(Hebden Unit 4 page 109 182) 182) We will cover the following topics: 1. Definition of Acids and Bases 2. Bronsted-Lowry Acids and Bases 2 1
Arrhenius Definition of Acids and Bases An acid is a substance that reacts with water to produce hydronium ions, H 3 O +. A base is a substance that produces hydroxide ions, OH,in water. Hydroxide ion OH - When an acid and base reacts, an ionic compound is produced as one of the products. We call ionic compounds salts. 3 Arrhenius Theory of Acid and Bases Support for the Arrhenius concept comes from measurements of H of neutralization of any strong acid and base. Consider the following reactions: NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O() LiOH(aq) + HBr(aq) LiBr(aq) + H 2 O() H = -56 kj/mol H = -56 kj/mol If we write the net ionic equation for both of these reactions we see that they are the same H + (aq) + OH (aq) H 2 O() 4 2
Monoprotic and Polyprotic acids 1. Monoprotic acids capable of ionizing 1 H + ion per molecule of acid. Strong acid = 100% dissociation into ions Example: HCl (aq) + H 2 O () H 3 O + (aq) + Cl (aq) CH 3 COOH (aq) + H 2 O () H 3 O + (aq) + CH 3 COO (aq) Weak acid = partial dissociation into ions 5 Monoprotic and Polyprotic acids 2. Polyprotic acids capable of ionizing more than one H + ion per molecule of acid. Example of a Diprotic acid: H 2 SO 4 (aq) + H 2 O () H 3 O + (aq) + HSO 4 (aq) HSO 4 (aq) + H 2 O () H 3 O + (aq) + SO 42 (aq) Example of a Triprotic acid: H 3 PO 4 (aq) + H 2 O () H 3 O + (aq) + H 2 PO 4 (aq) H 2 PO 4 (aq) + H 2 O () H 3 O + (aq) + HPO 42 (aq) HPO 2 4 (aq) + H 2 O () H 3 O + (aq) + PO 43 (aq) You should be able to write acid dissociation reactions. Pay attention to the charges on all the ions, and balance the reactions. 6 3
Monoprotic and Polyprotic bases 1. Monoprotic bases capable of ionizing 1 OH - ion per molecule of base. Example: Strong base = 100% dissociation into ions NaOH (aq) Na + (aq) + OH (aq) Organic nitrogen compounds, amines, are bases because they dissolve in water to produce OH - ions. NH 3 (aq) + H 2 O () NH 4+ (aq) + OH (aq) CH 3 NH 2 (aq) + H 2 O () CH 3 NH 3+ (aq) + OH (aq) Weak base = partial dissociation into ions 7 Monoprotic and Polyprotic bases 2. Polyprotic bases capable of ionizing more than one OH - ion per molecule of base. Example: Ca(OH) 2 (aq) Ca 2+ (aq) + 2 OH (aq) Al(OH) 3 (aq) Al 3+ (aq) + 3 OH (aq) You should be able to write base dissociation reactions. Pay attention to the charges on all the ions, and balance the reactions. 8 4
In general, 1. Ionization of an acid in water HA (aq) + H 2 O () H 3 O + (aq) + A (aq) 2. Ionization of a base in water B( (aq)+ho 2 () BH + (aq)+oh (aq) or B(OH) x (aq) B x+ (aq) + x OH (aq) 9 BrØnsted-Lowry Acids and Bases When neutralization ation reaction occurs in aqueous solution, Arrhenius definition of acid and base works well. HCl (aq) + NH 3 (aq) NH 4 Cl (aq) Hydrochloric ammonia ammonium acid (base) chloride OH. Recall that Arrhenius Definition of Acid and Base An acid as a substance that reacts with water to produce H 3 O +. A base that reacts with water to produce 10 5
BrØnsted-Lowry Acids and Bases This same neutralization ation reaction can occur when it is not in an aqueous medium. Therefore, Arrhenius s definition of an acid and a base is too restrictive. Example: Neutralization reaction of reactants in gas phase HCl (g) + NH 3 (g) NH 4 Cl (s) Hydrogen ammonia smoke containing chloride gas tiny crystals gas of ammonium chloride We need a broader definition! 11 BrØnsted-Lowry Acids and Bases What is in common between een the gaseous s and aqueous reactions? 1. Both involve a transfer of a H + from one reactant to another. In solution, HCl is completely ionized and the H + ion is transferred from HCl to NH 3. NH 3 (aq) + H 3 O + (aq) + Cl (aq) NH 4 + (aq) + Cl (aq) + H 2 O () In gas phase, NH 3 (g) + HCl (g) NH 4+ Cl (s) 12 6
BrØnsted-Lowry Acids and Bases View acid-base reactions as simply the transfer of H + (protons) from one species to another. BrØnsted-Lowry Acids and Bases Definitions Acid is a proton donor. Base is a proton acceptor. HCl (g) + NH 3 (g) NH 4+ Cl (s) This species donates a proton to NH 3, therefore, it is an acid. This species accepts a proton, therefore, it is a base. 13 BrØnsted-Lowry Acids and Bases In general, HA + B BH + + A H + donor ACID H + acceptor BASE H + donor ACID H + acceptor BASE CONJUGATE ACID-BASE PAIRS Two species that differ by H + are called a conjugate pair. There is a whole table of CHEM BrØnsted-Lowry 0012 Lecture Notes Acids and Bases. 14 7
BrØnsted-Lowry Acids and Bases WEAK Stren ngth of acid STRONG 15 WEAK Strength of base STRONG BrØnsted-Lowry Conjugate Acids-Bases Pairs Acid H 2 O NH 4 + HCO 3 H 3 O + N 2 H 4 H 3 O + NH 3 H 2 CO 3 HPO 42 -H + +H + Base OH NH 3 CO 32 H 2 O N 2 H 3 H 2 O NH 2 HCO 3 PO 43 16 8
We will cover the following topics: 1. Amphiprotic Substance 2. Amphoteric Compounds 17 1. Amphiprotic Substance A substance that can act either as a proton acceptor or a proton donor. All amphiprotic substances contain a hydrogen atom. Example 1 Water, H 2 O H 2 O () + HCl (aq) H 3 O + (aq) + Cl (aq) base acid H 2 O () + NH 3 (aq) NH 4+ (aq) + OH (aq) acid base 18 9
1. Amphiprotic Substance A substance that can act either as a proton acceptor or a proton donor. All amphiprotic substances contain a hydrogen atom. Example 2 Bicarbonate, HCO 3 - You should be able to write these equations. HCO 3 (aq) + H 3 O + (aq) H 2 CO 3 (aq) + H 2 O () base acid HCO 3 (aq) + OH (aq) CO 32 (aq) + H 2 O () acid base 19 1. Amphiprotic Substance A substance that can act either as a proton acceptor or a proton donor. All amphiprotic substances contain a hydrogen atom. Example 3 hydrogen sulfate ion, HSO 4 - You should be able to write these equations. HSO 4 (aq) + H 3 O + (aq) H 2 SO 4 (aq) + H 2 O () base acid HSO 4 (aq) + OH (aq) SO 42 (aq) + H 2 O () acid base 20 10
2. Amphoteric Substance A compound that can react as either an acid or a base. Ampho- is Greek, it means both. Metals have amphoteric oxides. (eg ZnO, Al(OH) 3 ) Example 1 Zinc oxide, ZnO Reacts with acid: ZnO(aq) + 2H 3 O + (aq) Zn 2+ (aq) + 3H 2 O() acts as a base Reacts with base: ZnO (aq) + 2 OH (aq) + H 2 O() [Zn(OH) 4 ] 2 (aq) acts as an acid zincate ion 21 2. Amphoteric Substance A compound that can react as either an acid or a base. Ampho- is Greek, it means both. Metals have amphoteric oxides. (eg ZnO, Al(OH) 3 ) Example 2 Water, H 2 O Reacts with acid: H 2 O () + HCl (aq) H 3 O + (aq) + Cl (aq) acts as a base Reacts with base: H 2 O () + NH 3 (aq) NH 4+ (aq) + OH (aq) acts as an acid 22 11
2. Amphoteric Substance A compound that can react as either an acid or a base. Ampho- is Greek, it means both. Metals have amphoteric oxides. (eg ZnO, Al(OH) 3 ) Example 3 Aluminum hydroxide, Al(OH) 3 Reacts with acid: Al(OH) 3 (aq) +3HCl (aq) AlCl 3 (aq) +3HO 2 () acts as a base Reacts with base: Al(OH) 3 (aq) + OH (aq) [Al(OH) 4 ] (aq) acts as an acid aluminate ion 23 Strong acids An acid which dissociates 100% into its ions in a water solution. Strong acid = 100% dissociation into ions HCl (aq) + H 2 O () H 3 O + (aq) + Cl (aq) The [H + ] (or [H 3 O + ]) concentration of a 0.10 M HCl solution is 0.10 M. For monoprotic strong acids [acid] initial = [H + ] A common notation to express [H + ] is ph, where ph = -log [H + ] = -log 0.10 = 1.00 24 12
Recall the ph scale ph = - log [H + ] 0.1 M 0.01 M 10-10 M 0.001 M 10-9 M 10-11 M For an increase in 1 ph unit, the solution s [H + ] decreases 10x. 25 Strong acids K a These are strong acids. Dissociation is a single arrow to produce ions. 26 13
Weak acids An acid which dissociates less than100% into its ions in a water solution. Weak acid = < 100% dissociation into ions. Use double arrows to indicate this. CH 3 COOH (aq) + H 2 O () H 3 O + (aq) + CH 3 COO (aq) The [H + ] (or [H 3 O + ]) concentration of a 0.10 M CH 3 COOH solution is LESS THAN 0.10 M. Therefore, ph >1.00 In weak acids [acid] initial [H + ], [acid] initial > [H + ] 27 Weak acids An acid which dissociates less than100% into its ions in a water solution. HA (aq) + H 2 O () H 3 O + (aq) + A (aq) 28 14
Weak acids K a... These are weak acids except the bottom two. These last two are too weak to act as acids in water. Note the single arrows pointing backward because the forward dissociation never occurs. 29 Illustration of Strong and Weak Acids Graphic representation of the behavior of acids of different strengths in aqueous solution. (a) A strong acid. (b) A weak acid. 30 15
Illustration of Strong and Weak Acids (a) (a) (b) A strong acid HA is completely ionized in water. (b) A weak acid HB exists mostly as undissociated HB molecules in water. Note that the water molecules are not shown in this figure. 31 Strong bases A base which dissociates 100% into its ions in a water solution. Strong base = 100% dissociation into ions NaOH (aq) Na + (aq) + OH (aq) The [OH ] concentration of a 0.10 M NaOH solution is 0.10 M. For NaOH, [base] initial = [OH ] A common notation to express [OH ] is poh, where poh = -log [OH ] = -log 0.10 = 1.00 32 16
ph and poh scales NEUTRAL [H + ] > [OH ] [H + ] = [OH ] [OH ] > [H + ] acidic basic Convert between ph and poh: ph + poh = 14 at 25 C 33 Strong bases These are strong bases. Any substance which dissociates completely to produce OH, O 2, or NH 2 is a strong base. Examples: Group 1 hydroxides: LiOH, NaOH, KOH, RbOH, CsOH Group 2 hydroxides: Mg(OH) 2, Ba(OH) 2, Sr(OH) 2 34 17
Strong bases For NaOH, [OH ] = [base] For Ba(OH) 2, [OH ] = 2 x [base] Salts which produce O 2 are strong bases. Example: CaO (s) Ca +2 (aq) + O 2 (aq) O 2 (aq) + H 2 O () 2 OH (aq) 35 Weak bases A base which dissociates less than100% into its ions in a water solution. Weak bases = < 100% dissociation into ions. Use double arrows to indicate this. NH 3 (aq) + H 2 O () NH 4+ (aq) + OH (aq) The [OH ] concentration of a 0.10 M NH 3 solution is LESS THAN 0.10 M. Therefore, poh >1.00 In weak bases [base] initial [OH ], [base] initial > [OH ] 36 18
Weak bases A base which dissociates less than100% into its ions in a water solution. B (aq) + H 2 O () BH + (aq) + OH (aq) 37 Weak bases K a Conjugate bases of strong acids are VERY weak bases.... These are weak bases, found on the right side of the table. 38 19
Amphiprotic Species Found on both sides of the table (eg HCO 3 ) Can act as acids or as bases When found on the left, it is an acid. K a Acid Strength increases HCO 3 is a weaker acid than phenol and a stronger acid than hydrogen peroxide. 39 Amphiprotic Species Found on both sides of the table (eg HCO 3 ) Can act as acids or as bases When found on the right, it is a base. K a Base Strength increases HCO 3 is a weaker base than citrate ion and a stronger base than Al(H 2 O) 5 (OH) 2+. 40 20
BrØnsted-Lowry Acids and Bases WEAK Stren ngth of acid STRONG 41 WEAK Strength of base STRONG We will work with the reference table RELATIVE STRENGTHS OF BRØNSTED-LOWRY ACIDS AND BASES to determine the Acid-Base Equilibria and Relative Strengths of Acids and Bases. By examining the relative strengths of the acids on both sides of the equilibrium, we can determine which side of the equilibrium is favoured. 42 21
1. When H 2 CO 3 and SO 32 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? H 2 2 CO 3 (aq) + SO 32 (aq) HCO 3 (aq) +HSO 3 (aq) This can only donate a H + ion. It is an ACID. This ion can only act as a BASE, accept a H + ion. HCO 3 is the conjugate base of H 2 CO 3. As a result HSO 3 is the conjugate acid of SO 32. 43 1. When H 2 CO 3 and SO 32 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? H 2 CO 3 (aq) +SO 32 (aq) HCO 3 (aq) + HSO 3 (aq) acid base conjugate base conjugate acid Compare the K a of the two acids H 2 CO 3 is a slightly stronger acid than HSO 3. The equilibrium will favour the products. 44 22
1. When H 2 CO 3 and SO 32 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? H 2 2 CO 3 (aq) + SO 32 (aq) HCO 3 (aq) + HSO 3 (aq) acid base conjugate base conjugate acid What is the value of K for this equilibrium? Since K >1, equilibrium favors products 45 2. When H 2 PO 4 and CO 32 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? H 2 2 2 PO 4 (aq) + CO 32 (aq) HCO 3 (aq) + HPO 42 (aq) This is an amphiprotic ion. It can donate or accept a H + ion. After considering, in this mixture, H 2 PO 4 will act as an acid. This ion can only act as a BASE, accept a H + ion. As a result HCO 3 is the HPO 2 42 is the conjugate conjugate acid of base of CO 32. H 2 PO 4. 46 23
2. When H 2 PO 4 and CO 32 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? H 2 2 2 PO 4 (aq) + CO 32 (aq) HCO 3 (aq) + HPO 42 (aq) acid base conjugate acid conjugate base Compare the K a of the two acids H 2 PO 4 is a stronger acid than HCO 3. The equilibrium will favour the products. 47 3. When HSO 4 and NH 3 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? HSO + 2 4 (aq) + NH 3 (aq) NH 4 (aq) + SO 42 (aq) This is an amphiprotic ion. It can donate or accept a H + ion. This can only act as a BASE, accept a H + ion. As a result After considering, in this mixture, HSO 4 will act as an acid. NH 4+ is the conjugate acid of NH 3. SO 42 is the conjugate base of HSO 4. 48 24
3. When HSO 4 and NH 3 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? HSO + 2 4 (aq) + NH 3 (aq) NH 4+ (aq) + SO 42 (aq) acid base conjugate acid conjugate base Compare the K a of the two acids HSO 4 is a stronger acid than NH 4 +. The equilibrium will favour the products. 49 4. When HSO 4 and H 2 PO 4 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? Both HSO 4 and H 2 PO 4 are amphiprotic ions. Which will play the role of an acid? Let s compare their K a values. HSO 4 is a stronger acid than H 2 PO 4. HSO 4 will act as an acid donating protons to H 2 PO 4. After considering, in this mixture, HSO 4 will act as an acid. H 2 PO 42 will act as the base. 50 25
4. When HSO 4 and H 2 PO 4 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? HSO +H 2 4 (aq) 2 PO 4 (aq) H 3 PO 4 (aq) + SO 42 (aq) K a of HSO 4 is bigger. It has the higher tendency to donate H +. K a of H 2 PO 4 is smaller. Comparatively, it has less tendency to donate H +. After considering, in this mixture, HSO 4 will act as an acid. H 3 PO 4 is the conjugate acid of H 2 PO 4. As a result SO 42 is the conjugate base of HSO 4. 51 4. When HSO 4 and H 2 PO 4 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? HSO 2 4 (aq) + H 2 PO 4 (aq) H 3 PO 4 (aq) + SO 42 (aq) acid base conjugate acid conjugate base Compare the K a of the two acids HSO 4- is a stronger acid than H 3 PO 4. The equilibrium will favour the products. 52 26
5. When CH 3 COOH and NH 3 are mixed, an equilibrium is set up. Which side of the equilibrium is favoured? CH 3 COOH (aq) +NH 3 (aq) CH 3 COO (aq) +NH 4+ (aq) acid base conjugate base conjugate acid Compare the K a of the two acids CH + +CHCOO - 3 COOH H 3 COO 18x10 1.8x10-5 NH 4+ H + + NH 3 5.6x10-10 CH 3 COOH is a stronger acid than NH 4+. The equilibrium will favour the products. 53 6. Consider an amino acid, an equilibrium is set up between the amine group, R-NH 2, and the carboxylic acid group, R-COOH. Which side of the equilibrium is favoured? H 2 NCHRCOOH (aq) + H 3 NCHRCOO (aq) base acid conjugate acid conjugate base Compare the K a of the two acids CH + 3 COOH H +CHCOO 3 COO 18x10 1.8x10-5 NH 4+ H + + NH 3 5.6x10-10 CH 3 COOH is a stronger acid than NH 4+. The equilibrium will favour the products (i.e. a double charged species). 54 27
Direction of Reaction Example Predict the direction favored in each of the following acid-base reactions. That is, does the reaction tend to go more in the forward or in the reverse direction? a) NH 4+ + OH H 2 O + NH 3 b) HSO 4 + NO 3 HNO 3 + SO 42 c) HSO 3 + CH 3 COO SO 32 + CH 3 COOH d) CH 3 COOH + Cr(H 2 O) 5 (OH) 2+ Cr(H 2 O) 3+ 6 + CH 3 COO e) HNO 2 + ClO 4 HClO 4 + NO 2 f) H 2 CO 3 + CO 32 HCO 3 + HCO 3 Forward Reverse Reverse Reverse Reverse Forward 55 In aqueous solutions, H 3 O + (aq) & OH (aq) are the strongest acids and bases that exist. Stronger bases react with water to form OH (aq) O 2 (aq) + H 2 O () OH (aq) + OH (aq) Stronger acids react with water to form H + 3 O (aq). HCl (aq) + H 2 O () H 3 O + (aq) + Cl (aq) This is known as the Leveling Effect 56 28
The Self-Ionization of Water 57 Ion Product of Water H 2 O()+ H 2 O() H 3 O + (aq) + OH (aq) K w = H 3 O + ][OH ] K w = [H 3 O + ][OH - ] = 1x10-14 at 25 C Consider HA(aq) + H 2 O() A - (aq) + H 3 O + (aq) K a K a K b A - (aq) + H 2 O() HA(aq) + OH (aq) A H O 3 HA HAOH A H O OH 3 K K b w K w K K a b 58 29
ph and poh The potential of the hydrogen ion was defined in 1909 as the negative of the logarithm of [H + ]. ph = -log[h 3 O + ] poh = -log[oh ] K W = [H 3 O + ][OH ] = 1.0x10-14 at 25 C -logk =-log[h + ]-log[oh ]= -log(1-14 W 3 O ] log(1.0x10 ) pk W = ph + poh= -(-14) pk W = ph + poh = 14 at 25 C Concentration and ph and poh Examples at 25 C [H + ] [OH ] ph poh Acidic/Basic ph = 5.52 0.15 M HNO 3 0.0100 M Ca(OH) 2 1.0x10 10 M Ca(OH) 2 30
Concentration and ph and poh Examples at 25 C [H + ] [OH ] ph poh Acidic/Basic ph = 5.52 3.0x10 6 M 3.3x10 9 M 5.52 8.48 Acidic 0.15 M HNO 3 0.15 M 6.7x10 14 M 0.82 13.18 Acidic 0.0100 M Ca(OH) 2 5.00x10 13 M 2.00x10 2 M 12.301 1.699 Basic 1.0x10 10 M Ca(OH) 2 1x10 7 M 1x10 7 M 7 7 Neutral Strong Acid and Bases Examples 1. Calculate the ph of a solution of 0.400 g of HI in 500. ml of solution. 31
Strong Acid and Bases Examples 1. Calculate the ph of a solution of 0.400 g of HI in 500. ml of solution. H 0.400 g HI ph log mol HI mol H 127.9 g HI mol HI 6.25x10 0.500 L H log6.25x10 3 2. 204 3 M Strong Acid and Bases Examples 2. Calculate the mass of Sr(OH) 2 which must be dissolved in 600. ml of solution to make a ph of 12.00. 32
Strong Acid and Bases Examples 2. Calculate the mass of Sr(OH) 2 which must be dissolved in 600. ml of solution to make a ph of 12.00. poh ph 14.00 OH 10 10 n OH g Sr( OH ) g Sr( OH ) 10 2.00 2 3 1.0 x10 M 0.600 6.0x10 2 2 6.0x10 3 mol OH 0.36 g Sr( OH ) 2 2 1.0x10 M mol OH mol Sr( OH ) 2 mol OH 2 121.6 g Sr( OH ) mol Sr( OH ) 2 2 Weak Acid and Bases Examples 1. Butyric acid, HC 4 H 7 O 2 (or CH 3 CH 2 CH 2 COOH) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC 4 H 7 O 2 is found to have a ph of 2.72. Determine K A for butyric acid. 33
Weak Acid and Bases Examples 1. Butyric acid, HC 4 H 7 O 2 (or CH 3 CH 2 CH 2 COOH) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC 4 H 7 O 2 is found to have a ph of 2.72. Determine K A for butyric acid. HC 4 H 7 O 2 +H 2 O C 4 H 7 O 2 + H 3 O + I 0.250 0 0 C 10 2.72 10 2.72 10 2.72 E 0.250 10 2.72 10 2.72 10 2.72 K a C H O H O 2.72 2.72 10 10 5 4 7 2 3 1.5x10 2.72 HC4H 7O2 0.250 10 Weak Acid and Bases Examples 2. Calculate the ph of a 0.20 M solution of KF, K b for F is 1.5x10 11. 34
Weak Acid and Bases Examples 2. Calculate the ph of a 0.20 M solution of KF, K b for F is 1.5x10 11. F + H 2 O HF + OH I 0.20 0 0 C x x x E 0.20 x x x HF OH xx K b K b F 11 1.5x10 0.20 x is small x is probably small, thus 0.20 x 0.20 1.5x10 so x 11 poh ph 14.00 2 2 x x 0.20 x 0.20 or x 6 OH 1.7x10 0.20 log log 6 OH 1.7x10 ph 14.00 5.77 8. 23 11 0.201.5 x10 assumption is OK 5.77 1.7x10 6 Hydrolysis is the reaction between water and the ions contained in the water solution. When this occurs, this could result in producing an acidic or basic solution. Spectator ions do NOT undergo hydrolysis. They are: 1. Group 1 (alkali metals): Li +, Na +, K +, Rb +, Cs +, Fr + 2. Group 2 (alkaline-earth metals): Be 2+, Mg 2+, Ca 2+, Ba 2+, Sr 2+ 3. Conjugate bases of strong acids: ClO 4, I, Br, Cl, NO 3 70 35
Cation that undergo hydrolysis NH 4+, ammonia Hydrolysis equation: NH 4+ (aq) + H 2 O () H 3 O + (aq) + NH 3 (aq) 71 Hydrated cations that undergo hydrolysis Transition metals have smaller ions and bigger charges This attracts t H 2 O molecules l Example Fe 3+ or iron (III) ions forms Fe 3+ (aq) + 6 H 2 O () Fe(H 2 O) 3+ 6 (aq) Hexaaquoiron, or iron (III) ion is a weak acid. Other hydrated ions are chromium (III) and aluminum (III) ions 72 36
Anions that undergo hydrolysis... All ions in this section can undergo hydrolysis. Anions that are not amphiprotic will act as weak bases in water. 73 Anions that undergo hydrolysis Examples of net ionic hydrolysis equations are: IO 3 (aq) + H 2 O () HIO 3 (aq) + OH (aq) NO 2 (aq) + H 2 O () HNO 2 (aq) + OH (aq) CH 3 COO (aq)+ho 2 () CH 3 COOH (aq)+oh (aq) 74 37
Anions that undergo hydrolysis These 2 ions act as STRONG Bases. They undergo 100% hydrolysis to form OH ions. 75 Anions that do not undergo hydrolysis These 5 ions do NOT undergo hydrolysis. They are spectators. 76 38
Amphiprotic anions that undergo hydrolysis These anions start with H and have a - charge. Examples: HSO 4, HSO 3, H 2 PO 4, HPO 42, HS etc. Amphiprotic anions hydrolyze as acids to produce H 3 O +, but They also hydrolyze as bases to produce OH. Compare the K a and K b values to determine the predominant hydrolysis. Find the K a of the ion by looking for the ion on the LEFT side of the acid table. Find the K b of the ion by looking for the ion on the RIGHT side of the acid table. (K b = K w /K a (conjugate acid) ) 77 1. Is the salt NaCl acidic, basic or neutral in water? NaCl (s) Na + (aq) + Cl (aq) Na + and Cl ions are spectator ions. As a result No hydrolysis reaction occurs. Solution is neutral. 78 39
2. Is the salt NaF acidic, basic or neutral in water? What is the net ionic equation? NaF (s) Na + (aq) + F (aq) Na + ions are spectator ions. F ions are found on the right side of the acid table. It is a weak base. As a result F undergoes hydrolysis reaction. F (aq) + H 2 O () HF (aq) + OH (aq) Solution is basic. 79 3. Is the salt NH 4 NO 3 acidic, basic or neutral in water? What is the net ionic equation? NH 4 NO 3 (s) NH 4+ (aq) + NO 3 (aq) 4 3 ( ) 4 ( q) 3 ( q) NO 3 ions are spectator ions. NH 4+ ions are found on the left side of the acid table. It is a weak acid. As a result NH 4+ undergoes hydrolysis reaction. NH 4+ (aq) + H 2 O () NH 3 (aq) + H 3 O + (aq) Solution is acidic. 80 40
4. Is the salt NaHC 2 O 4 acidic, basic or neutral in water? What is the net ionic equation? NaHC 2 O 4 (s) Na + (aq) + HC 2 O 4 (aq) 2 4 ( ) ( q) 2 4 ( q) Na + ions are spectator ions. HC 2 O 4 ions are found on the right and left side of the acid table. It can either be a weak acid or a weak base. As a result, compare K a and K b values of HC 2 O 4. K a K a > K b. HC 2 O 4 (aq) + H 2 O () C 2 O 42 (aq) + H 3 O + (aq) -5 6.4x10 Solution is acidic. ; K b 14 10 5.9x10-2 1.7x10 81-13 5. Is the salt NH 4 NO 2 acidic, basic or neutral in water? Write the net ionic equation. NH 4 NO 2 (s) NH 4+ (aq) + NO 2 (aq) 4 2 ( ) 4 ( q) 2 ( q) NH 4+ ions are found on the left side of the acid table. NO 2 ions are found on the right side of the acid table. As a result, compare K a of NH 4+ K b of NO 2. K a 5.6x10-10 ; K b 14 10 4.6x10-4 2.2x10-11 K a > K b. Solution is acidic. 82 41
6. Is the HCO 3- ion acidic, basic or neutral in water? HCO 3 3 ions are found on the right and left side of the acid table. As a result, compare K a and K b of HCO 3. 11 K a 5.6x10 ; Kb 10 14 7 4.3x 10 2.3x 10 8 K b > K a. HCO 3 (aq) + H 2 O () H 2 CO 3 (aq) + OH (aq) Solution CHEM 0012 is Lecture basic. Notes 83 Salt Examples Predict whether the following solutions are acidic or basic? a) NaCN Na + no hd hydrolysis, CN is conjugate base of HCN. Therefore solution is basic. b) KCl K + no hydrolysis, Cl no hydrolysis. Therefore solution is neutral. c) NH 4 Br Br no hydrolysis y (HBr is strong acid), ) 4 d) NH 4 CN NH 4+ is a weak acid. Therefore solution is acidic. Both NH 4 + and CN hydrolyze in aqueous solutions. Must compare the values of K a (NH 4+ ) = 5.6x10-10 and K b (CN ) = 2.0x10-5. Since K b >K a the solution is basic. 42
Is the salt Na 2 CO 3 acidic, basic or neutral in water? Write the net ionic equation. Na 2CO 3 (s) 2 Na + (aq) + CO 32 (aq) Na + ions are spectator ions. CO 32 ions are found on the right side of the acid table. It is a weak base. As a result CO 32 undergoes hydrolysis reaction. The net ionic equation is CO 32 (aq) + H 2 O () HCO 3 (aq) + OH (aq) Solution is basic. 86 43
Calculate the ph of 0.30 M Na 2 CO 3. Step 1: Identify the ions involved. Na 2 CO 3 (s) 2 Na + (aq) + CO 32 (aq) Spectator ion Found on the right side of the acid table; undergo base hd hydrolysis 87 Calculate the ph of 0.30 M Na 2 CO 3. Step 2: Write the net ionic equation for the hydrolysis. Set up ICE table. CO 32 (aq) + H 2 O () HCO 3 (aq) + OH (aq) CO 3 2 (aq) + H 2 O () HCO 3 (aq) + OH (aq) [I] 0.30 0 0 [C] x + x + x [E] 0.30 x x x Stoichiometric ratio 88 44
Calculate the ph of 0.30 M Na 2 CO 3. Step 3: Determine the K b for the base hydrolysis. CO 32 (aq) + H 2 O () HCO 3 (aq) + OH (aq) Look up in the acid table the K a for HCO 3 a 3-14 K w 1.0 x 10 K b 1.786 x 10-11 K (HCO ) 5.6 x 10-4 89 Calculate the ph of 0.30 M Na 2 CO 3. Step 4: Write the K b expression for the base hydrolysis. CO 32 (aq) + H 2 O () HCO 3 (aq) + OH (aq) K b [HCO 3 2 3 [CO ] [OH ] - ] 90 45
Calculate the ph of 0.30 M Na 2 CO 3. Step 5: Substitute in the equilibrium concentrations from the ICE table. CO 3 2 (aq) + H 2 O () HCO 3 (aq) + OH (aq) [I] 0.30 0 0 [C] x + x + x [E] 030 0.30 x x x 3 ] 2 3 - [HCO [OH ] x K b 1.786 x 10 [CO ] 0.30 - x 2-4 91 Calculate the ph of 0.30 M Na 2 CO 3. Step 6: Solve for x, where x = [OH ]. K b 2 [HCO [CO 3 ] 2 3 - [OH ] ] 2 x 0.30 - x 1.786 x 10 x -4 1.786 x 10 Use assumption 0.30 x 0.30 030 0.30 because [0.30] > 1000 x 1.786 x 10 4-4 [OH - ] = x = 7.319 x 10-3 M 92 46
Calculate the ph of 0.30 M Na 2 CO 3. Step 7: From the [OH ], calculate the poh. poh = -log (7.319 x 10-3 ) = 2.1355 Step 8: Convert poh to ph. ph = 14 poh =14 2.1355 = 11.86 Step 9: Check that your answer makes sense. We expect CO 32 ions will undergo base hydrolysis. ph > 7. 93 Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. 94 47
Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 1: Determine the concentration of the solution. 45.0 g 53.5 g/mole 0.841 mole 0.841 mole 0.561M NH4 Cl 1.50 L 95 Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 2: Identify the ions involved. NH 4 Cl (s) NH 4+ (aq) + Cl (aq) Found on the left side of the acid table; undergo acid hydrolysis Spectator ion 96 48
Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 3: Write the net ionic equation for the hydrolysis. Set up ICE table. NH 4+ (aq) + H 2 O () NH 3 (aq) + H 3 O + (aq) NH 4+ (aq) + H 2 O () NH 3 (aq) + H 3 O + (aq) [I] 0.561 0 0 [C] x + x + x [E] 0.561 x x x Stoichiometric ratio 97 Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 4: Determine the K a for the acid hydrolysis. NH 4+ (aq) + H 2 O () NH 3 (aq) + H 3 O + (aq) Look up in the acid table the K a for NH 4 + K a 98 49
Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 5: Write the K a expression for the acid hydrolysis. NH 4+ (aq) + H 2 O () NH 3 (aq) + H 3 O + (aq) [NH ] [H O 3 3 K a [NH4 ] ] 99 Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 6: Substitute in the equilibrium concentrations from the ICE table. NH 4+ (aq) + H 2 O () NH 3 (aq) + H 3 O + (aq) [I] 0.561 0 0 [C] x +x +x [E] 0.561 x x x K a [NH3] [H O [NH 3 4 ] ] 2 x 0.561 - x 5.6x10-10 100 50
Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 7: Solve for x, where x = [H 3 O + ]. K a [NH3] [H O [NH 3 4 ] ] 2 x 0.561 - x Make assumption: 0.561 x 0.561 K a 2 x 0.561 5.6x10-10 5.6x10-10 Check assumption: [Initial] > 1000 x K a [H 3 O + ] = x = 1.8 x 10-5 M Assumption is valid because 0.561 > 1000 x 5.6 x 10-10 101 Dissolve 45.0 g of NH 4 Cl in 1.50 L of water. What is the ph of this solution? Molar mass of NH 4 Cl is 53.5 g/mole. Step 8: Convert to ph. [H 3 O + ] = x = 1.8 x 10-5 M ph = - log (1.8 x 10-5 ) = 4.74 102 51
Titrations Equivalence point: The point in the reaction at which both acid and base have been consumed. Neither acid nor base is present in excess. End point: The point at which the indicator changes color. Titrant: The solution added to the solution in the flask. Titration Curve: A plot of ph of the solution being analyzed as a function of the amount of titrant added. 103 Acid-Base Indicators Color of some substances depends on the ph. HIn(aq) + H 2 O() Inˉ(aq) + H 3 O+(aq) >90% acid form the color appears to be the acid color >90% base form the color appears to be the base color Intermediate color is seen in between these two states. Complete color change occurs over 2 ph units Marks the end point of a titration i by changing color. The equivalence point is not necessarily the same as the end point. 104 52
The acid and base forms of the indicator phenolphthalein. In the acid form (Hln), the molecule is colorless. When a proton (plus H 2 O) is removed to give the base form (lnˉ), the color changes to pink. 105 106 53
Titration of a Strong Acid with a Strong Base The ph has a low value at the beginning. The ph changes slowly until just before the equivalence point. The ph rises sharply perhaps 6 units per 0.1 ml addition of titrant. The ph rises slowly again. Any Acid Base Indicator will do. As long as color change occurs between ph 4 and 10. 107 The ph curve for the titration of 100.00 ml of 0.10 M HCl with 0.10 M NaOH. Note: At the equivalence point ph = 7 Either indicator will give a fairly accurate result 108 54
For the titration of a strong base with a strong acid the titration curve is just flipped over. The ph curve for the titration of 100.0 ml of 0.50 M NaOH with 1.0 M HCl. 109 Weak Acid - Strong Base Titration Step 1 A stoichiometry problem reaction is assumed to run to completion then determine remaining species. Step 2 An equilibrium problem determine position of weak acid equilibrium and calculate late ph. 110 55
Weak Acid - Strong Base Titration The ph curve for the titration of 50.0 ml of 0.100 M HC 2H 3O 2 with 0.100 M NaOH. Note: At the equivalence point ph > 7 Choice of indicator is important pk a = ph at the point halfway to the equivalence point The ph is fairly constant t around halfway to the equivalence point i.e it is a buffer solution 111 Comparison of strong and weak acid titration curves. 112 56
The ph curves for the titrations of 50.0-mL samples of 0.10 M acids with various K a values with 0.10 M NaOH. 113 For the titration of a weak base with a strong acid the titration curve is just flipped over. The ph curve for the titration of 100.0 ml of 0.050 M NH 3 with 0.10 M HCl. Note at equivalence point ph < 7 114 57
Titration of a Weak Polyprotic Acid NaOH NaOH NaOH H 3 PO 4 H 2 PO 4ˉ HPO 4 ²ˉ PO 4 ³ˉ 3 4 2 4 4 4 115 Buffer Solutions Two component systems that change ph only slightly on addition of acid or base. The two components must not neutralize each other but must neutralize strong acids and bases. A weak acid and it s conjugate base. A weak kbase and it s conjugate acid After addition of strong acid or base, deal with stoichiometry first, then equilibrium. 116 58
How A Buffer Works 117 The Henderson-Hasselbalch Equation A variation of the ionization constant expression. Consider a hypothetical weak acid, HA, and its salt NaA: HA(aq) H+(aq) + Aˉ(aq) K a pk a H A HA A ph log HA ph pk a log or A K log H log a ph pk conjugate base log conjugate acid a A log HA HA 118 59
The Henderson-Hasselbalch Equation Only useful when you can use initial concentrations of acid and salt. This limits the validity of the equation. Limits can be met by: A 10 0.1 HA and both A 10K a and HA 10K a 119 Characteristics of the Henderson-Hasselbalch Equation ph depends only on the ratio [Aˉ]/[HA] This means that the ph of a buffer solution does not change on dilution. When [Aˉ] = [HA] then ph = pk a. This is also the point where the ratio changes least upon the addition of either acid or base. Therefore a system is best at buffering at a ph = pk a. ph changes as the Log of the ratio [Aˉ]/[HA] ]/[HA]. Therefore ph is not very sensitive to the value of [Aˉ]/[HA] The larger the value of [Aˉ] and [HA] the more acid or base that can be added without significantly changing the ph (i.e. the greater the buffering capacity of the solution). 120 60
Buffer Solutions Consider the case of acetic acid CH 3 COOH(aq) + H 2 O() H 3 O+ (aq) + CH 3 COOˉ(aq) pk a = 4.75 [CH 3 COOˉ]/[CH 3 COOH] 0.1 05 0.5 1 2 10 ph 3.75 445 4.45 4.75 5.05 5.75 121 Preparing a Buffer Solution of a Desired ph What mass of NaC 2 H 3 O 2 must be dissolved in 0.300 L of 0.25 M HC 2 H 3 O 2 (K a = 1.8x10 5 ) to produce a solution with ph = 5.09? (Assume that the solution volume is constant at 0.300 L) HC 2 H 3 O 2 H + (aq) + C 2 H 3 O 2ˉ(aq) 122 61