Acids & Bases in Aqueous Solution Chemical Equilibrium Acids & Bases in Aqueous Solution

Acids and Bases Definitions

Arrhenius Definition A protonic acid or Arrhenius acid is a substance which in water solution produces an excess of H+ ions HCl(aq) H+(aq) + Cl-(aq) Perchloric acid HClO4 Nitric acid HNO3 Sulfuric acid H2SO4

Base Arrhenius base is a substance which directly or indirectly, forms excess OH- ion in solution. NaOH, KOH , Ca(OH)2

Bronsted-Lowry Concept An acid-base reaction is one in which there is a proton transfer from one species to another. The species which gives up, or donates, the proton is an acid. The molecule or ion which accepts the proton is a base.

The Lewis Concept An acid accepts an electron pair BCl3, AlCl3 BF3 A base donates an electron pair. NH3, H2O, F- All act as Bronsted bases

Protic Acids Protic = Number of ionizable H Monoprotic = 1 HCl , HNO3, HCN, HC2H3O2 Diprotic = 2 H2CO3 , H2SO4 Triprotic = 3 H3PO4

Examples HF(aq) H+(aq) + F-(aq) HF(aq) + H2O H3O+(aq) + F-(aq) acid base acid base HF = proton donor H2O = proton acceptor

Examples NH4+(aq) + H2O H3O+(aq) + NH3(aq) Acid Base Acid Base F-(aq) + H2O HF(aq) + OH-(aq) Base Acid Acid Base NH3(aq) + H2O NH4+(aq) + OH-(aq) Base Acid Acid Base

Example NH4+(aq) + H2O H3O+(aq)+ NH3(aq) Acid Base Acid Base Note: H2O can act as an acid and as a base. Such species are said to be amphoteric.

Lewis Base There is an unshared pair of electrons. Utilizes the unshared pair of electorns to accept a proton NH3, H2O, and F- are Lewis bases since they possess an unshared electron pair which can be donated to an acid.

Examples H+(aq) + H2O H3O+(aq) Acid Base Acid H+(aq) + NH3(aq) NH4+(aq) Acid Base Acid Zn2+(aq) + 4H2O Zn(H2O)42+(aq) Acid Base Acid

Acid-Base Pair HC2H3O2(aq H+(aq) + C2H3O2- (aq) Acetic Acid Acetate ion (Acid) (Base) The acetate ion is referred to as the conjugate base of acetic acid. Acetic acid is the conjugate acid of the acetate ion. This term can be applied to all weak acid – weak base pairs.

Strong Acids Strong acids ionize completely in water to produce hydrogen ion and an anion. HCl(aq) H+(aq) + Cl-(aq) Other examples: HBr, HI, HNO3 , HClO4, H2SO4

Strong Bases Strong bases completely dissociate into a cation and OH- ion. NaOH(s) Na+(aq) + OH-(aq) Other examples: Hydroxides of the 1A metals - LiOH, NaOH, KOH, RbOH, CsOH Hydroxides of the 2A metals – Mg(OH)2, Ca(OH)2, Sr(OH)2, Ba(OH)2

Strong Acid For 0.5 M HCl soln: Before Dissociation AfterDissociation HCl(aq) H+(aq) + Cl- Before Dissociation AfterDissociation [HCl] = 0.5 M [HCl] =0 [H+] = 0.0 M [H+] = 0.5 M [Cl-] = 0.0 M [Cl-] = 0.5 M

Strong Base For NaOH (0.8M) NaOH(s) Na+(aq) + OH-(aq) Before Dissociation After Dissociation [NaOH] = 0.8 M 0.0 M [Na+] = 0.0 M 0.8 M [OH-] = 0.0 M 0.8 M

Weak Acids and Bases A weak acid or a weak base in solution forms H+ or OH- only to a very small extent. HF(aq) H+(aq) + F-(aq) Conc of HF > conc of H+ or conc of F- after dissociation. NH3(aq) + H2O NH4+(aq) + OH-(aq) The conc of NH3 > the conc of NH4+ and OH- after

Weak Acids and Bases For a weak acid and base, Concentration Before Dissociation = Concentration After Dissociation If a = original conc, and x = amount dissociated (a-x) a

pH Instead of using [H+] or [OH-] to describe how acidic or basic a solution is, the term pH is preferred. pH = -log10[H+]

Acidic OR Basic? Neutral soln: [H+] = 10-7; pH = 7.0 Acidic soln: [H+] > 10-7; pH = or < 7.0 Basic soln:[H+] < 10-7; pH = or > 7.0

pOH pOH is used to describe the amount of OH- in aqueous solution. pOH = -log10[OH-]

Relationship between pH, pOH and Kw Kw is called water dissociation constant Kw = [H+][OH-] = 1.0 x 10-14 pH + pOH = 14

Buffers A solution whose pH changes relatively little on addition of acid or base is said to be buffered. Usually a buffer consists of a mixture of an acid and its conjugate base.

K is K is K is K No matter what type of reaction you are talking about – equilibrium properties remain the same. Kc, Kp, Ka, Kb, Kw, Ksp, Kf The subscripts refer to certain specific TYPES of equilibria, but…

Acid Dissociation Reactions This is just a specific type of reaction. Referring to Bronsted-Lowry acids: proton donors An acid is only an acid when in the presence of a base Water is the universal base

Chemical Equilibrium Important Equilibria in Analytical Chemistry Base Dissociation: NH3 + HOH  NH4+ + OH- Kb(NH3) = [NH4+][OH-]/[NH3] = 1.75 x 10-5 Hydrolysis of Salts: NH4Cl(s)  NH4+ + Cl- NH4+ + HOH  NH3 + H3O+ Ka(NH4+) = Kw/Kb(NH3) = 10-14/1.75 x 10-5 Ka(NH4+) = 5.7 x 10-10

Chemical Equilibrium Important Equilibria in Analytical Chemistry Autoprotolysis: HOH + HOH  H3O+ + OH- Ke = [H3O+][OH-]/[HOH]2 Ke[HOH]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 @ 25oC In pure water @ 25oC [H3O+] = [OH-] = 10-7 Acid Dissociation: H2CO3 + HOH  H3O+ + HCO3- Ka = [H3O+][HCO3-]/[H2CO3] = 4.3 x 10-7

General Ka Reaction The general form of this reaction for any generic acid (HA) is: HA(aq) + H2O (l) A-(aq) + H3O+(aq) Acid Base Conjugate Conjugate base acid

Shorthand Notation Sometimes the water is left out: HA(aq) A-(aq) + H+(aq) This is simpler, but somewhat less precise. It looks like a dissociation reaction, but it doesn’t look like an acid/base reaction.

Equilibrium Constant Expression Ka = [H3O+][A-] [HA] NOTE: This is just a Keq, there is nothing new here. It is just a specific type of reaction. So, ICE charts, quadratic formula, etc. all still apply!

Old Familiar solution 1st we need a balanced equation: HOAc (aq) + H2O (l)↔ H3O+ (aq) + OAc- (aq) The we need to construct an ICE chart

ICE ICE Baby ICE ICE ??? HOAc (aq) + H2O (l)↔ H3O+ (aq) + OAc- (aq) I What do we know, what do we need to know? ???

What is the pH of a 0.100 M HOAc solution? The Ka of HOAc = 1.8 x 10-5 What do we know? What do we need to know?

. What is the pH of a 0.100 M HOAc solution? The Ka of HOAc = 1.8 x 10-5 What do we know? The INITIAL CONCENTRATION of HOAc What do we need to know? The EQUILIBRIUM CONCENTRATION of H3O+ (Recall, that’s what pH is: pH = - log [H3O+]

ICE ICE Baby ICE ICE HOAc (aq) + H2O (l)↔ H3O+ (aq) + OAc- (aq) I C E How do we solve for x? 0.100 M - -x +x 0.100 – x x

Use the Equilibrium Constant Expression Ka = 1.8x10-5 = [H3O+][A-] [HA] 1.8x10-5 = [x][x] [0.100-x] How do we solve this?

2 Possibilities 1.8x10-5 = [x][x] [0.100-x] Assume x <<0.100 Don’t assume x<<0.100 and use quadratic formula

The long way x = - b +/- SQRT(b2-4ac) 2a 1.8 x 10-5 = (x)(x)/(0.1-x) = x2/0.1-x x2 = 1.8 x 10-5 (0.1-x) =1.8x10-6 – 1.8x10-5 x x2 + 1.8x10-5 x – 1.8 x 10-6 = 0 Recall the quadratic formula: x = - b +/- SQRT(b2-4ac) 2a

The long way x2 + 1.8x10-5 x – 1.8 x 10-6 = 0 x = - b +/- SQRT(b2-4ac) x = - 1.8x10-5 +/- SQRT((1.8x10-5 )2-4(1)(– 1.8 x 10-6 )) 2(1) x = [-1.8x10-5 +/- SQRT (7.200x10-6)]/2 x = [-1.8x10-5 +/- 2.68 x 10-3]/2

2 roots - only 1 makes sense x = [-1.8x10-5 +/- 2.68 x 10-3]/2 The negative root is clearly non-physical x = 1.33x10-3 M We can now put this back into the ICE chart

ICE ICE Baby ICE ICE HOAc (aq) + H2O (l)↔ H3O+ (aq) + OAc- (aq) I C E 0.100 M - -x = -1.33x10-3 M +x=x = 1.33x10-3 M 0.100 M – 1.33x10-3 = 0.0997 M 1.33x10-3 M

pH = - log [H3O+] pH = - log [H3O+] = - log (1.33x10-3) = 2.88 Was all of that work necessary? Let’s look at making the assumption!

Assume x<<0.100 1.8x10-5 = [x][x] [0.100-x] If x<<0.100, then 0.100-x≈0.100 [0.100] 1.8x10-6 = [x][x] = x2 x = 1.34x10-3 M

Base Dissociation Reactions Acids and bases are matched sets. If there is a Ka, then it only makes sense that there is a Kb The base dissociation reaction is also within the Bronsted-Lowry definition Water now serves as the acid rather than the base.

General Kb Reaction The general form of this reaction for any generic acid (B) is: B(aq) + H2O (l)  HB(aq) + OH- (aq) Base Acid Conjugate Conjugate acid base

Kb It is, after all, just another “K” Kb = [HB][OH-] [B] And this gets used just like any other equilibrium constant expression.

Water, water everywhere Both Ka and Kb reactions are made possible by the role of water. Water acts as either an acid or a base. Water is amphiprotic. If water is both an acid and a base, why doesn’t it react with itself?

Water does react with itself Autoionization of water: H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)

Autoionization of water: H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) This is, in fact, the central equilibrium in all acid/base dissociations This is also the connection between Ka and Kb reactions.

The Equilibrium Constant Expression Kw H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) Kw = [H3O+][OH-] = 1.0 x 10-14 K IS K IS K IS K – this is just another equilibrium constant. Let’s ICE

ICE ICE Baby ICE ICE H2O (l) + H2O (lH3O+ (aq) + OH- (aq) I C E ???

Evaluating Kw Kw = [H3O+][OH-] = 1.0 x 10-14 [x] [x] = 1.0 x 10-14

ICE ICE Baby ICE ICE - H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) I C E What’s the pH? - +x =1.0x10-7 +x=1.0x10-7 1.0x10-7

pH = - log [H3O+] pH = - log (1.0x10-7) pH = 7 This is why “7” is considered neutral pH. It is the natural pH of water. Neutral water doesn’t have NO acid, it has the EQUILIBRIUM (Kw) amount!!!

Kb, Ka, and Kw It is the Kw of water (1.0 x 10-14­) which is responsible for the observation that: pOH + pH = 14 Since we’ve already established that pure water has 1x10-7 M concentrations of both H+ and OH- In an aqueous solution, this relationship always holds because Kw must be satisfied even if there are other equilibria that also must be satisfied.

Kb, Ka, and Kw The general Ka reaction involves donating a proton to water. HA + H2O ↔ H3O+ + A- where A- is the “conjugate base” to HA, and H3O+ is the conjugate acid to H2O. The general Kb reaction involves accepting a proton from water. A- + H2O ↔ HA + OH-

Writing the K for both reactions Ka = [H3O+][A-] [HA] Kb = [HA][OH-] [A-] If you multiply Ka by Kb: Ka*Kb = [H3O+][A-] [HA][OH-] [HA] [A-] = [H3O+][OH-] =Kw So, if you know Kb, you know Ka and vice versa because: Ka*Kb=Kw

Ka and Kb refer to specific reactions. For example, consider the acid dissociation of acetic acid: HOAc (aq) + H2O (l)↔ H3O+ (aq) + OAc- (aq) This reaction has a Ka, it does not have a Kb. BUT, its sister reaction is a base dissociation that has a Kb: OAc- (aq) + H2O (l)↔ OH- (aq) + HOAc (aq) It is this reaction that you are calculating the Kb for if you use the relationship Kw = Ka*Kb

Kw = AH+ AOH– = [H+ ]fH+ [OH–]fOH– = 1.008 ×10–14 (25oC) Ionization and pH of water / Ion product constant for water H2O = H+ + OH– Kw = AH+ AOH– = [H+ ]fH+ [OH–]fOH– = 1.008 ×10–14 (25oC) [H+] = [OH–] = 1.0 ×10–7 pH = – log AH+ = – log[H+ ]fH+ = 7.00 pKw = pH + + pOH– Ex. [H+] in 0.10M KCl at 25 oC ionic strength = 0.10M Kw = AH+ AOH– = [H+ ]fH+ [OH–]fOH– = 1.008 ×10–14 = xfH+ xfOH– = x (0.83) x (0.76) = 1.008 ×10–14 x = 1.26 ×10–7 AH+ = [H+ ]fH+ = (1.26 ×10–7 )(0.83) = 1.05 ×10–7 pH = – log AH+ = – log[H+ ]fH+ = 6.98 matrix effect

acid dissociation constant : Ka HA + H2O = A– + H3O+ Weak acids and bases : small Ka or Kb acid dissociation constant : Ka HA + H2O = A– + H3O+ Ka = AA– AH3O+ /AHA  [A–] [H3O+] / [HA] base dissociation constant : Kb B + H2O = BH+ + OH– Kb = ABH+ AOH– /AB  [BH+] [OH–] / [B]

Common types of weak acids and bases All carboxylic acids are weak acids, and all carboxylate anions are weak bases. RCOOH = RCOO– + H+ Amines are weak bases, and ammonium ions are weak acids. RNH2 = RNH3+ + OH– R2NH = R2NH2+ + OH– R3N = R3NH+ + OH– ex. CH3COOH = CH3COO– + H+ Ka = 1.75×10–5 CH3NH2 = CH3NH2+ + OH– Kb = 4.4 ×10–4

HA = A– + H+ Ka = [A–] [H3O+] / [HA] pKa pKa = –log ( AA– AH3O+ /AHA )  –log ( [A–] [H3O+] / [HA] ) pKb  –log ( [BH+] [OH–] / [B] ) Relation between Ka and Kb HA = A– + H+ Ka = [A–] [H3O+] / [HA] A– + H2O = HA + OH– Kb = [HA] [OH–] / [A–] H2O = H+ + OH– Kw Ka Kb = Kw Ka1 Kb2 = Kw Ka2 Kb1 = Kw Ka1 Kb3 = Kw Ka2 Kb2 = Kw Ka3 Kb1 = Kw

Ex. Finding Kb for the conjugate base HAC Ka = 1.75×10–5 AC– Kb = ? Kb = Kw/ Ka = 1.0×10–14 / 1.75×10–5 = 5.7×10–10 Ex. Finding Ka for the conjugate acid methylamine Kb = 4.4×10–4 methylammonium ion pKa = ? Ka = Kw/ Kb = 2.3×10–11 pKa = 10.64

Acidic OR Basic? Neutral soln: [H+] = 10-7; pH = 7.0 Acidic soln: [H+] > 10-7; pH = or < 7.0 Basic soln:[H+] < 10-7; pH = or > 7.0

pH = - log [H3O+] pH = - log (1.0x10-7) pH = 7 This is why “7” is considered neutral pH. It is the natural pH of water. Neutral water doesn’t have NO acid, it has the EQUILIBRIUM (Kw) amount!!!

Polyprotic acids and bases Polprotic acids and bases are compounds that can donate or accept more than one proton. Monoprotic : HA = A– + H+ B = BH+ + OH– Diprotic : H2A = HA– + H+ Ka1 (COOH)2 , H2CO3 HA– = A–2 + H+ Ka2 Triprotic : H3A = H2A– + H+ Ka1 H3PO4 H2A– = HA–2 + H+ Ka2 HA– = A–3 + H+ Ka3

Ex. Calculation of the pH of weak bases. 1) 0.0500 M sodium benzoate pH=? Kb = Kw / Ka = 1.00×10–14 / 6.28×10–5 = 1.59×10–10  x = [OH–] = KbF = 2.82×10–6 pOH= 5.55 pH = 14.00 – 5.55 = 8.45 COO – Na+ Sodium benzoate Ka = 6.28 ×10–5 COO – OH 2) 0.0500 M salicylic acid pH=? Kb = Kw / Ka = 1.00×10–14 / 1.07×10–3 = 9.35×10–12  x = [OH–] = KbF = 6.84×10–7 pOH= 6.17 pH = 14.00 – 6.17 = 7.83 3) 0.0500 M p-hydroxybenzoic acid pH=? Kb = Kw / Ka = 1.00×10–14 / 2.63×10–5 = 3.80×10–10  x = [OH–] = KbF = 4.36×10–6 pOH= 5.36 pH = 14.00 – 5.36 = 8.64 Salicylic acid Ka1 = 1.07×10–3 Ka2 = 1.82×10–14 COO – HO p-Hydroxybenzoic acid Ka1 = 2.63×10–5 Ka2 = 8×10–10

Fraction of dissociation :  value The fraction that is in the dissociated form of weak acid. HA = H+ + A– F–x x x  = [A–] / ([HA] + [A–]) = x / {(F–x)+ x} = x / F Ex. 0.0500 M benzoic acid Ka = [A–][H+] / [HA] = (x)(x) / (F –x)  x = [H+] = [A–] = 1.77 × 10–3   = x / F = 1.77 × 10–3 / 0.05 = 0.0354 = 3.54 % The fraction of dissociation of weak acids increases as the acid is diluted.

The fraction of dissociation of weak electrolyte increases as the electrolyte is diluted. Stronger acid is more dissociated than the weaker acid at all concentrations.

1) Calculate the pH of 0.05 M aqueous solution of CH3COONa Ex. pH of weak base solution 1) Calculate the pH of 0.05 M aqueous solution of CH3COONa Assume Ka for acetic acid is equal to 1.75×10–5 Solution : NaOAC  Na+ + OAC– completely dissociate OAC– + H2O = HOAC + OH– initial 0.05 0 0 final 0.05–x x x Kb = Kw / Ka = 1.00×10–14 / 1.75×10–5 = 5.71×10–10 = [OH–] [HOAC] / [OAC–] = x2 / (0.05 – x) 0.05>> x , 0.05 – x  0.05  x  [OH–] = KbF = 5.71×10–10 × 0.05 = 5.34 ×10–6 M pOH = – log 5.34 ×10–6 = 5.27  pH = 14.00 – 5.27 = 8.73

pOH pOH is used to describe the amount of OH- in aqueous solution. pOH = -log10[OH-]

Relationship between pH, pOH and Kw Kw is called water dissociation constant Kw = [H+][OH-] = 1.0 x 10-14 pH + pOH = 14