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 E 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 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? E 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 2 SO 4 HSO 4 H 2 SO 4 NH 3 E H 3 OOH 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 E None of the above H 2 SO 4 H 2 O H 3 O + HSO 4 E 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 + 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 + E cid ase Neither oth None of the above E cid ase oth Neither None of the above Slide 21 / 208 Slide 22 / 208 8 Which of the following cannot act as a Lewis base? l 9 In the reaction F 3 + F F 4 F 3 acts as a/an acid. NH 3 F 3 N E H 2 O E rrhenius ronstedlowry Lewis rrhenius, ronstedlowry, and Lewis rrhenius and ronstedlowry Slide 23 / 208 Slide 24 / 208 onjugate cids and ases onjugate cid ase Pairs 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 + Return to the Table of contents 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 cids in Water What Happens When an cid issolves in Water? Which is the acid? Which is the base? 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 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 lo 4+, HS, PH 3, O 3 both an acid and a base lo 4, HS, PH 3, O 3 2 the conjugate acid of NH 3 E the reducing agent the oxidizing agent lo 4 2, HS 2, PH 3 3, O 3 2 lo 4, H 2S, PH 3, O 3 2 Question from the ollege oard 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 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 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. Slide 33 / 208 cid and ase Strength Hl (aq) + H 2O (l) H 3O + (aq) + l (aq) acid base conj. acid conj. base Slide 34 / 208 cid and ase Strength onsider this equilibrium between acetic acid and acetate ion: In this example, H 2O is a much stronger base than l, so the proton moves from Hl to H 2O onversely, Hl is a much stronger acid than the hydronium ion, so equilibrium lies very far to the right K >>1 100% 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 H 3OOH (aq) + H 2O (l) If you look for the stronger acid: Equilib 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: Equilib favors this base accepting a proton. Strong Weak Neg Slide 35 / 208 cid and ase Strength H 3OOH (aq) + H 2O (l) 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 2 H2PO4 HPO4 + NH4 NH3 2 HO3 O3 2 3 HPO4 PO4 H2O OH OH O 2 Strong Weak Negligible H2 H H4 H3 ase strength increases H 3O + (aq) + H 3OO (aq) 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. The double headed arrow is used only in weak acid or weak base dissociation equations. H 3OOH (aq) + H 2O (l) single arrow is used for strong acid or strong bases which dissociate completely. NaOH H 3O + (aq) + H 3OO (aq) 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 aqueous solutions of Hr contain equal concentrations of H + and OH does not dissociate at all when it is dissolved in water negative cannot be neutralized by a base 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 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 2 SO4 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 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 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 mphoteric Substances Return to the Table of ontents
Slide 43 / 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 Slide 44 / 208 mphoteric Substances 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. E autosomal conjugated amphoteric saturated miscible Slide 46 / 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. Slide 47 / 208 Slide 48 / 208 Strong cids Recall, strong acids completely ionize in solution. The seven strong acids are: Strong cids and ases Return to the Table of ontents 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 Memorize this list.
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 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? 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 0.44 0.88 E Not enough information. 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 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? nswer
Slide 55 / 208 Slide 56 / 208 utoionization of Water 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 61 / 208 24 The ionproduct constant for water, K w is represented by [H 2O] 2 [H 3O + ] x [OH ] [H 3O + ] + [OH ] [H 3O + ] [OH ] 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 1 x 10 5 1 x 10 5 1.23 x 10 5 E 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 E 1 x 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 5 E 8. 13 x 10 6 Slide 65 / 208 Slide 66 / 208 ph 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. Return to the Table of contents 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 Slide 68 / 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. 1.0 x 10 10 10 Is the relationship between [H + ] and ph a direct or an inverse one? Slide 69 / 208 Slide 70 / 208 28 The molar concentration of hydronium ion, [H 3O + ], in pure water at 25 is. 0 1 7 10 7 E 10 14 Slide 71 / 208 29 solution with ph = 3 has a hydrogen ion concentration that is than a solution with ph = 5. Slide 72 / 208 30 solution with ph = 14 has a hydrogen ion concentration that is than a solution with ph = 11. 2x more 2x less 100x more 100x less 3x more 3x less 1000x more 1000x less
Slide 73 / 208 Slide 74 / 208 ph Therefore, in pure water ph = log [H 3O + ] = 7.00 ph = log (1.0 # 10 7 ) = 7.00 n acid has a higher [H 3O + ] than pure water, so its ph is <7. I [H+] > [OH] There are excess hydrogen ions in solution. ph SE [H+] < [OH] There are excess hydroxide ions in solution. 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 Slide 75 / 208 Slide 76 / 208 These are the ph values for several common substances. More basic More acidic ph 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Slide 77 / 208 leach Soapy water mmonia Milk of Magnesia aking soda Sea water istilled water Urine lack coffee Tomato juice Orange juice Lemon juice Gastric acid 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 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. ph = 3.21 ph = 7.00 ph = 8.93 ph = 12.6
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 ph =11.4 2 6 7 9 E 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 10 2 12 2 10 12 E 13 Slide 83 / 208 38 solution with the ph of 5.0 is basic has a hydrogen ion concentration of 5.0M is neutral has a hydroxideion concentration of 1x10 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 + nswer: 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 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 41 Slide 90 / 208 What is the ph of an aqueous solution at 25.0 in which [H + ] is 0.0025 M? E 3.4 2.6 2.6 3.4 2.25
Slide 91 / 208 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. Slide 92 / 208 42 What is the ph of a solution whose hydronium ion concentration is 7.14 x 10 3 M? 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 93 / 208 43 What is the ph of a solution whose hydronium ion concentration is 1.92 x 10 9 M? Slide 94 / 208 44 What is the hydronium 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? 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 pk a = log K a pk b = log K b
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 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 2.11 E 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 E 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 E 4.03x10 7 Slide 101 / 208 Slide 102 / 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 Weak cids and ases 5.67x10 8 9.07x10 9 E 4.88x10 7 Return to the Table of contents
Slide 103 / 208 Slide 104 / 208 cid issociation onstants, K a cid issociation onstants, K a H (aq) + H 2O (l) (aq) + H 3O + (aq) The greater the value of K a, the stronger is the acid. 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] 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 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 Equilibrium 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 E 6.78 x 10 12
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% E.6889% 1.3 x 10 5 9.9 x 10 2 1.2 x 10 8 4.2 x 10 7 E 4.6 x 10 4 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% E.6889% 4.6 x 10 3 7.45 x 10 1 7.48 x 10 5 9.0 x 10 4 E 1.2 x 10 6 Slide 119 / 208 Slide 120 / 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% 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. First, we write the dissociation equation for acetic acid 0.2% 1% 2% E 4% H 2H 3O 2 (aq) + H 2O (l) H 3O + (aq) + 2H 3O 2 (aq) 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 IE 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 Equilibrium 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 IE 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 EIML 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 nswer 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 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 1 x 10 5 1 x 10 3 1 x 10 1 1.15 4.22 3.21 2.13 E 6.78 Slide 129 / 208 Weak ases ases react with water to produce a hydroxide ion. Slide 130 / 208 Weak ases + H 2O H + OH The equilibrium constant expression for this reaction is K b = [H][OH ] [ ] Even 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. where K b is the basedissociation constant. 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 Slide 132 / 208 61 Which base has the smallest base dissociation constant, K b? 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 This is a strong base. ammonia n aqueous solution of this would be acidic.
Slide 133 / 208 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 Slide 134 / 208 alculating K b from the ph poh = 14 ph poh = 14 10.9 poh = 3.1 H 2NNH 2(aq) + H 2O (l) OH (aq) + H 2NNH 3+ (aq) 3.1 = log [OH ] From the dissociation equation, we obtain the equilibrium constant expression: K b = [OH ][H 2NNH 3+ ] [H 2NNH 2] 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 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: K b = [OH ][H 2NNH 3+ ] [H 2NNH 2] Initially 0.20 0 0 hange 7.94 x 10 4 +7.94 x 10 4 +7.94 x 10 4 K b = [7.94 x 10 4 ] [7.94 x 10 4 ] [0.20] t Equilibrium 0.20 7.94 x 10 4 0.20 7.94 x 10 4 7.94 x 10 4 K b = 3.15 x 10 6 Slide 137 / 208 alculating Percent Ionization What would be the analogous formula to calculate percent ionization for a base? Percent Ionization = [OH ] eq # 100% [ase] initial 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 E 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. 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 4.80 x 10 23 4.00 x 10 6 2.35 x 10 3 nswer 1.22 x 10 4 E 6.10 x 10 5 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. Slide 142 / 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 K b = [NH 4+ ][OH ] [NH 3] =1.8 x 10 5 hange x +x +x t Equilibrium 015 x 0.15 x x Slide 143 / 208 Slide 144 / 208 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! alculating ph from the K b In this case, "x" is [OH ]. Therefore, [OH ] = 1.6 x 10 3 M poh = log (1.6 x 10 3 ) poh = 2.80 So, now solving for ph: ph = 14.00 2.80 ph = 11.20
Slide 145 / 208 66 What is the OH concentration of a 0.550M solution of the weak base aniline, 6H 5NH 2? (K b = 4.3 x 10 10 ) Slide 146 / 208 67 What is the ph of a 0.550M solution of the weak base aniline, 6H 5NH 2? (K b = 4.3 x 10 10 ) 2.37 x 10 11 1.54 x 10 5 7.69 x 10 10 1.30 x 10 11 2.37 x 10 11 1.54 x 10 5 7.69 x 10 10 1.30 x 10 11 Slide 147 / 208 68 What is the ph of a 0.135M solution of trimethylamine, (H 3) 3N? (K b = 6.4 x 10 5 ) Slide 148 / 208 69 What is the ph of a 0.112M solution of ethylamine ( 5H 5NH 2)? (K b = 6.4 x 10 4 ) Students type their answers here 8.94 11.47 2.53 5.06 E 3.45 Slide 149 / 208 Slide 150 / 208 Polyprotic cids Polyprotic cids Polyprotic acids are characterized by having more than one acidic proton. Here are some examples: Sulfuric acid H 2SO 4 Phosphoric acid H 3PO 4 arbonic acid H 2O 3 Return to the Table of contents Oxalic acid H 2 2O 4
Slide 151 / 208 Polyprotic cids Polyprotic acids have a K a value for each proton that can be removed. For example, consider carbonic acid, H 2O 3. The first ionization equation for carbonic acid is H 2O 3 + H 2O HO 3 + H 3O + Write the Ka expression for this equation. This is referred to as K a1. Slide 152 / 208 Polyprotic cids Now, we examine both dissociation constants. The second ionization equation for carbonic acid is HO 3 + H 2O O 3 2 + H 3O + Write the Ka expression for this equation. This is referred to as K a2. Slide 153 / 208 Polyprotic cids Notice that the bicarbonate ion, HO 3, appears in each expression; in the numerator for K a1 and in the denominator for K a2. [HO 3 ][H 3O + 2 ][O 3 ][H 3O + ] K a1 x K a2 = [H 2O 3] [HO 3 ] [HO 3 ][H 3O + 2 ][O 3 ][H 3O + ] K a1 x K a2 = [H 2O 3] [HO 3 ] Slide 154 / 208 Polyprotic cids The equation for the complete ionization of carbonic acid is H 2O 3 + H 2O 2 O 3 + 2 H 3O + and the K a expression for this reaction is [O 2 3 ][H 3O + ] 2 K = [H 2O 3] [HO 3 ][H 3O + 2 ][O 3 ][H 3O + ] K a1 x K a2 = [H 2O 3] [HO 3 ] K a1 x K a2 = overall K So, the product of K a1 x K a2 for a diprotic acid yields the overall K for complete dissociation. Slide 155 / 208 Polyprotic cids Generally, the ph of polyprotic acids depends only on the removal of the first proton. Slide 156 / 208 70 The K a of carbonic acid is 4.3 x 10 7 H 2O 3 H + + HO 3 This means that H 2O 3 is a. This holds true when the difference between the K a1 and K a2 values is at least 10 3. good hydrogenion acceptor good hydrogenion donor poor hydrogenion acceptor poor hydrogenion donor
Slide 157 / 208 71 diprotic acid, H 2X, has the following dissociation constants: K a1 = 2.0 x 10 4 K a2 = 3.0 x 10 6 What is the overall K value for this acid? 5.0 X 10 10 6.0 X 10 10 5.0 X 10 24 6.0 x 10 24 E 6.0 x 10 24 Slide 158 / 208 72 diprotic acid, H 2X, has the following dissociation constants: K a1 = 2.0 x 10 4 K a2 = 3.0 x 10 6 What is the ph of a 0.10 M solution of H 2X? 5.11 6.11 10.22 2.11 E 5.21 Slide 159 / 208 73 bsorbic acid, H 2 2H 6O 6 has the following dissociation constants: K a1 = 8.0 x 10 5 K a2 = 1.6 x 10 12 What is the ph of a 0.20 M solution of bsorbic acid? 2.40 8.30 3.00 4.20 E 11.2 Slide 160 / 208 The Relationship etween K a and K b Return to the Table of ontents Slide 161 / 208 K a and K b For a conjugate acidbase pair, K a and K b are related in a special way. Write the ionization equations for the following: 1) reaction of ammonia (NH 3) and water NH 3 + H 2O NH 4 + + OH Slide 162 / 208 K a and K b NH 4 + + H 2O NH 3 + H 3O + Write the corresponding equilibrium constant expression for each of these equations. 2) reaction of ammonium ion (NH 4+ ) and water Which of these expressions is referred to as "K a"? Which is referred to as "K b"?
Slide 163 / 208 K a and K b Slide 164 / 208 K a and K b NH 3 + H 2O NH 4 + + OH NH 4 + + H 2O NH 3 + H 3O + K b x K a = [NH 4+ ][OH ] [NH 3] [NH 3][H 3O + ] [NH 4+ ] K b = [NH4+ ][OH ] [NH 3] K a = [NH3][H3O+ ] [NH 4+ ] K b x K a = [NH 4+ ][OH ] [NH 3] [NH 3][H 3O + ] [NH 4+ ] What do these expressions have in common? K b x K a = [OH ][H 3O + ] = K w So, the product of K a x K b for any conjugate acidbase pair yields the ionproduct constant, K w. Slide 165 / 208 Slide 166 / 208 K a and K b 74 For the acid HN, what is the equation on which the K a expression is based? For a specific conjugate acidbase pair, K a and K b are related in this way: K a x K b = K w Therefore, if you know the value of one of them, you can calculate the other. HN <> OH + HN HN <> OH + N HN + H 2O <> N + H 3O + HN + H 2 O <> OH + N Slide 167 / 208 Slide 168 / 208 75 The K a for an acid, HX, is 2.0 x 10 4. What is the K b for its conjugate base, X? No calculator. 5.0 X 10 11 2.0 X 10 10 5.0 X 10 10 8.0 x 10 10 E 2.0 x 10 4 76 The K b for NH 3 is 1.8 x 10 5. What is the K a for the ammoninum ion, NH 4+? 2.5 X 10 3 2.5 X 10 4 2.5 X 10 5 4.3 x 10 4 E 5.6 x 10 11
Slide 169 / 208 Slide 170 / 208 77 Which of the substances below is the strongest acid? HlO HO 3 H 2S NH 2H + 3 cidase Properties of Salt Solutions E H 2S and HlO Return to the Table of ontents Slide 171 / 208 Neutralization Reactions Neutralization reactions are a special class of doublereplacement reactions that occur between an acid and a base. Recall from last year that the general formula for a doublereplacement reaction is: Slide 172 / 208 Neutralization Reactions The general formula for any acidbase neutralization reaction is: acid + base > salt + water Note that the term "salt" refers to any ionic compound that does not include H + or OH. + > + oublereplacement reactions are also known as ionexchange or precipitation reactions. Slide 173 / 208 cidase Properties of Salt Solutions acid + base > salt + water When an acid and base react together, the resulting solution is not always neutral. The ph of the resulting mixture depends on the relative strengths of the acid and of the base. The ph of a salt solution depends on the ability of the salt ions to hydrolyze. Hydrolyze means to interact with water molecule and dissociate. Slide 174 / 208 nion Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. Each salt contains a positive ion, a cation, and a negative ion, the anion. Let's consider the anion first which can be denoted, X. X is the conjugate base of the acid in the neutralization reaction. This anion may undergo hydrolysis as shown below: X + H 2O Predict the products. o you think the resulting solution will be acidic, basic or neutral? Justify your response.
Slide 175 / 208 nion Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. The conjugate base (l ) of a strong acid (Hl) will not hydrolyze, it will not react with water because Hl dissociates completely. In fact, none of the anions from the strong seven acids will hydrolyze. In other words, these anions will not affect ph. List here the conjugate bases of the strong seven acids. Slide 176 / 208 nion Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. The conjugate base (NO 2 ) of a weak acid (HNO 2) will hydrolyze, it will react with water because HNO 2 does not dissociate completely and a dynamic equilibrium exists. Write the equilibrium expression for the reaction of NO 2 and water. The conjugate bases of strong acids will not undergo hydrolysis and will not affect ph. ll conjugate bases of weak acids will hydrolyze and the resulting solutions will be basic. Slide 177 / 208 nion Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. The conjugate base (HSO 3 ) of a weak acid (H 2SO 3) is a special case because HSO 3 is amphoteric. It can behave as either an acid or as a base. How it behaves is determined by the relative magnitude of K a for its acid and K b for its base. If K a> K b it will behave as an acid if K b> K a then it will behave as a base. Slide 178 / 208 ation Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. The metal cation (K + ) of a strong base (KOH) will not hydrolyze. In fact, none of the cations from the strong bases will hydrolyze. In other words, these metal cations in solution will not affect ph. List here the metal cations from Groups 1 and 2 that form strong bases. When amphoteric anions hydrolyze they will make acidic solutions if their K a>k b or make basic solutions if their K b>k a. None of the cations of strong bases will undergo hydrolysis and affect ph. Slide 179 / 208 ation Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. The conjugate acid (NH 4+ ) of a weak base (NH 3) will hydrolyze and the resulting solution will be acidic. Write the equilibrium expression for the reaction of NH 4 + and water. Slide 180 / 208 ation Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. Metal cations other than Group 1 and heavy Group 2 cations can hydrolyze to affect ph. This effect is greatest for small, highly charged cations like Fe 3+ and l 3+. The effect of these cations increases as ionic charge increases. The K a for Fe 2+ is 3.2 x 10 10 and for Fe 3+ is 6.3 x 10 3. These values are in the range for some weak acids we are familiar with like acetic acid K a = 1.8 x 10 5. ll cations that are conjugate acids of weak bases will undergo hydrolysis and will affect ph.
Slide 181 / 208 cidase Properties of Salt Solutions We have observed that metal cations (hydrated in solution) lower the ph of a solution. Lone pairs on oxygen are attracted to the cation. The shift of electron density in water makes the OH bond more polar. Hydrogen atoms become more acidic, and thus more likely to become H + ions in solution. Shift in electron density away from hydrogen atoms. Slide 182 / 208 cidase Properties of Salt Solutions We have observed that metal cations (hydrated in solution) lower the ph of a solution. ompare the following: attraction between a 1 + cation and H 2O molecule. attraction between a 3 + cation and H 2O molecule. The cation with the greater charge will make the solution more acidic. Slide 183 / 208 ation Effect on ph The ph of a salt solution depends on the ability of the salt ions to hydrolyze. onsider the ability of the aluminum Fe 3+ to hydrolyze: Fe 3+ + H 2O Fe(OH) 3 + H + The increased H + concentration reduces the ph. Small positively charged cations will hydrolyze and form acidic solution. Slide 184 / 208 ombined Effect of ation and nion If a salt contains an anion that does not react with water like l and a cation that does not react with water like Na + the solution will be neutral and have a ph of 7. Why do l and Na + not react with water? l is the conjugate base of a strong acid and Na + is the cation of a strong base. Neither would react with water. Slide 185 / 208 ombined Effect of ation and nion If a salt contains an anion that reacts with water to produce hydroxide ions and a cation that does not react with water then the resulting solution will be basic. an you give two examples of this type of salt? Slide 186 / 208 ombined Effect of ation and nion If the salt contains an cation that reacts with water and an anion that does not, the resulting solution will acidic. This results when the cation is the conjugate acid of a weak base or a small metal ion with a 2+ or greater charge. an you give two examples of this type of salt? Li 3PO 3, a(lo 2) 2, Na 2H 3O 2 NH 4l, Fe(NO 3) 3,l(lO 4) 3
Slide 187 / 208 ombined Effect of ation and nion If a salt contains a cation that reacts with water and an anion that reacts with water then the solution's ph will be determined by comparing the cation's K a to the anion's K b. If K a> K b then the solution will be acidic if K b<k a then the solution will be basic. For example, let's consider a salt solution of Fe(lO) 3, K a for Fe 3+ 6.3 x 10 3 is the K b for lo is 3.33 x 10 7 This solution would be acidic because K a>k b. Slide 188 / 208 78 The ph of a solution of 0.1M solution of Nal will be approximately. 2 5 7 11 E 13 Slide 189 / 208 79 The ph of a solution of 0.1 M solution of K 3PO 4 will be approximately. 2 5 7 11 E 13 Slide 190 / 208 80 The ph of a solution of 0.1M solution of LiHO 3 at 25 o will be approximately. 2 5 7 11 E 13 Slide 191 / 208 81 0.1M aqueous solution of will have a ph of 7.0 at 25 o. NaOl ar 2 NH 4 l Slide 192 / 208 82 0.1M aqueous solution of has a ph of less than 7.0 at 25 o. Na 2S KF NaNO 3 a( 2 H 3 O 2 ) 2 E none of these E NH 4l NaF
Slide 193 / 208 83 n aqueous solution of will produce a basic solution at 25 o. Slide 194 / 208 84 1molar solution of which of the following salts has the highest ph? NH 4 lo 4 Kr Nal ao 3 E Fe(lO 3) 2 NaNO 3 Na 2O 3 NH 4l NaHSO 4 E Na 2SO 4 Question from the ollege oard Slide 195 / 208 Slide 196 / 208 85 n aqueous solution of will produce a neutral solution. NH 4 lo 4 lr 3 Fel 3 SrO 3 E al 2 86 Of the following substances, an aqueous solution of will form basic solutions. NH 4l, u(no 3) 2, K 2O 3, NaF NH 4 l and u(no 3 ) 2 NH 4 l and K 2 O 3 E K 2 O 3 and NaF NaF only NH 4 l only Slide 197 / 208 cidase Properties of Salt Solutions Summary: nions The conjugate bases of strong acids will not undergo hydrolysis and not affect ph. (Examples: l,i, NO 3 ) The conjugate bases of weak acids will hydrolyze and will raise ph by creating excess OH ions. (Examples: NO 2, lo, O 3 2 ) mphoteric anions will either act acidic or basic based on the K a and K b values of the acid and base. (Examples: HO 3, H 2PO 4 ) Slide 198 / 208 cidase Properties of Salt Solutions Summary: ations ations from strong bases will not undergo hydrolysis and will not affect ph. (Examples, Na +, Li +, a 2+ ) The conjugate acids of weak bases will hydrolyze and will lower ph by creating excess H + ions. (Examples: NH 4+, 2H 5NH 3+ ) Small metal cations from weak bases will hydrolyze and lower ph by creating excess H + ions. The higher the charge on the cation the greater the affect on ph. (Examples Fe 2+, Fe 3+, l 3+ )
Slide 199 / 208 Slide 200 / 208 Factors ffecting cid Strength Factors ffecting cid Strength The more polar the HX bond and/or the weaker the HX bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group. Return to the Table of ontents Slide 201 / 208 Factors ffecting cid Strength In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. K a = 3.0x10 8 K a= 2.3x10 11 In HOl, the electron density will be shifted to the more electronegative l atom weakening the OH bond. The strength of the acid will be in the order Slide 202 / 208 Factors ffecting cid Strength mong oxyacids, the more oxygen atoms in the molecule, the stronger the acid would be. The electron density will be more towards the oxygen atoms and the oxidation number increases, the strength of the acid increases. For a series of oxyacids, acidity increases with the number of oxygen atoms. HOl > HOr > HOI Slide 203 / 208 Factors ffecting cid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic. Slide 204 / 208 87 Which of the following would be the strongest acid: Hr, Hl or HI? Hl Hr HI they would all be equally strong acids
Slide 205 / 208 88 H 2S is a stronger acid than H 2O because. The strength of the HX bond is greater in H 2O The strength of the HX bond is less than in H 2O The HX bond is more polar in H 2O than in H 2S The HX bond is more polar in H 2S than in H 2O Slide 206 / 208 89 Order the following acids from weakest to strongest: H 2O, HF, NH 3, and H 4. HF<H 2O<NH 3<H 4 HF<NH 3<H 2O<H 4 H 4<HF<NH 3<H 2O NH 3<H 4<H 2O<HF E H 4<NH 3<H 2O<HF nswer Slide 207 / 208 90 rrange the following compounds in order of decreasing acid strength: H 2SO 4, H 2SeO 3, H 2SeO 4. Justify your Students type their answers here response. Slide 208 / 208