph but not a very high ph. strong base would have a much higher ph 2 6 7 9 13
Slide 81 () / 208 36 For a 1.0M solution of a strong acid, a reasonable ph would be. Slide 82 / 208 37 The ph of a solution with a concentration of 0.01M hydrochloric acid is 2 6 7 9 reasonable ph for a strong acid is 2. 10 2 12 2 10 12 13 Slide 82 () / 208 37 The ph of a solution with a concentration of 0.01M hydrochloric acid is 38 solution with the ph of 5.0 Slide 83 / 208 10 2 12 2 10 12 ph = log [H + ] ph = log (0.010) ph = 2 is basic has a hydrogen ion concentration of 5.0M is neutral has a hydroxideion concentration of 1x10 9 38 solution with the ph of 5.0 is basic Slide 83 () / 208 has a hydrogen ion concentration of 5.0M is neutral has a hydroxideion concentration of 1x10 9 The hydrogen ion concentration of a solution with a ph of 5.0 is 1x10 5. [OH] = Kw/[H + ] [OH ] =1 x 10 14 /1 x 10 5 = 1 x 10 9. Slide 84 / 208 How o We Measure ph? For less accurate measurements, one can use Litmus paper Red litmus paper turns blue above ~ph = 8 lue litmus paper turns red below ~ph = 5 Or an indicator (usually an organic dye) such as one of the following: ph range for color change 0 2 4 6 8 10 12 14 Methyl violet Thymol blue Methyl orange Methyl red romothymol blue Phenolphthalein lizarin yellow R
For more accurate measurements, one uses a ph meter, which measures the voltage in the solution. Slide 85 / 208 How o We Measure ph? Slide 86 / 208 How o We alculate ph? Recall that ph is defined as the negative base10 logarithm of the concentration of hydronium ion (or hydrogen ion). ph = log [H 3O + ] or ph = log [H + ] Slide 87 / 208 How o We alculate ph? What is the ph of the solution with hydrogen ion concentration of 5.67 x 10 8 M (molar)? ph = log [H + ] First, take the log of 5.67 x 10 8 = 7.246 Now, change the sign from to + : ph = 7.246 Slide 88 / 208 39 What is the ph of a solution with a hydrogen ion concentration of 1 x 10 5 M? 1 x 10 5 5 5 9 If you take the log of 5.67 x 10 8, you will end up with an incorrect answer. The order of operations: 1. Take the log 2. Switch the sign Slide 88 () / 208 39 What is the ph of a solution with a hydrogen ion concentration of 1 x 10 5 M? Slide 89 / 208 40 What is the ph of a solution with a hydroxide ion concentration of 1 x 10 12 M? 1 x 10 5 5 5 ph = log H + ph = log (1 x 10 5 ) ph = 5 1 x 10 2 12 2 9 12
Slide 89 () / 208 40 What is the ph of a solution with a hydroxide ion concentration of 1 x 10 12 M? 1 x 10 2 12 2 12 ph = 2 Kw = [H + ][OH ] [H + ] = 1 x 10 14 /[OH ] [H + ] = 1 x 10 14 / 1 x 10 12 [H + ] = 1 x 10 2 ph = log H + ph = log (1 x 10 2 ) 41 Slide 90 / 208 What is the ph of an aqueous solution at 25.0 in which [H + ] is 0.0025 M? 3.4 2.6 2.6 3.4 2.25 Slide 90 () / 208 Slide 91 / 208 41 What is the ph of an aqueous solution at 25.0 in which [H + ] is 0.0025 M? 3.4 2.6 2.6 3.4 2.25 ph = log H + ph = log (0.0025) ph = 2.60 dditional ph alculations If you are given the ph and asked to find the [H + ] (or [H 3O + ]) in a solution, use the inverse log. Since ph = log [H + ], then [H + ] = 10 ph What is the hydrogen ion concentration (M) in a solution of Milk of Magnesia whose ph = 9.8? [H + ] = 10 9.8 [H + ] = 1.58 x 10 10 M or mol/liter Slide 92 / 208 Slide 92 () / 208 42 What is the ph of a solution whose hydronium ion concentration is 7.14 x 10 3 M? 42 What is the ph of a solution whose hydronium ion concentration is 7.14 x 10 3 M? ph = log [H + ] ph = log (7.14x 10 3 M) ph = 2.15
Slide 93 / 208 43 What is the ph of a solution whose hydronium ion concentration is 1.92 x 10 9 M? Slide 93 () / 208 43 What is the ph of a solution whose hydronium ion concentration is 1.92 x 10 9 M? ph = log [H + ] ph = log (1.92 x 10 9 M) ph = 8.72 Slide 94 / 208 44 What is the hydronium ion concentration in a solution whose ph = 4.29? Slide 94 () / 208 44 What is the hydronium ion concentration in a solution whose ph = 4.29? [H + ] = 10 ph [H + ] = 10 4.29 [H + ] = 5.13 x 10 5 M Slide 95 / 208 45 What is the hydroxide ion concentration in a solution whose ph = 4.29? Slide 95 () / 208 45 What is the hydroxide ion concentration in a solution whose ph = 4.29? [H + ] = 10 ph [H + ] = 10 4.29 [H + ] = 5.13 x 10 5 M K w = [H + ] [OH ] [OH ] =K w/ [H + ] [OH ] = 1x10 14 / 5.13 x 10 5 M [OH ] = 1.95 x 10 10
Slide 96 / 208 Other p Scales The p in ph tells us to take the negative base10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are poh = log [OH] pk w = log K w Slide 97 / 208 Relationship between ph and poh ecause [H 3O + ] [OH ] = Kw = 1.0 x 10 14, we know that log [H 3O + ] + log [OH ] = log K w = 14.00 or, in other words, ph + poh = pk w = 14.00 pk a = log K a pk b = log K b Slide 98 / 208 46 n aqueous solution of a base has a poh 6.12 what is its ph? Slide 98 () / 208 46 n aqueous solution of a base has a poh 6.12 what is its ph? 4.22 8.88 7.88 4.22 8.88 7.88 ph + poh = 14 ph = 14 poh ph = 14 6.12 ph = 7.88 2.11 7.59 x 10 7 2.11 7.59 x 10 7 Slide 99 / 208 47 n aqueous solution of an acid has a hydrogen ion concentration of 2.5x10 4. What is the poh of this solution? 10.4 6.13 8.4 7.5 9.4 Slide 99 () / 208 47 n aqueous solution of an acid has a hydrogen ion concentration of 2.5x10 4. What is the poh of this solution? ph = log[h + ] 10.4 ph = log (2.5 x10 4 M) ph = 3.60 6.13 ph + poh = 14 poh = 14 3.60 8.4 poh = 10.4 7.5 9.4
Slide 100 / 208 48 n aqueous solution of an acid has ph of 4.11 what is the [OH] concentration of this solution? 4.55x10 11 5.78x10 11 4.25x10 8 1.29x10 10 4.03x10 7 Slide 100 () / 208 48 n aqueous solution of an acid has ph of 4.11 what is the [OH] concentration of this solution? 4.55x10 11 5.78x10 11 4.25x10 8 1.29x10 10 4.03x10 7 ph + poh = 14 poh = 14 ph poh = 14 4.1 poh = 9.89 [OH ] = 10 poh [OH ] = 10 9.89 [OH ] = 1.29 x 10 10 Slide 101 / 208 49 n aqueous solution of an base has poh of 3.33 what is the [H + ] concentration of this solution? 4.68x10 4 2.14x10 11 5.67x10 8 9.07x10 9 4.88x10 7 Slide 101 () / 208 49 n aqueous solution of an base has poh of 3.33 what is the [H + ] concentration of this solution? 4.68x10 4 2.14x10 11 5.67x10 8 9.07x10 9 4.88x10 7 ph + poh = 14 ph = 14 poh ph = 14 3.33 ph = 10.67 [H + ] = 10 ph [H + ] = 10 10.67 [H + ] = 2.14 x 10 11 Slide 102 / 208 Slide 103 / 208 cid issociation onstants, K a H (aq) + H 2O (l) (aq) + H 3O + (aq) Weak cids and ases Return to the Table of contents For a generalized acid dissociation, the equilibrium expression is K c, K K a = [H 3O + ] [ ] [H] This equilibrium constant is called the aciddissociation constant, Ka. K a = [H3O+ ][ ] [H]
Slide 104 / 208 cid issociation onstants, K a The greater the value of K a, the stronger is the acid. cid Proton Transfer K a Value Hydrochloric acid Hl + H 2O H 3O + + l Large Sulfuric acd H 2SO 4 + H 2O H 3O + + HSO 4 Large Nitric acid HNO 3 + H 2O H 3O + + NO bout 20 Hydrofluoric acid HF + H 2O H 3O + + F 1.2 x 10 4 arbonic acid H 2O 3 + H 2O H 3O + + HO 3 4.3 x 10 7 Hydrogen cyanide HN + H 2O H 3O + + N 4.9 x 10 16 Slide 105 / 208 50 The acid dissociation constant (K a) of HF is 6.7 x 10 4. Which of the following is true of a 0.1M solution of HF? [HF] is greater than [H + ][F ] [HF] is less than [H + ][F ] [HF] is equal to [H + ][F ] [HF] is equal to [H ][F + ] Slide 105 () / 208 50 The acid dissociation constant (K a) of HF is 6.7 x 10 4. Which of the following is true of a 0.1M solution of HF? [HF] is greater than [H + ][F ] [HF] is less than [H + ][F ] [HF] is equal to [H + ][F ] ecause the Ka value is so small it means that the amount of reactants is much greater than the amount of products therefore [HF] is greater than [H + ][F ] [HF] is equal to [H ][F + ] Slide 106 / 208 alculating K a from the ph The ph of a 0.10 M solution of formic acid, HOOH, at 25 is 2.38. alculate K a for formic acid at this temperature. The dissociation equation for formic acid may be written as a reaction with water HOOH + H 2O HOO + H 3O + or, without water HOOH HOO + H + Slide 107 / 208 alculating K a from the ph The ph of a 0.10 M solution of formic acid, HOOH, at 25 is 2.38. alculate K a for formic acid at this temperature. HOOH + H 2O HOO + H 3O + From this dissociation equation, write the K a expression: K a = [H 3O + ][HOO ] [HOOH] To calculate K a, we need the equilibrium concentrations of all three species. We know the concentration of HOOH, how do we determine the concentration of H 3O +? Slide 108 / 208 alculating K a from the ph ph = log [H 3O + ] 2.38 = log [H 3O + ] 2.38 = log [H 3O + ] 10 2.38 = 10 log [H3O+] = [H 3O + ] 4.2 x 10 3 M = [H 3O + ] = [HOO ] Note that this is a monoprotic acid, so [acid] = [conjugate base]
Slide 109 / 208 alculating K a from the ph The ph of a 0.10 M solution of formic acid, HOOH, at 25 is 2.38. alculate K a for formic acid at this temperature. K a = [H 3O + ][HOO ] [HOOH] [HOOH], M [HOO ], M [H 3O + ], M Initially 0.10 0 0 hange 4.2 x10 3 +4.2 x10 3 +4.2 x10 3 Slide 110 / 208 alculating K a from the ph The ph of a 0.10 M solution of formic acid, HOOH, at 25 is 2.38. alculate K a for formic acid at this temperature. Important note: In the case of weak acids and bases, we assume little ionization and therefore we ignore the amount of ionization and write 0.10 and not 0.10 x. If you find that the % ionization is greater than 5% you can't ignore the x term and you must use the quadriatic equation to solve. t quilibrium 0.10 4.2 x 10 3 0.10 4.2 x 10 3 4.2 x 10 3 Slide 111 / 208 alculating K a from the ph Now, we substitute values into the Ka expression and solve: [H 3O + ][HOO ] K a = [HOOH] K [4.2 # 10 3 ] [4.2 # 10 3 a = ] [0.10] K a = 1.8 # 10 4 Slide 112 / 208 alculating Percent Ionization One way to compare the strength of two acids is by the extent to which each one ionizes. This is done by calculating percent ionization, or the ratio of [H + ] ions that are produced, compared to the original acid concentration. [H 3O + ] eq Percent Ionization = # 100% [H] initial Slide 113 / 208 alculating Percent Ionization Slide 114 / 208 51 What is the K a of a 0.125M solution of hypobromous acid (HrO) that has a ph of 4.74. [H 3O + ] eq Percent Ionization = # 100% [H] initial In this example [H 3O + ] eq = 4.2 # 10 3 M [HOOH] initial = 0.10 M Percent Ionization = 4.2 x 10 3 # 100% 0.10 = 4.2 % 2.42 x 10 4 2.65 x 10 9 3.71 x 10 10 2.13 x 10 5 6.78 x 10 12
Slide 114 () / 208 51 What is the K a of a 0.125M solution of hypobromous acid (HrO) that has a ph of 4.74. Slide 115 / 208 52 What is % ionization of a 0.125M solution of hypobromous acid (HrO) that has a ph of 4.74. 2.42 x 10 4 1.45% 2.65 x 10 9 3.71 x 10 10 2.13 x 10 5 K a = x 2 /0.125 x = [H + ] = 10 4.74 x = 1.82 x 10 5 K a = (1.82 x 10 5 ) 2 / 0.125 K a = 2.65 x 10 9 68.89%.0145%.00145% 6.78 x 10 12.6889% Slide 115 () / 208 52 What is % ionization of a 0.125M solution of hypobromous acid (HrO) that has a ph of 4.74. Slide 116 / 208 53 What is the K a of a 0.20M solution of nitrous acid (HNO 2) that has a ph of 2.02? 1.45% 68.89%.0145%.00145% [H + ] = 10 4.74 = 1.82 x 10 5 % ionization = [H + ]/[HrO] x 100 % = 1.82 x 10 5 / 0.125 x 100 % =.0145% 1.3 x 10 5 9.9 x 10 2 1.2 x 10 8 4.2 x 10 7.6889% 4.6 x 10 4 Slide 116 () / 208 53 What is the K a of a 0.20M solution of nitrous acid (HNO 2) that has a ph of 2.02? Slide 117 / 208 54 What is % ionization of a 0.20M solution of nitrous acid (HNO 2) that has a ph of 2.02? 1.3 x 10 5 9.9 x 10 2 1.2 x 10 8 K a = x 2 /0.20 x = [H + ] = 10 2.02 x = 9.55 x 10 3 K a = (9.55 x 10 3 ) 2 / 0.20 K a = 4.6 x 10 4.0477 4.77% 5.99% 4.2 x 10 7.0599% 4.6 x 10 4.6889%
Slide 117 () / 208 54 What is % ionization of a 0.20M solution of nitrous acid (HNO 2) that has a ph of 2.02? Slide 118 / 208 55 What is the K a of a 0.115M solution of a weak acid that has a 2.55% ionization?.0477 4.77% 5.99%.0599%.6889% [H + ] = 10 2.02 = 9.55 x 10 3 % ionization = [H + ]/[HNO 2] x 100 % = 9.55 x 10 3 / 0.20 x 100 % = 4.77% 4.6 x 10 3 7.45 x 10 1 7.48 x 10 5 9.0 x 10 4 1.2 x 10 6 Slide 118 () / 208 55 What is the K a of a 0.115M solution of a weak acid that has a 2.55% ionization? 4.6 x 10 3 7.45 x 10 1 7.48 x 10 5 9.0 x 10 4 1.2 x 10 6 % ionization = [H + ]/[H] x 100 % [H + ] = % ionization/100 x [H] [H + ] = 2.55/100 x 0.115 [H + ] = 2.93 x 10 3 Ka = [H + ] 2 / [H] Ka = (2.93 x 10 3 ) 2 / 0.115 Ka = 7.48 x 10 5 Slide 119 / 208 56 If the acid dissociation constant, K a for an acid H is 8x10 4 at 25, what percent of the acid is dissociated in a 0.50molar solution of H? 0.08% 0.2% 1% 2% 4% Question from the ollege oard Slide 119 () / 208 56 If the acid dissociation constant, K a for an acid H is 8x10 4 at 25, what percent of the acid is dissociated in a 0.50molar solution of H? Slide 120 / 208 alculating ph from K a alculate the ph of a 0.30 M solution of acetic acid, H 2H 3O 2, at 25. K a for acetic acid at 25 is 1.8 x 10 5. 0.08% 0.2% 1% 2% K a = [H + ] 2 / [H] [H + ]= K a x [H] [H + ] = 8 x 10 4 x 0.50 [H+] = 0.02 % ionization = [H + ]/[H] x 100% % ionization = 0.02 / 0.50 x 100 % % ionization = 4 % First, we write the dissociation equation for acetic acid H 2H 3O 2 (aq) + H 2O (l) H 3O + (aq) + 2H 3O 2 (aq) 4% Question from the ollege oard
Slide 121 / 208 alculating ph from K a alculate the ph of a 0.30 M solution of acetic acid, H 2H 3O 2, at 25. K a for acetic acid at 25 is 1.8 x 10 5. From the dissociation equation, we obtain the equilibrium constant expression: Slide 122 / 208 alculating ph from K a alculate the ph of a 0.30 M solution of acetic acid, H 2H 3O 2, at 25. K a for acetic acid at 25 is 1.8 x 10 5. We next set up an I chart... [H 2H 3O 2], M [H 3O + ], M [2H3O2], M K a = [H 3O + ][ 2H 3O 2 ] [H 2H 3O 2] Initial 0.30 M 0 0 hange x + x + x quilibrium about 0.30 M x x We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. Slide 123 / 208 Slide 124 / 208 alculating ph from K a alculate the ph of a 0.30 M solution of acetic acid, H 2H 3O 2, at 25. K a for acetic acid at 25 is 1.8 x 10 5. Now, substituting values from the I chart into the K a expression yields 1.8 # 10 5 = (x) 2 (0.30) (1.8 # 10 5 ) (0.30) = x 2 5.4 x 10 6 = x 2 2.3 x 10 3 = x Remember what your "x" is! In this case, it's [H 3O + ], but other times it's [OH ]. alculating ph from K a alculate the ph of a 0.30 M solution of acetic acid, H 2H 3O 2, at 25. K a for acetic acid at 25 is 1.8 x 10 5. ph = log [H 3O + ] ph = log (2.3 x 10 3 ) ph = 2.64 Significant figure rules for ph on the P exam: The calculated ph value should have as many IML places as the [H + ] has sig figs. So if the [H + ] has 2 sig figs, report the ph to the 0.01 place value. If the [H + ] has 3 sig figs, report the ph to the 0.001 place value. Slide 125 / 208 57 What is the hydrogen ion concentration of a weak acid that has a dissociation constant is 1 x 10 6 and a concentration of 0.01M? 1 x 10 6 1 x 10 5 1 x 10 4 1 x 10 3 Slide 126 / 208 58 What is the ph of an acid that has an acid dissociation constant K a of 3.2 x 10 4 and the acid concentration is 0.122M? 2.00 2.20 2.50 2.17
Slide 126 () / 208 58 What is the ph of an acid that has an acid dissociation constant K a of 3.2 x 10 4 and the acid concentration is 0.122M? Slide 127 / 208 59 What is the concentration of the acid if the ph is 4 and the K a is 1 x 10 7? 2.00 2.20 2.50 2.17 K a = [H + ] 2 / [H] 3.2 x 10 4 = x 2 /0.122 x 2 = 3.2 x 10 4 x 0.122 x 2 = 3.904 x 10 5 x = 6.25 x 10 3 ph = log (6.25 x 10 3 ) ph = 2.20 1 x 10 7 1 x 10 5 1 x 10 3 1 x 10 1 Slide 127 () / 208 59 What is the concentration of the acid if the ph is 4 and the K a is 1 x 10 7? Slide 128 / 208 60 What is the ph of a 0.05M solution of acetic acid (H 2H 3O 2) that has a % ionization of 1.22%? 1 x 10 7 K a = [H + ] 2 /[H] 1 x 10 5 [H] = [H + ] 2 / K a [H + ] = 10 ph 1 x 10 3 [H + ] = 10 4 [H] = (1 x 10 4 ) 2 /1 x 10 7 1 x 10 1 [H] = 1 x 10 1 1.15 4.22 3.21 2.13 6.78 Slide 128 () / 208 60 What is the ph of a 0.05M solution of acetic acid (H 2H 3O 2) that has a % ionization of 1.22%? 1.15 4.22 3.21 2.13 6.78 % ionization = [H + ] /[H] x 100% [H+] = % ionization/100 x [H] [H+] = 1.22/100 x 0.05M [H+] = 6.1 x 10 4 ph = log [H + ] ph = =3.21 Slide 129 / 208 Weak ases ases react with water to produce a hydroxide ion. ven though NH 3 does not have the hydroxide ion, OH, in its formula, it is a base according to both the rrenhius and ronstedlowry definitions.
Slide 130 / 208 Weak ases + H 2O H + OH The equilibrium constant expression for this reaction is [H][OH ] K b = [ ] where K b is the basedissociation constant. Slide 131 / 208 61 Which base has the smallest base dissociation constant, K b? potassium hydroxide sodium hydroxide calcium hydroxide ammonia Just as for K a, the stronger a base is, it will have a higher K b value. In fact, since the strong bases dissociate 100%, their K b values are referred to as "very large". K b can be used to find [OH ] and, ultimately, ph. Slide 131 () / 208 61 Which base has the smallest base dissociation constant, K b? Slide 132 / 208 62 base has a dissociation constant, K b = 2.5 x 10 11. Which of the following statements is true? potassium hydroxide This is a concentrated base. sodium hydroxide This base ionizes slightly in aqueous solution. calcium hydroxide mmonia is the only weak base so ammonia its K b is very small. This is a strong base. n aqueous solution of this would be acidic. Slide 132 () / 208 Slide 133 / 208 62 base has a dissociation constant, K b = 2.5 x 10 11. Which of the following statements is true? This is a concentrated base. aqueous solution. Its Kb value is This base ionizes slightly in aqueous solution. This is a strong base. This base ionizes slightly in very small which means the amount of reactants is greater than the amount of the products. n aqueous solution of this would be acidic. alculating K b from ph What is the K b of a 0.20 M solution of hydrazine H 2NNH 2 at 25 that has a ph of 10.9? First, we write the dissociation equation for hydrazine H 2NNH 2(aq) + H 2O (l) OH (aq) + H 2NNH 3+ (aq) From the dissociation equation, we obtain the equilibrium constant expression: K b = [OH ][H 2NNH 3+ ] [H 2NNH 2]
Slide 134 / 208 alculating K b from the ph poh = 14 ph poh = 14 10.9 poh = 3.1 3.1 = log [OH ] 3.1 = log [OH ] 10 3.1 = 10 log [OH] = [OH ] 7.94 x 10 4 = [OH ] = [H 2NNH 3+ ] Slide 135 / 208 alculating K b from the ph What is the K b of a 0.20 M solution of hydrazine H 2NNH 2 at 25 that has a ph of 10.9? H 2NNH 2(aq) + H 2O (l) OH (aq) + H 2NNH 3+ (aq) [H 2NNH 2], M [OH ], M [H 2NNH 3+ ], M Initially 0.20 0 0 hange 7.94 x 10 4 +7.94 x 10 4 +7.94 x 10 4 t quilibrium 0.20 7.94 x 10 4 0.20 7.94 x 10 4 7.94 x 10 4 Slide 136 / 208 alculating K b from the ph What is the K b of a 0.20 M solution of hydrazine H 2NNH 2 at 25 that has a ph of 10.9? Now, we substitute values into the K b expression and solve: Slide 137 / 208 alculating Percent Ionization What would be the analogous formula to calculate percent ionization for a base? K b = K b = [OH ][H 2NNH 3+ ] [H 2NNH 2] [7.94 x 10 4 ] [7.94 x 10 4 ] [0.20] Percent Ionization = [OH ] eq # 100% [ase] initial K b = 3.15 x 10 6 Slide 138 / 208 63 alculate the K b of a 0.450M solution of weak base solution with a poh of 4.98. Slide 138 () / 208 63 alculate the K b of a 0.450M solution of weak base solution with a poh of 4.98. 3.22 x 10 7 2.11 x 10 5 2.03 x 10 18 2.33 x 10 5 2.44 x 10 10 3.22 x 10 7 2.11 x 10 5 2.03 x 10 18 2.33 x 10 5 2.44 x 10 10 K b = [OH ] 2 /[] [OH] = 10 poh [OH ] = 10 4.98 [OH ] = 1.05 x 10 5 K b = (1.05 x 10 5 ) 2 / 0.450 K b = 2.44 x 10 10
Slide 139 / 208 64 alculate the K b of a 0.724M solution of hypobromite ion (ro ) that has a ph of 11.23. 4.80 x 10 23 4.00 x 10 6 2.35 x 10 3 1.22 x 10 4 6.10 x 10 5 Slide 139 () / 208 64 alculate the K b of a 0.724M solution of hypobromite ion (ro ) that has a ph of 11.23. [H + ] = 10 ph [H + ] = 10 11.23 4.80 x 10 23 [H + ] = 5.89 x 10 12 [OH ] = 1 x 10 14 / [H + ] [OH 4.00 x 10 ] = 1 x 10 14 / 5.89 x 10 12 6 [OH ] = 1.70 x 10 3 K b = [OH ] 2 /[ro ] 2.35 x 10 3 K b = (1.70 x 10 3 ) 2 x 0.724 K b = 4.00 x 10 6 1.22 x 10 4 6.10 x 10 5 Slide 140 / 208 65 What is the K b of 0.125M solution of a weak base that is 1.25% ionized? Students type their answers here Slide 141 / 208 alculating ph from the K b The same process we followed to determine K a from the ph we can use to determine K b from the ph. What is the ph of a 0.15 M solution of NH 3? NH 3 (aq) + H 2O (l) NH 4 + (aq) + OH (aq) First write the equilibrium expression for the dissociation equation. Obtain the K b value from an earlier page. K b = [NH 4+ ][OH ] [NH 3] =1.8 x 10 5 Slide 142 / 208 Slide 143 / 208 alculating ph from the K b What is the ph of a 0.15 M solution of NH 3? NH 3 (aq) + H 2O (l) NH + 4 (aq) + OH (aq) [NH 3], M [NH 4+ ], M [OH ], M Initially 0.15 0 0 hange x +x +x alculating ph from the K b 1.8 10 5 = (1.8 # 10 5 ) (0.15) = x 2 2.7 x 10 6 = x 2 1.6 x 10 3 = x (x) 2 (0.15) gain, remember what your "x" is! t quilibrium 015 x 0.15 x x