AP Chemistry – Chapter 16 Acid and Base Equilibrium HW:

16.1 – Brief Review Arrhenius – Acid:Substance that, when dissolved in water, increases the concentration of hydrogen ions. – Base:Substance that, when dissolved in water, increases the concentration of hydroxide ions.

16.2 – Bronsted-Lowry Brønsted–Lowry – Acid:Proton donor – Base:Proton acceptor

A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons.

If it can be either…...it is amphoteric. HCO 3 − HSO 4 − H2OH2O Add H+ to form:Remove H+ to form:

Conjugate Acids and Bases: From the Latin word conjugare, meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids.

Relative Strengths of Acids and Bases Strong Acid = Ionizes “completely” – Hydrochloric – Hydrobromic – Hydroiodic – Nitric – Chloric – Perchloric – Sulfuric (the first proton only) Reactions:

Strengths of Acids and Bases Weak Acid = Ionizes only a limited amount – Hydrofluoric – Acetic – Ammonium ion Reaction: At equilibrium:

Strengths of Acids and Bases Strong Base = Ionizes “completely” – Metal hydroxides Reactions:

Strengths of Acids and Bases Weak Base = Ionizes only a limited amount – Ammonia Reactions: At equilibrium:

Acid and Base Strength Strong acids are completely dissociated in water. – Their conjugate bases are quite weak. Weak acids only dissociate partially in water. – Their conjugate bases are weak bases.

Acid and Base Strength Substances with negligible acidity do not dissociate in water. – Their conjugate bases are exceedingly strong.

Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl − (aq) H 2 O is a much stronger base than Cl −, so the equilibrium lies so far to the right K is not measured (K>>1).

Acid and Base Strength Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K<1). C 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq)

General Rules 1.If an acid is strong, the conjugate base is very weak 2.Hydronium is the strongest acid that can exist in aqueous solution 3.Hydroxide is the strongest base that can exist in aqueous solution

16.3 – Acid and Base Properties of Water Water acts as a Brønsted–Lowry base and abstracts a proton (H + ) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed. In the presence of an acid:

Water In the presence of a base: Water can act as an acid and add a proton to a base NH 3 + H 2 O -> NH OH -

Water Water self-ionizes: 2 H 2 O (l)  H 3 O + (aq) + OH − (aq) OR H 2 O (l)  H + (aq) + OH − (aq) Ion-Product of Water: Kc = [H+][OH-] [H 2 O] Kw = [H+][OH-] (because [water] does not change)

Water For any aqueous solution at 25oC, Kw = 1 x = [H+][OH-] When NEUTRAL WATER [H+] = [OH-] = 1 x When acid or base – when one varies the other will by the Kw equation If [H+] > [OH-] = Acidic If [OH-]>[H+] = Basic

Example: [H+] = 1.0 x M Calculate hydroxide concentration

16.4 – The pH Scale More practical than concentration due to small values usually seen in concentrations. pH = −log [H 3 O + ] (Typical range 1 – 14, but can be outside of this)

pH Pure water, pH = −log (1.0  10 −7 ) = 7.00 An acid has a higher [H 3 O + ] than pure water, so its pH is <7 A base has a lower [H 3 O + ] than pure water, so its pH is >7. Neutral solution has pH  7.0

pH These are the pH values for several common substances.

Other “p” Scales The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). Some similar examples are – pOH −log [OH − ] – pK w −log K w

Because [H 3 O + ] [OH − ] = K w = 1.0  10 −14, −log [H 3 O + ] + −log [OH − ] = −log K w = or, in other words, pH + pOH = pK w = Algebra

Examples:

How Do We Measure pH? For less accurate measurements, one can use – Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 – An indicator

How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

Calculations using Strong Acid and Strong Base Monoprotic STRONG ACIDS: – NOT Equilibrium work! – Break apart completely Example – M HCl

Calculations using Strong Acid and Strong Base STRONG BASES: – NOT Equilibrium work! – Break apart completely Example – 0.02 M Ba(OH) 2

For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid- dissociation constant, K a. [H 3 O + ] [A − ] [HA] K c = HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq) 16.6 – Weak Acids and Ionization Constants

Dissociation Constants The greater the value of K a, the stronger the acid.

ICE Example Calculate the equilibrium concentrations and pH of a 0.5 M HF solution. Option instead of quadratic equation…

Accuracy of the approximation: =([H+] / [ ] initial ) x 100 Must be 5% or less! If not, redo with the quadratic equation.

Calculating K a from the pH Example: The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is Calculate K a for formic acid at this temperature. We know that [H 3 O + ] [COOH − ] [HCOOH] K a =

Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is Calculate K a for formic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three things. We can find [H 3 O + ], which is the same as [HCOO − ], from the pH.

Calculating K a from the pH pH = −log [H 3 O + ] 2.38 = −log [H 3 O + ] −2.38 = log [H 3 O + ] 10 −2.38 = 10 log [H 3 O + ] = [H 3 O + ] 4.2  10 −3 = [H 3 O + ] = [HCOO − ]

Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO − ], M Initially Change −4.2    10 −3 At Equilibrium 0.10 − 4.2  10 −3 = =  10 −3

Calculating K a from pH [4.2  10 −3 ] [0.10] K a = = 1.8  10 −4

Calculating Percent Ionization Percent Ionization =  100 In this example [H 3 O + ] eq = 4.2  10 −3 M [HCOOH] initial = 0.10 M A more dilute solution would have a higher % ionization, due to LeChatlier’s Principle and a shift to MORE ionization to counteract the low concentration. [H 3 O + ] eq [HA] initial

Calculating Percent Ionization Percent Ionization =   10 − = 4.2%

Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25°C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq) K a for acetic acid at 25°C is 1.8  10 −5.

Calculating pH from K a The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2 − ] [HC 2 H 3 O 2 ] K a =

Calculating pH from K a We next set up a table… [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 − ], M Initially Change−x−x+x+x+x+x At Equilibrium 0.30 − x  0.30 xx We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

Calculating pH from K a Now, (x) 2 (0.30) 1.8  10 −5 = (1.8  10 −5 ) (0.30) = x  10 −6 = x  10 −3 = x

Calculating pH from K a pH = −log [H 3 O + ] pH = −log (2.3  10 −3 ) pH = 2.64

Polyprotic Acids Have more than one acidic proton. Ionize one proton at a time. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation. [CB] of a second step is numerically equal to K a2.

Example –Example 16.14

Example – Practice Exercise (if needed)

Weak Bases Bases react with water to produce hydroxide ion. K b =

Weak Bases The equilibrium constant expression for this reaction is [HB] [OH − ] [B − ] K b = where K b is the base-dissociation constant.

Weak Bases K b can be used to find [OH − ] and, through it, pH.

pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? K b = 1.8  10 −5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH − (aq)

pH of Basic Solutions Tabulate the data. [NH 3 ], M[NH 4 + ], M[OH − ], M Initially At Equilibrium x  0.15 xx

pH of Basic Solutions (1.8  10 −5 ) (0.15) = x  10 −6 = x  10 −3 = x 2 (x) 2 (0.15) 1.8  10 −5 =

pH of Basic Solutions Therefore, [OH − ] = 1.6  10 −3 M pOH = −log (1.6  10 −3 ) pOH = 2.80 pH = − 2.80 pH = 11.20

16.8 – Relationship between K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other.

Proof: Write reactions for Ka of Acetic Acid and Kb for Acetate. Find that K a  K b does = K w.

Ka and Kb The stronger the acid, the higher the Ka and the weaker the base, the lower the Kb

16.9 – Acid-Base Properties of Salts Determined by the anion’s ability to react with water: X- + H 2 O = HX + OH- AND the cation’s ability to react with water: X+ + H 2 0 = X + H 3 O+

16.9 – Acid Base Properties of Salts 1. Neutral Salts Cation is Group 1 or 2 Anion is conjugate base of a strong acid Example – Sodium nitrate Think through the reaction of each ion in water: Sodium ion in water Nitrate in water

16.9 – Acid Base Properties of Salts 2. Salts that produce basic solutions Cation is a Group one or 2 Ion. Anion is a Conjugate Base of a WEAK Acid Example: Sodium acetate Think through the reaction of each ion in water. Sodium ion in water Acetate ion in water

16.9 – Acid Base Properties of Salts 2. Salts that produce basic solutions Cation is a Group one or 2 Ion. Anion is a Conjugate Base of a WEAK Acid Work:

16.9 – Acid Base Properties of Salts 3. Salts that produce acidic solutions Cation is a Conjugate Acid of a WEAK Base Anion is a Conjugate Base of a Strong Acid Example: Ammonium chloride Think through the reaction of each ion in water. Ammonium ion in water Chloride ion in water

16.9 – Acid Base Properties of Salts -Metal Ions Most metal cations that are hydrated in solution also lower the pH of the solution (more acidic). Most noticeable in small, highly charged metals. Example: Al +3 :

Salts in which both Cation and Anion hydrolyze Cation = Conjugate acid of a WEAK Base Anion = Conjugate base of a WEAK Acid Example: Ammonium fluoride If Kb > Ka then If Ka > Kb then (Table 16.2 and 16.4)

Summary - Effect of Cations and Anions 1.An anion that is the conjugate base of a strong acid will not affect the pH. 2.An anion that is the conjugate base of a weak acid will increase the pH. 3.A cation that is the conjugate acid of a weak base will decrease the pH.

Effect of Cations and Anions 4.Cations of the strong Arrhenius bases will not affect the pH. 5.Other metal ions will cause a decrease in pH. 6.When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the K a and K b values.

Factors Affecting Acid Strength Binary acids (HX) are mostly affected by BOND DISSOCIATION ENERGY. HF has the Highest bond dissociation energy, so it is the weakest acid. HF<<HCl<HBr<HI

Factors Affecting Acid Strength In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. Same basic formula… – Higher electronegative element is stronger acid

Factors Affecting Acid Strength For a series of oxyacids with the same central element, acidity increases with the number of oxygens due to larger pull on the electrons in the H-O bond.

Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids.

Lewis Bases Lewis bases are defined as electron-pair donors. Anything that could be a Brønsted–Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however.