Acids and Bases Chapters 14 and 15

Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions (H+ or H3O+) in aqueous solutions Bases produce hydroxide ions (OH-) in aqueous solutions Bronsted-Lowry Model Acids are proton (H+) donors Bases are proton (H+) acceptors Lewis Model Acids are electron pair acceptors Bases are electron pair donors

HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq) Example 1 Water acting as a base by accepting a proton, HCl acitng as an acid by donating a proton. HCl acting as an acid by producing H3O+ ions in solution. H+ ions provided by the HCl act as an acid by accepting a pair of electrons from the oxygen atom in the water (a base) to form a bond in H3O+. HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq) acid base conjugate conjugate acid base

NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) Example 2 Water acting as an acid by donating a proton, NH3 acting as a base by accepting a proton. NH3 acting as a base by producing OH- ions in solution. NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) base acid conjugate conjugate acid base

Example 3 Electron rich ammonia acting as a base by donating a pair of electrons to electron deficient Boron, that in turn acts as an acid by accepting the pair of electrons. NH3 + BF3  H3N-BF3

Conjugate Acid/Base Pairs Related by a hydrogen ion on either side of the equation Example 2 previously NH3 and NH4+

Strength of Acids and Bases Strong acids and bases are assumed to completely ionize in solution. For a strong acid (HA) and a strong base (B), the following reactions go to completion (no reverse reaction occurs) HA(aq) + H2O(l)  H3O+ + A-(aq) B(aq) + H2O(l)  BH+(aq) + OH-(aq)

Strength of Acids and Bases Weak acids and bases have very little ionization Equilibria are set up (reverse reactions are possible) The equilibrium positions for each lie heavily to the left hand side of the equation HA(aq) + H2O(l)  H3O+ + A-(aq) B(aq) + H2O(l)  BH+(aq) + OH-(aq)

Strength of Acids and Bases Weak Acids Organic acids, any acid not listed as a strong acid Strong Acids HBr, HCl, HI, HNO3, H2SO4, HClO4, HClO3 Weak Bases Ammonia and organic bases (ex. Methylamine) LiOH, NaOH, KOH, RbOH, CsOH, Ba(OH)2, Sr(OH)2, Ca(OH)2, Mg(OH)2

Define the following words: Arrhenius Acid/Base Bronsted-Lowry Acid/Base Lewis Acid/Base Strong Concentrated Corrosive Weak Dilute Equilibrium Indicator

Self Ionization of Water H2O + H2O  H3O+ + OH- At 25, [H3O+] = [OH-] = 1 x 10-7 Kw is a constant at 25 C: Kw = [H3O+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

Amphoterism Water can accept H+ or donate H+ This is called amphoterism Pure water does it by itself, although not much Water will make both protons and hydroxides without any outside work

Amphoterism Reaction 2H2O(l) H3O+(aq) + OH-(aq) We use the typical chemist’s shortcut to write this equation 2H2O(l) H+(aq) + OH-(aq) Kw = [H+][OH-] The value of Kw is known at room temp

Amphoterism If you increase either protons or hydroxides (by adding acid or base) you MUST decrease the other mathematically in order to maintain the constant value of Kw Example – calculate [H+] for a solution with 1.0x10-5 M OH- at 25oC. Is it neutral, acidic or basic?

The pH Scale Used to indicate the strength of an acid or base Ranges from 0 – 14 Acids are less than 7 Bases are greater than 7 7 is considered neutral

The pH Scale

The pH Scale Calculating pH, pOH Relationship between pH and pOH pH = -log10(H3O+) pOH = -log10(OH-) Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

pH pOH The pH Scale H+ OH- [OH-] = 1 x 10-14 [H+] [H+] = 1 x 10-14 [H+] = 10-pH pOH = -log[OH-] pH = -log[H+] [OH-] = 10-pOH pOH = 14 - pH pH = 14 - pOH

The pH Scale Examples Calculate the pH of a solution of 0.030M hydrochloric acid Calculate the pH of a 0.010M solution of calcium hydroxide Calculate the H3O+ concentration in a solution with a pH of 4.32 Calculate the pH of a solution made by dissolving 2.00g of potassium hydroxide in distilled water to a total volume of 250.mL

Ka – The Equilibrium Constant for Weak Acids For a weak acid (HA) dissociation is incomplete so an equilibrium must be set up HA(aq) + H2O(l) H3O+ + A-(aq) Initial X 0 0 Change -x +x +x Equilibrium X-x 0+x 0+x Can’t go directly to the pH equation because we don’t know how much acid is dissociated Derive the following equilibrium expression to help solve these problems Ka = [H3O+][A-] [HA]

Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #1: Write the dissociation equation HC2H3O2  C2H3O2 - + H+

Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step 2: ICE it! HC2H3O2  C2H3O2 - + H+ I 0.50 0 0 C -x +x +x E 0.50 – x x x

Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step 3: Set up the law of mass action HC2H3O2  C2H3O2 - + H+ 1.8 x 10-5 = (x)(x) = x2 (0.50-x) 0.50

Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step 4: Solve for x, which is also [H+] [H+] = 3.0 x 10-3 M

Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step 5: Convert [H+] to pH pH = -log(3.0 x 10-3) = 2.52

Kb – The Equilibrium Constant for Weak Bases For a weak base (B) dissociation is incomplete so an equilibrium must be set up B(aq) + H2O(l) BH+ + OH-(aq) Initial X 0 0 Change -x +x +x Equilibrium X-x 0+x 0+x Can’t go directly to the pOH equation because we don’t know how much base is dissociated Derive the following equilibrium expression to help solve these problems Kb = [BH+][OH-] [B]

Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #1: Write the equation for the reaction NH3 + H2O  NH4+ + OH-

Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step 2: ICE it! NH3 + H2O  NH4+ + OH- I 0.50 0 0 C -x +x +x E 0.50 – x x x

Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step 3: Set up the law of mass action NH3 + H2O  NH4+ + OH- 1.8 x 10-5 = (x)(x) = x2 (0.50-x) (0.50)

Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step 4: Solve for x, which is also [OH-] [OH-] = 3.0 x 10-3M

Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step 5: Convert [OH-] to pH pOH = -log(3.0 x 10-3M) = 2.52 pH = 14.00 – pOH = 11.48

Percent Dissociation % dissociation = amount dissociated(mol/L) x 100% initial concentration(mol/L) Specifies the amount of weak acid that has dissociated to achieve equilibrium

Example 0.500 M uric acid (HC5H3N4O4) is 1.6% dissociated. Find Ka and pH. What is the percent dissociation of 0.500 M HC2H3O2, Ka = 1.8x10-5?

HIn(aq)  H+(aq) + In-(aq) Indicators A substance that changes color according to the pH of the solution Often are weak acids where the ionized and unionized forms are different colors HIn(aq)  H+(aq) + In-(aq) color 1 color 2

Indicators Common indicators for acids and bases Litmus paper: red for acids, blue for bases Phenophthalein: colorless for acids, pink for bases Methyl orange: red for acids, yellow for bases

Selection of Indicators

Some Acid-Base Indicators

Strong Acid/Strong Base Titrations

Strong Acid/Strong Base Example Calculate the pH after these volumes of 0.2500M HCl are added to 50.00 mL of 0.1500M NaOH. a) 0.00 mL b) 4.00 mL c) 29.50 mL d) 30.00 mL e) 30.50 mL f) 40.00 mL

Strong Acid/Strong Base Example Notice the RAPID drop in pH. 1.00 mL between points c and e moves us from pH 11.20 to 2.81 (9 orders of magnitude; a billion times difference!)

Weak Acid/Strong Base Titration

Strong Acid/Weak Base Titrations

Weak Acid/Strong Base Example Find pH after these volumes of 0.400 M NaOH are added to 50.00 mL of 0.200 M HCOOH (Ka = 1.8x10-4) a) 0.00 mL e) 25.00 mL b) 5.00 mL f) 25.50 mL c) 12.50 mL g) 40.00 mL d) 24.50 mL

Weak Acid/Strong Base Example Notice weak acid/strong base has endpoint (equivalence point) in the BASIC region, not at pH of 7.00 like in strong acid/strong base. The exact same logic applies (and the same math is used) for strong ACID/weak BASE

Monoprotic and Polyprotic Acids Monoprotic acids contain only one “ionizable” proton Polyprotic acids contain more than one Usually only the first H+ comes off easily as a strong ion Each comes H+ off in a stepwise manner, each with its own Ka

Polyprotic acids Example: Carbonic Acid Stage 1: H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq) Ka1 = [H3O+][HCO3-] [H2CO3] Stage 2: HCO3-(aq) + H2O(l)  H3O+(aq) + CO3-2(aq) Ka2 = [H3O+][CO3-2] [HCO3-] Overall: H2CO3(aq) + 2H2O(l)  2H3O+(aq) + CO3-2(aq) Ka = [H3O+]2[CO3-2] = (Ka1 )(Ka2)

Example Find [PO4-3], pH and [OH-] in 6.0 M H3PO4 Ka1 = 7.5x10-3 Ka2 = 6.2x10-8 Ka3 = 4.8x10-13 Since Ka1 >> Ka2 >> Ka3 we can ignore the H+ in reactions 2 and 3 To find [PO4-3] we must work our way all the way to the last equation

Buffered Solutions A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt

Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH) Weak Acid Formula of the acid Example of a salt of the weak acid Hydrofluoric HF KF – Potassium fluoride Formic HCOOH KHCOO – Potassium formate Benzoic C6H5COOH NaC6H5COO – Sodium benzoate Acetic CH3COOH NaH3COO – Sodium acetate Carbonic H2CO3 NaHCO3 – Sodium bicarbonate Propanoic HC3H5O2 NaC3H5O2 – Sodium propanoate Hydrocyanic HCN KCN – Potassium cyanide

Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3) Base Formula of the base Example of a salt of the weak base Ammonia NH3 NH4Cl – ammonium chloride Methylamine CH3NH2 CH3NH2Cl - Methylammonium chloride Ethylamine C2H5NH2 C2H5NH3NO3 - Ethylammonium nitrate Aniline C6H5NH2 C6H5NH3Cl - Aniline hyrdrochloride Pyridine C5H5N C5H5NHCl - Pyridine hydrochloride

Buffered Solutions - Example A 0.100M solution of ethanoic acid (Ka = 1.8 x 10-5) is mixed with a solution of 0.100M potassium ethanoate. Calculate the pH of the resulting solution.

Henderson-Hasselbalch Equation pH = pKa + log [A-] = pKa + log [salt] [HA] [acid] pOH = pKb + log [BH+] = pKb + log [salt] [B] [base]

Acid-Base Properties of Salts These salts simply dissociate in water: KCl(s)  K+(aq) + Cl-(aq) Type of Salt Examples Comment pH of solution Cation is from a strong base, anion from a strong acid KCl, KNO3, NaCl, NaNO3 Both ions are neutral neutral

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is from a strong base, anion from a weak acid NaC2H3O2KCN, NaF Cation is neutral, anion is basic basic The basic anion can accept a proton from water: C2H3O2- + H2O HC2H3O2 + OH-

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is the conjugate acid of a weak base, anion is from a strong acid NH4Cl, NH4NO3 Cation is acidic, anion is neutral acidic The acidic cation can act as a proton donor: NH4+ NH3 + H+

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid NH4C2H3O2 NH4CN Cation is acidic, anion is basic See below IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is a highly charged metal ion; anion is from a strong acid Al(NO3)2; FeCl3 Hydrated cation acts as an acid; anion is neutral acidic Step #1: AlCl3(s) + 6H2O  Al(H2O)63+(aq) + Cl-(aq) Salt water Complex ion anion Step #2: Al(H2O)63+(aq)  Al(OH)(H2O)52+(aq) + H+(aq) Acid Conjugate base Proton