Chapter 14: Part 1 Acids and Bases PDF

Chapter 14: Part 1 Acids and Bases Table of Common Acids 2 Table of Common Bases 3  Common Strong Acids and Bases There are three different concepts of acids and bases: Arrhenius Bronsted-Lowry Lewis...

Chapter 14: Part 1 Acids and Bases Table of Common Acids 2 Table of Common Bases 3  Common Strong Acids and Bases There are three different concepts of acids and bases: Arrhenius Bronsted-Lowry Lewis 4  Definitions of Acids and Bases Arrhenius definition (simplest and most restrictive) Acids are substances that when dissolved in water produce a hydronium, H3O+ (hydrogen ion H+). Bases are substances that when dissolved in water produce a hydroxide ion, OH–. Brønsted–Lowry definition (based on reactions in water) Acids are substances that when dissolved in water donate hydronium, H3O+ (hydrogen ion H+). Bases are substances that accept a hydronium, H3O+ (hydrogen ion H+). Lewis definition (most expansive and based on electron donors and acceptors) Acids are substances that accept or need an electron pair. Bases are substances that donate an electron pair to another substance. 5  Arrhenius Concept H +ag) + H20 , 1) - > HzOt Acids produce H+ ions in aqueous solution. HCl(aq) → H+(aq) + Cl−(aq) 6  Arrhenius Concept Bases produce OH− ions in aqueous solution. NaOH(aq) → Na+(aq) + OH−(aq) 7  Arrhenius Acid-Base Reactions The H+ from the acid combines with the OH− from the base to make a molecule of H2O. Think of H2O as H—OH. The cation from the base combines with the anion from the acid to make a salt. acid + base → salt + water exothermic rxn Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (acid) (base) (salt) formula (water) strongbases and Natag) + Clag) + H20(l) ions # (ag) + Cl Lagh+ Natag) + OHjag) > - * espectator 8 H+ (a) +OH-1ag) - > Heo(e) * Brønsted–Lowry Concept of Acids and Bases Viewed as proton (H+)-transfer reactions Acid: the species donating a proton in a proton-transfer reaction Base: the species accepting the proton in a proton-transfer reaction Any reaction involving an H+ (proton) that transfers from one molecule to another is an acid–base reaction, regardless of whether it occurs in aqueous solution or if there is OH− present. All reactions that fit the Arrhenius definition also fit the Brønsted–Lowry definition. 9  Brønsted–Lowry Theory The acid is an H+ donor. The base is an H+ acceptor. Base structure must contain an atom with an unshared pair of electrons. In a Brønsted–Lowry acid–base reaction, the acid molecule donates an H+ to the base molecule. H—A + :B → :A– + H—B+ (acid) (base) (conjugate) (conjugate) HA A- , base acid B , HBt 10  Conjugate acid-base pairs Consists of two species in an acid-base reaction, one acid and one base, that differ by the loss or gain of a proton. Each reactant and the product it becomes is called a conjugate acid-base pair. H20 + NHz - NyttOH- HHzOt 11 base ad  Brønsted–Lowry Acids Brønsted–Lowry acids are H+ donors. Any material that has H can potentially be a Brønsted–Lowry acid. Because of the molecular structure, often one H in the molecule is easier to transfer than others. When HCl dissolves in water, the HCl is the acid because HCl transfers an H+ to H2O, forming H3O+ ions. Water acts as base, accepting H+. HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) (acid) (base) (conjugate) (conjugate) base acid 12  Brønsted–Lowry Bases Brønsted–Lowry bases are H+ acceptors. Any material that has atoms with lone pairs can potentially be a Brønsted–Lowry base. Because of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer than others. When NH3 dissolves in water, the NH3(aq) is the base because NH3 accepts an H+ from H2O, forming OH–(aq). Water acts as acid, donating H+. NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq) (base) (acid) (conjugate) (conjugate) acid base 13  Brønsted–Lowry Concept Summary A base is a species that accepts protons. OH- is only one example of a base. Acids and bases can be molecular substances or ions. Acid-base reactions are not restricted to aqueous solutions. Some species are amphiprotic, the species can act as either an acid or a base, depending on what the other reactant is. 14  Lewis Concept of Acids and Bases Lewis Acid: a species that can form a covalent bond by accepting an electron pair from another species They are electron deficient Lewis Base: a species that can form a covalent bond by donating an electron pair to another species Must have a lone pair of electrons on it that it can donate 15  Strong Acid/Base vs Weak Acid/Base A strong acid is a strong electrolyte. Complete or near complete ionization of acid molecule in water A weak acid is a weak electrolyte. Only a small percentage or partial ionization of the acid molecule in water A strong base is a strong electrolyte. Complete or near complete ionization of base molecule in water Produces OH– ions, either through dissociation or reaction with water A weak base is a weak electrolyte. Only a small percentage or partial ionization of the base molecule in water Produces OH– ions, either through dissociation or reaction with water 16  Relative Strengths of Acids and Bases The strongest acids have the weakest conjugate bases. The strongest bases have the weakest conjugate acids. Reactions proceed in the direction of the weaker acid or weaker base. with certain · Only groups 132 paired 17 bases  Molecular Structure and Acid Strength Binary acids (H—Y) have acidic hydrogens attached to a nonmetal atom. – Example: HCl and HF Going across a row of elements, the electronegativity increases, the more polarized the bond (δ+H—Xδ−), the more acidic the bond. Going down a column of elements, the size of atom X increases, the H—X bond strength decreases, the acid strength increases. Binary acid strength increases to the right across a period. Acidity: H—C < H—N < H—O < H—F Binary acid strength increases down the column. Acidity: H—F < H—Cl < H—Br < H—I 18  Strengths of Oxoacids, H-O-Y- For a series of oxoacids of the same structure, differing only in atom Y, the acid strength increases with the electronegativity of Y HIO < HBrO < HClO For a series of oxoacids, (HO)mYOn, the acid strength increases with n, the number of O atoms bonded to Y HClO < HClO2 < HClO3 < HClO4 19  Autoionization of Water Pure water is considered a nonelectrolyte, a nonconductor of electricity Measurements show a very small conduction Results from autoionization A reaction in which two like molecules react to give ions + − H2 O 𝑙 + H2 O 𝑙 ⇌ H3 O 𝑎𝑞 + OH (𝑎𝑞) All aqueous solutions contain both H3O+ and OH–. The concentrations of H3O+ and OH– are equal in water. [H3O+] = [OH–] = 10−7 M at 25 °C 20  Equilibrium Constant Expressions Write an equilibrium constant expression for: reactants products H2 O 𝑙 + H2 O 𝑙 ⇌ H3 O+ 𝑎𝑞 + OH − 𝑎𝑞 Kot][O] 21  Ion-Product constant for Water, Kw Kw: the equilibrium value of the ion product [H3O+][OH –] H2 O 𝑙 + H2 O 𝑙 ⇌ H3 O+ 𝑎𝑞 + OH − (𝑎𝑞) Kw = [H3O+][OH –] = 1.0 ×10-14 at 25 °C The product of the H3O+ and OH– concentrations is always the same number at room temperature, 1 × 10–14. If acid or base is added to water, [H3O+] and [OH –] will no longer be equal, but we can still use Kw = [H3O+][OH –] For example, if [H3O+] increases, the [OH–] must decrease so that the product stays constant. Inversely proportional 22  Solutions of a Strong Acid Strong acids donate practically all their hydrogen atoms. – Near 100% ionization of acid molecule occurs in water – Strong electrolyte Example: Calculate the [OH-] hydroxide-ion concentration in 0.10 M HCl. (Hint: The autoionization equilibrium still kw = [H +] [OH-] exists.) CH ] 0 10 M + =. [H-] = 1 0x18-13M. 23  Example 15.4 CH +] /Ez0 ] + Calculate the concentrations of hydronium ion and hydroxide ion at 25 °C in: a) 0.15 M HNO3 b) 0.010 M Ca(OH)2 0 010 mol Cata Imol 0 15 M b)(pH ]. - [y0 ] = a) - + =. L Imol = G 020 mol/ OH- H-]M. & [HzO ] =+ = 5. 0x10-13 [OH] = 6 7x 18. - M 0. 020M 24  Solutions of a Strong Acid or Base All aqueous solutions contain both H3O+ and OH– ions. Neutral solutions have equal [H3O+] and [OH–]. [H3O+] = [OH–] = 1.00 × 10−7 M 7 Acidic solutions have a larger [H3O+] than [OH–]. [H3O+] > 1.00 × 10−7; [OH–] < 1.00 × 10−7 M1 6 - M Basic solutions have a larger [OH–] than [H3O+]. [H3O+] < 1.00 × 10−7; [OH–] > 1.00 × 10−7 M M 8-14 25  The pH of a Solution The acidity of basicity of a solution depends on the hydronium-ion concentration and is expressed as pH. pH: the negative of the base 10 logarithm of the molar hydronium-ion concentration pH = – log[H3O+] Example: A solution has a hydronium-ion concentration of 1.0 x 10-3 M. What is the pH? pH = 3 00. 26  The pH scale pH < 7 is acidic, pH > 7 is basic; pH = 7 is neutral to Normal range of pH is 0 to 14 pH goes two decimal places 27  What about the hydroxide-ion? Can calculate the pOH (a measure of hydroxide-ion concentration) similar to pH. pOH = log [OH -] - Since Kw = [H3O+][OH –] = 1.0 ×10-14 at 25 °C, we can determine that pH + pOH = 14 00. 28  Strong Acid and Strong Base Solutions There are two sources of H3O+ in an aqueous solution of a strong acid—the acid and the water. There are two sources of OH− in an aqueous solution of a strong acid—the base and the water. For a strong acid or base, the contribution of the water to the total [H3O+] or [OH−] is negligible. The [H3O+]acid is so large that [H3O+]water is too small to be significant. Except in very dilute solutions, generally < 1 × 10−4 M 29  Finding the pH of a Strong Acid There are six strong acids: Five are monoprotic and one is diprotic. –Monoprotic: HCl, HBr, HI, HClO4, and HNO3 –Diprotic: H2SO4 For a monoprotic strong acid, the acid concentration equals the hydronium concentration. –[H3O+] = [HAcid] –Example: 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00 30  Chapter 14 Part 2: Acid-Base Equilibria  Solutions of a Weak Acid Weak acids donate a small fraction (partial ionization) of their hydrogen atoms. Weak acid molecules do not donate much of their hydrogens to water. Much less than 1% ionized in water [H3O+] HCI + OH H2O - ↳ (1 - + H + - An anion that is the conjugate base of a weak acid is basic. F−(aq) + H2O(l) > HF-tOH- - 45  Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base. Others are the counterions of a strong base. Therefore, some cations can potentially be acidic. The stronger the base, the weaker the conjugate acid. A cation that is the counterion of a strong base is pH neutral. NORXN Na+(aq) + H2O(l) -> NaOH t Ht A cation that is the conjugate acid of a weak base is acidic. NH4+(aq) + H2O(l) - > NHz HzOt 46  Rules for Classifying Salt Solutions as Neutral, Acidic, or Basic 1) A salt of a strong base and a strong acid. No hydrolysable ions, so it gives a neutral aqueous solution. 2) A salt of a strong base and a weak acid. The anion of the salt is the conjugate of the weak acid. The anion hydrolyzes to give a basic solution. 3) A salt of a weak base and a strong acid. The cation of the salt is the conjugate of the weak base. The cation hydrolyzes to give an acidic solution. 4) A salt of a weak base and a weak acid. Both ions hydrolyze. The classification is dependent upon the relative acid-base strengths of the two ions. 47  pH of a Salt Solution Many reference books give tables of only Ka values since Kb values can be found from them by using the relationship: kw [H30 ] [OH-] = + Ka × Kb = Kw - When you add equations, you / - / multiply the K’s. / I S 2201) HzOHagikw = Kak 30+ ](H ] = - 48  Example Find the pH of a 0.100 M NaCHO2 solution. The salt completely dissociates into Na+(aq) and CHO2-(aq), and the Na+ has no acid or base properties. Ka = 1.8 x 10 -4 -5 6x10. - 1 CHO-2cag) + H20(1) HCHO2cag) + OH-lag) O d E I 0 106M. +X Y # + X - 0 10GM - Y Y X. 49  0 100 M x10-11. (5 6x 10 1) (0 100) x2 -. =. x = 5. 6x10-12 2 366x10 60H ] - - x =. polt 5 62b =. pl = 8 374.  Common-Ion Effect The shift in an ionic equilibrium caused by the addition of a solute that provides an ion that takes part in the equilibrium HC2H3O2(aq) + H2O(l) ⇌ C2H3O2-(aq) + H3O+(aq) Addition of HCl Provides the H3O+ ion; according to Le Chatelier’s principle, the equilibrium composition shifts to the left. HC2H3O2(aq) + H2O(l) ⟵ C2H3O2-(aq) + H3O+(aq) The degree of ionization of acetic acid is decreased. 50  Common Ion Effect & HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq) Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left. Example: A solution is prepared to be 0.10 M acetic acid, HC2H3O2, and 0.10 M sodium acetate, NaC2H3O2. What is the pH of this solution at 298 K? Ka for acetic acid is 1.8 x 10-5. 51. IOM HA 0 O IOM Ac. HAcF Ac HAct H2O F Ac tHzOt O IOM O F 3 0 10 M.. +X +X Y - 0 18. M- 0. IGME + X neg neg 5 , 556100 [c-] [HySt] Ka 1. 8x10-5 [Hz0 ] + 1 8x10 5 - X = =. 4 74 pH -log(zot] = =.  Buffers A solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. The components of the buffer act by neutralizing the acid or ↑ base that is added to the buffered solution. Buffer solutions contain either significant amounts of a weak acid and its conjugate base; or significant amounts of a weak base and its conjugate acid. Example of a buffer system in nature: & Blood has a mixture of H2CO3 and HCO3−. 52  HAcFAC add SB : (OH-] OH - +HAc > Ac - + H2O(e) 3 1 5 ma. Smal Smal - 1 5mal-1 Smol. +1. Smal. O 3 Smal. 6 5mol. add strong acid : (Ht) H+ + Ac--HAc  Making an Acidic Buffer Solution An acidic buffer solution must contain significant amounts of both a weak acid and its conjugate base. Buffer buffers Given HAc + NaAc 0 20M HA given ·. O IOM HAC + O IOM NaOH.. #0 10 MAc. · HActOH- > Ac- + H2O 0 20M G IOM O &.. - 0 IOM. O. IOM to Com. 0. 20m J 0. IOM 53  Nassume An Acidic Buffer Solution equilibrium Adding a strong base to an acidic buffer: If a strong base is added to the buffer solution, the added base is neutralized by the weak acid component (HC2H3O2) in the buffer. NaOH(aq) + HC2H3O2(aq) → H2O(l) + NaC2H3O2(aq) If the amount of NaOH added is less than the amount of acetic acid present, the pH change is small. Adding a strong acid to an acidic buffer: If a strong acid is added, the added acid is neutralized by the conjugate base component (NaC2H3O2) in the buffer. HCl(aq) + NaC2H3O2(aq) → HC2H3O2(aq) + NaCl(aq) If the amount of HCl is less than the amount of NaC2H3O2 present, the pH change is small. 54 Action of a Buffer/How a Buffer Works 55  Henderson-Hasselbalch Equation [base] conj pH = pKa + log [acid] Conj It is an equation derived from the Ka expression that allows us to calculate the pH of a buffer solution. The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate base, as long as the “x is small” approximation is valid. 56  Deriving the Henderson-Hasselbalch equation Consider the equilibrium expression for a generic acidic buffer: Solving for H3O+: 57  Deriving the Henderson-Hasselbalch equation Taking the logarithm of both sides and expanding since log[AB] = log A + log B, we obtain the following: 58  Deriving the Henderson-Hasselbalch equation Multiplying both sides by –1 and rearranging, the following equation is obtained: –log[H3O+] = –logKa + log([A–]/[HA]) Since pH = –log[H3O+] and pKa = logKa pH = pKa + log([A–]/[HA]) Then the equation can be generalized as follows: [base] pH = pKa + log [acid] 59  ② S PH BH20FHBt tOH- ] [OH-] #B + kn = [B] ② Bt H + - HBt  Buffering effectiveness and capacity A good buffer should be able to neutralize moderate amounts of added acid or base. However, there is a limit to how much can be added before the pH changes significantly. The buffering range is the pH range at which the buffer can be effective. The buffering capacity is the amount of acid or base a buffer can neutralize before giving a significant pH change. The effectiveness of a buffer depends on two factors: 1. The relative amounts of acid and base 2. The absolute concentrations of acid and base 60  Buffering Range A buffer will be most effective when 0.1 < [base]:[acid] < 10. Substituting into the Henderson–Hasselbalch equation, we can calculate the maximum and minimum pH at which the buffer will be effective. [base] pH = pKa + log [acid] t Lowest pH Highest pH pH = pKa + log(0.10) pH = pKa + log(10) pH = pKa – 1 pH = pKa + 1 Therefore, the effective pH range of a buffer is pKa ±1. When choosing an acid to make a buffer, choose one whose pKa is closest to the pH of the buffer. 61  Buffering Capacity A concentrated buffer can neutralize more added acid or base than a dilute buffer. 62  Effectiveness of Buffers A buffer will be most effective when the [base]:[acid] = 1. Equal concentrations of acid and base A buffer will be effective when 0.1 < [base]:[acid] < 10. A buffer will be most effective when the [acid] and the [base] are large. 63  Buffering capacity Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH. The buffering capacity increases with increasing absolute concentration of the buffer components. As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves. Buffers that need to work mainly with added acid generally have [base] > [acid]. Buffers that need to work mainly with added base generally have [acid] > [base]. 64  Titration In an acid–base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete. When the reaction is complete, we have reached the endpoint of the titration. An indicator may be added to determine the endpoint. An indicator is a chemical that changes color when the pH changes. # When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point. 65 Titration set-up for an acid-base reaction 66  Titration Curve It is a plot of pH versus the amount of added titrant. The inflection point of the curve is the equivalence point of the titration. Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH. The pH of the equivalence point depends on the pH of the salt solution. Equivalence point of neutral salt: pH = 7 Equivalence point of acidic salt: pH < 7 Equivalence point of basic salt: pH > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH. 67  / titrant &vacea Titration Curve Comparison + HBt Deguir Bt H. - Point Strong Acid versus Strong Base Weak Base versus gegHA + OH - -A - + H0 Strong Acid B+ H + - HBt E o O X EP HA + OH - A tHeO O O * Weak Acid versus Strong Base E 68.Base titrated with Strong Acid Weak > buffer region ① ② in exact amount pim moles of acidadecompletley reacts will all weak base t excess [B] V = OmL Volume H+ D : B Ha HBt tOH- [B] [OH-] [OH-] solve kp = j Use [B] pOH -log[OH] = f buffer regionwill be zer Final - > + step PH = 14 00-pOH ② BtH - HB. E Both B3 HB + present If you have kn use kw Kaki , equation = Use H-I PH pa 1 So it Ka = * = Mo -logka pra =. Dequivalance point 3 [JM t HB++ Haf BtHzOt E (mal) HB Intial BH Final - 55 5 - 5 + O 5 E ↓ Convert ] to molarity Ka =  Point after the Equivalance ⑪ moltH + - > HBt 7 O 3 5 - 5-d - pH -log[H ] + = Weak Acid titrated with strong Base ④ I V = OmL - pH 2 I I :38eP Volume [] ① HAt HzOA tHzOt VH-] OmL = ka = ]CH30t] = CHA] ②Buffer region HA + OH - > A- + O H20 Have both HA3 I A- use Henderson * end step equationV convert B + HBt PH = pka] - back to molarity HA + A- use HAtOH - A henderson H + H20 -. 5. 5 O in i = + pratlogm) between 0 weak acid/base. 1 se ↑ 3 I 10 Strong acid/base 3. EP mol HAtOH- > A- tH20 · E 550 +5 - 5j5 5 ↳ onvert to molarity A-t20FHA. toh- kn =. Past 4 EP molHA tOH-A-tHad g i+ g 5 pOH -log[OH-] = HAtOH- VHA = 20 OmL. [OH] =0. 125 M CHAJO 105M. zomHat Com) = 16 8 m.  Base titrated with Strong Acid Strong Sg I excess base& I 1 ph m Beguir point ⑪ 'excess I Stad Volume 1. VH+ = OmL [OH-I known pOH -log[CH-] = 2molOH-tH + , HaO(e) B 52 3 2 2 ↳ - - 3 O 3 Steps I 2 [OH -] - POH ; PH 3 Eguir. Point. (mol) gH + H + - H20- BS 5 I  R I [M] H2etH27, HgOt + OH- O O I X X 14 K= [H30 +][0H] x2 1 0 x 10 - = =. X = [+z0+]1. 0x 107 M PH = T 00. 4. excess + moCH - + H+ - > H20 B 5. 65 S. 5 is - o pH = - log Eguations 100mL sample of 0 10MNH3. 0. 135MHNO3 of 150 mi after the addition mol + > NHy + H - NHz 0 010 0 02025 O i.. 0 010 +0 010 6 010 - -... O 0. 01025 0 010. pH -log[H ] 70 250 0 041.. 0 04. = + PH = 1 39. [mol] NHztHt-.010 0 0 016 , NHym O = 00. 0. o O 0 010 00maH(man). =. [] + 0 0741L' NHytag) +H20 # NHzt Hz0t. O O E 0 0574. + X +X - Y X X - X 56x10-10-panes.. 64948-6= X 5 5. X= 5 25.