CHAPTER 13(16)

ACID BASE EQUILIBRIA

ContentsDefinition of Acid and Base : Arrhenius Brnsted-Lowry , conjugate acid-base pairsLewisStrength of Acid and Base : Strong acid weak acidStrong base weak baseIonization of Acid and BaseConcepts of pH, pOH, pKa, pKb Dissociation Constant : Ka, Kb, Kw

Learning OutcomesAble to differentiate and calculate acid dissociation constants for weak and strong acids (applicable to bases)Calculate pH and pOH

Arrhenius DefinitionArrhenius base contains significant OH- ionArrhenius acid never contains H+ ion, instead it contains H atom covalently bonded that ionize in waterE.g: Arrhenius acid HCl, HNO3 , HCNE.g: Arrhenius base NaOH, KOH, Ba(OH)2

Whenever an acid dissociates (ionizes) in water, solvent molecules participate in the reactionHA (g or aq) + H2O(l) A- (aq) + H3O+(aq)The H3O+ is called hydronium ionThe terms hydrogen ion = proton = H+ are used interchangeablyArrhenius Definition

Strong acids: dissociate/ionize completely into ions in water. Eg:HNO3 (aq)+ H2O(l) H3O+ (aq) + NO3- (aq)

Weak acids: dissociate/ionize very slightly into ions in water. Eg:HCN(aq) + H2O (l) H3O+(aq) + CN- (aq)

This classification correlates with classification of electrolytes: strong electrolytes dissociate/ionize completely, and weak electrolytes dissociate/ionize partially.Arrhenius Definition

Brnsted-Lowry Acid-Base DefinitionAcidproton donor. Any species that donates an H+ such as HCl, HNO3, H3PO4 All Arrhenius acids are Brnsted-Lowry acidsBaseproton acceptor. Any species that accepts an H+, must contain lone pair of electrons to bind the H+, such as NH3, CO32-, F- and OH-Brnsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases contain Brnsted-Lowry base OH-. -

HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)HCl acts as a Brnsted-Lowry acid ( it donates a proton to H2O)H2O acts as a Brnsted-Lowry base ( it accepts a proton from HCl)

Brnsted-Lowry Acid-Base Definition

NH3 (aq) + H2O (l) NH4+ (aq) +OH-(aq)H2O acts as an acid, it donates the H+ NH3 acts as a base, it accepts H+.Thus, H2O is ampotheric: it acts as a base in one case and as an acid in the other.

Brnsted-Lowry Acid-Base Definition

NH3 (aq) + H2O (l) NH4+ (aq) +OH-(aq)

Brnsted-Lowry Acid-Base Definition

ExampleDetermine acid and base in the following reactions:

HCO3- (aq) + HF (aq) H2CO3 (aq) + F- (aq)HCO3- (aq) + OH- (aq) CO32- (aq) + H2O (aq)SO32- (aq) + NH4+ (aq) HSO3- (aq) + NH3 (aq)HSO3- (aq) + NH3 (aq) SO32- (aq) + NH4+ (aq)

Answer Acid BaseHF (aq) HCO3- (aq)HCO3- (aq) OH- (aq) NH4+ (aq) SO32- (aq) HSO3- (aq) NH3 (aq)

Conjugate Acid-Base PairsConsider the reaction of acid HA:HA (aq) + H2O(l) A-(aq) + H3O+(aq)Forward reaction: HA is the acid, H2O is the baseReverse reaction: H3O+ is the acid, A- is the baseHA and A- : differ in the presence or absence of a proton (base) a conjugate acid-base pair.

Add H+Remove H+Conjugate Acid-Base Pairs

Acids such as HCl , HNO3 , or H2SO4 ionize in water to form a hydrated proton and a species called the conjugate base of the acid; Conjugate base of HCl results from the loss of one proton.Conjugate Acid-Base Pairs

Bases such as NH3 react with water to form a hydrated hydroxide ion and a species called the conjugate acid of base. Conjugate acid of NH3 results from the gain of one proton.Conjugate Acid-Base Pairs

In NH3 and H2ONote:Conjugate Acid-Base Pairs

a) What is the conjugate base of the following acids: HClO4 ; HCO3- - remove one proton ( H+) from the formula.ClO4- ; CO32-b) What is the conjugate acid of the following bases: CN- ; H2O ; HCO3- - add one proton to the formulaHCN ; H3O+ ; H2CO3Note: Hydrogen Carbonate ion, HCO3- is amphoteric: a substance that can act as an acid or as a base.Example

ExampleDetermine the conjugate acid-base pairs in the following:

HCO3- (aq) + HF (aq) H2CO3 (aq) + F- (aq)HCO3- (aq) + OH- (aq) CO32- (aq) + H2O (aq)SO32- (aq) + NH4+ (aq) HSO3- (aq) + NH3 (aq)HSO3- (aq) + NH3 (aq) SO32- (aq) + NH4+ (aq)

AnswerAcid Base Conj. acidConj. Base

HF (aq) HCO3- (aq) H2CO3 (aq) F- (aq) HCO3- (aq) OH- (aq) H2O (l)CO32- (aq) NH4+ (aq) SO32- (aq) HSO3- (aq) NH3 (aq)HSO3- (aq) NH3 (aq) NH4+ (aq) SO32- (aq)

Other example: HSO3-AcidHSO3- (aq) + H2O(l) SO32-(aq) + H3O+Remove H+Add H+BaseHSO3- (aq) + H2O(l) H2SO3(aq) + OH- (aq)Add H+Remove H+

Conjugate pairs in acid-base reactions

Lewis Acid-Base DefinitionOH-, H2O , an amine - are electron-pair donorsA base in the Brnsted-Lowry ( a proton acceptor) also base in the Lewis (an electron pair donor )In the Lewis theory: a base can donate its electron pair to species other than H+AcidElectron pair acceptorBaseElectron pair donor

Lewis Acids with Electron-Deficient Atoms:Eg: Reaction between NH3 and BF3.BF3 has a vacant orbital in its valence shell- acts as an electron pair acceptor (a Lewis acid) toward NH3. NH3 donates the electron pair.

Lewis Acid-Base Definition

Lewis acid - molecules have an incomplete octet of electrons.Eg. Cations Fe3+ interacts with cyanide ions;

Fe3+ ion has vacant orbitals - accept the electron pairs donated by the CN- ions.Lewis Acid-Base Definition

Note: Lewis acids and bases do not need to contain protons.Therefore, the Lewis definition is the most general definition of acids and bases.Lewis acids generally have an incomplete octet (eg. BF3).Transition metal ions are generally Lewis acids.Lewis acids must have a vacant orbital (into which the electron pairs can be donated).Lewis Acid-Base Definition

Metal Cation as Lewis Acid

Definitions Comparison

Strength of Acid and Base:ObjectivesStrength of acid-baseStrong acid - strong base / weak acid - weak baseStrength in conjugate acid base determine the direction of acid-base reactionWater auto-ionizationConcept of pH

Variation in Acid StrengthStrong acids: Ionize completely into ions in water.Eg:HNO3 (aq)+ H2O(l) H3O+ (aq) + NO3- (aq)

Weak acids: ionize very slightly into ions in water.Eg:HCN(aq) + H2O (l) H3O+(aq) + CN- (aq)

This classification correlates with classification of electrolytes: strong electrolytes dissociate/ionize completelyweak electrolytes dissociate/ionize partially.

Strong AcidsHCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)Strong acid will fully dissociate/ionize in water

Weak acids

ExampleHF (aq) + H2O (l) H3O+ (aq) + F- (aq)CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)

Weak acid will partially ionize in water

Strong BaseSoluble compounds containing O2- or OH-The cations are usually those of the most active metalsM2O or MOH; Where M= Group IA metal(Li, Na, K, Rb, Cs)MO or M(OH)2 ; Where M= Group IIA metal(Be, Mg, Ca, Sr, Ba)Ionic hydrides and nitridesH- (aq) + H2O (l) H2 (g) + OH- (aq)N3- (aq) + 3 H2O (l) NH3 (aq) + 3OH- (aq)

Weak BasesCompounds with an electron-rich nitrogen are weak bases (none are Arrhenius bases).The common structural feature is an N atom that has a lone pairAmmonia : NH3Amines such as CH3CH2 : NH2C5H5N:

Relative Strengths of Acid-BaseSome acids are better proton donors than others.The stronger the acid, the weaker its conjugate base.Some bases are better proton acceptors.The stronger the base, the weaker its conjugate acidThe strength of an acid tells the strength of its conjugate base.Strong acids: completely transfer their protons 100% dissociation Their conjugate bases have a negligible tendency to be protonated.(E.g: HCl + H2O H3O+ + Cl-)

Consider the proton transfer in this:

HA(aq) + H2O (l) H3O+(aq) + A- (aq)If HA is a stronger acid than H3O+;HA will transfer its proton to H2O ; equilibrium lies to the rightIf H3O+ is a stronger acid than HA (if HA is a weak acid);equilibrium lies to the leftIn every acid-base reaction, the position of the equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base.

Example: CN- reacts with water as follows:CN- (aq) + H2O(l) HCN (aq) + OH- (aq)A Cl- ion does not react with water to form HCl. Which acid is stronger, HCl (aq) or HCN (aq)? Why? HCl and HCN are the conjugated acids of the bases Cl- and CN-.The information provided tells us that Cl- is a weaker base than CN-.CN- accepts a proton from H2O to form HCN.Principle: The weaker a base, the stronger its conjugated acid, can conclude that HCl is a stronger acid than HCN

ExamplePredict the product of the following acid-base reactions and also predict whether the equilibrium lies to the left or to the right of the equation:Cl- (aq) + H3O+(aq) H2PO4- (aq) + H2O (l) HS- (aq) + HC2H3O2 (aq) HCO3- (aq) + OH (aq) NH4+ (aq) + H2PO4- (aq) HCl (aq) + H2O(aq) Lies to the leftHPO42- (aq) +H3O+(aq)Lies to theleftH2S (aq) + C2H3O2- Lies to theright CO32- (aq) + H2O (l) Lies to the right NH3 (aq) + H3PO4 (aq)Lies to the left

Autoionization of waterWater ionises into H+ and OH- (v. small extent). This process is called the autoionization of water.

The Ion-Product Constant for Water, KwWe can write an equilibrium constant expression for the autoionization of water: 2H2O(l) H3O+(aq) + OH-(aq)

Because H2O(l) is a pure liquid, the expression can be simplified:

The Ion-Product Constant for Water, KwKw is called the ion-product constant.At 25C the ion-product of water is always:

In neutral solution:[H3O+] = [OH-] In acidic solution: [H3O+] exceeds [OH-] In basic solution:[OH-] exceeds [H3O+] Note: The ion -product constant Kw is unaffected by whether a solution is acidic or basic.

Calculate the concentration of H3O+ (aq) in a solution in which [OH-] is 2.0 x 10-9 M at 25C.Example

The pH scaleExpress H+ in terms of pH.For [H+] = 1.0 x 10-3 M.pH = -log (1.0 x 10-3 ) = -(- 3.00 ) = 3.00pH = -log [H+ ] = -log [H3O+]

Acidic Neutral Basic

Example

Given [OH-] = 2.0 x 10-3, calculate [H+] (or [H3O+]) and pH at 25C.

pH = -log ( 5.0 X 10-12 ) = 11.3

The concentration of OH- can be expressed as pOHExample:A solution has a pH of 5.6 . Calculate the hydrogen ion concentration.pH = -log [H+], log [H+] = -5.6[H+] = 2.5 10-6 MOther p scalepOH = -log [OH- ]pH + pOH = pKw= -log Kw = 14.00

Calculate [H+] for a solution with pOH of 4.75pOH is defined as log [OH-]The pH and the pOH are related: pH + pOH = 14.00pH = 14.00 - pOH = 14 - 4.75 = 9.25pH = 9.25log [H+] = -pH = -9.25 [H+] = 5.6 x 10-10 M.Example

Acid-base indicator pH paperpH meterMethods for measuring pH

Strong AcidsStrong acids are strong electrolytes and they ionize completely in solution. HNO3 (aq) + H2O (l) H3O+ (aq) + NO3-(aq)

ExampleWhat is the pH of a 0.04 M solution HClO4?HClO4 is completely ionized: [H+] = 0.04 MHence, pH= -log (0.040) = 1.40

Strong BasesMost ionic hydroxides are strong bases (eg. NaOH, KOH and Ca(OH)2.)

Strong bases are strong electrolytes and dissociate completely in solution:NaOH (aq) Na+ (aq) + OH-(aq)

The pOH of a strong base is given by the initial molarity of the base.

ExampleWhat is the pH of a 0.011 M solution of Ca(OH)2?

Ca(OH)2 is a strong base.Ca(OH)2 Ca2+ + 2OH-0.011M0.011M 2 0.011MpOH = -log (0.022) = 1.66pH + pOH = 14.00 pH = 14.00 - pOH = 14.00 - 1.66 = 12.34

Acid Dissociation Constant, KaHA (g or aq) + H2O(l) A- (aq) + H3O+(aq)

Stronger acid larger KaWeaker acid Smaller Kalower % HA ionizes

Ka values for weak acidsAcidKa% HA dissociated

1 M HClO21.12 x10-210 %

1 M CH3COOH1.8 x10-50.42 %

1 M HCN6.2 x10-100.0025 %

Solving Problems Involving Weak-Acid Equilibria1. Given equilibrium concentrations, find Ka.2. Given Ka and other info, find other equilibrium concentration.

HA (aq) + H2O(l) H3O+ (aq) + A- (aq)Ionization is incomplete and some significant amount of undissociated acid remains at equilibrium.

Problem-Solving Approach : Problems Involving Weak-Acid Equilibria1. Write the balanced equation and Ka expression2. Define x as unknown concentration that changes during the reaction.3. Construct a reaction table that incorporates the unknown.4. Make assumptions that simplify the calculations.5. Substitute the values into the Ka expression and solve for x6. Check that the assumptions are justified.

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)If f is the formal concentration of acid , x be the equilibrium concentration of H3O+. Then, HA (aq) + H2O (l) H3O+(aq) + A- (aq) f - x x x

Problem-Solving Approach : Problems Involving Weak-Acid Equilibria

The equation can be solved using quadratic eqn.Iff>>x, then x2 Kaf x = If x is 5% of the initial concentration, we can use the assumption.If x is 5% of the initial concentration, it maybe best to solve the quadratic equation.Problem-Solving Approach : Problems Involving Weak-Acid Equilibria

Percent ionisation is another method to assess acid strength. % ionisation

The higher the percent ionisation, the stronger the acid.Problem-Solving Approach : Problems Involving Weak-Acid Equilibria

A student prepared 0.10 M solution of formic acid HCHO2.The pH= 2.38 at 25C.

Calculate KaPercentage of ionisation.Example

Calculate Ka

HCHO2(aq) H+(aq) + CHO2-(aq)Example

Answer HCHO2(aq) H+(aq) + CHO2-(aq)

Initial

ChangeEqm

(0.10 4.2 10-3)M 0.10 M

-4.2 10-3M 0.10 M0 0+4.2 10-3M+4.2 10-3M4.2 10-3M4.2 10-3M(0.10 - 4.2 10-3)M

The percent ionisation of a weak acid decreases as its concentration increases.

Calculate the percentage of HF (Ka = 6.8 x 10-4) molecules ionised in:

a. 0.10M HF solutionb. 0.010 M HF solutionExample

Acids have more than one ionizable H atomExample: Sulfurous acid, H2SO3. H2SO3 (aq) H+(aq) + HSO3- (aq) Ka1 = 1.7 x 10-2 HSO3- (aq) H+(aq) + SO32- (aq) Ka2 = 6.4 x 10-8

Ka2 is much smaller than Ka1 ; easier to remove the first proton from polyprotic acid than the secondKa1 is much larger than Ka2 ; can estimate the pH by considering only Ka1.Polyprotic Acid

ExampleCO2 dissolved in water at 25C and 0.1 atm to form H2CO3 with conc. 0.0037 M. What is the pH ?CO2(aq) + H2O(l) H2CO3(aq)[H2CO3] = 0.0037 M

H2CO3 is a polyprotic acid:H2CO3(aq) H+(aq) + HCO3-(aq)ka1= 4.3 10-7HCO3-(aq) H+ (aq) + CO32- (aq)ka2= 4.710-11ANSWER = 4.40

Weak BasesThere is an equilibrium between the base and the resulting ionsWeak base + H2O(l) conjugate acid + OH- (aq)Eg: NH3(aq) + H2O NH4+(aq) + OH-(aq)The base-dissociation constant , Kb, is defined as

The larger the Kb the stronger the base

ExampleCalculate the concentration of OH- in a 0.15 M solution of NH3. (Kb = 1.8 x 10-5)NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) 0.15 -x x x

x = [NH4+] =[OH-] = 1.6 x10-3 M

ExampleA solution is made by adding Sodium Hypochlorite, NaClO, to water (2.0 L soln). The pH of solution is 10.50. How many moles of NaClO were added?Note:NaClO is an ionic compound, strong electrolyte.NaClO Na+ + ClO-ClO-(aq) + H2O HClO(aq) + OH-(aq)Kb = 3.3 x 10-7.

Unit in molarity pOH + pH = 14 pOH = - log [OH-] pH = - log [ H+]Given: pH = 10.50 ; base. Whats the [OH-]Knowing the [OH-] then we know the Molarity of ClO-. Molarity of NaClO mole ( in the same volume: 2.0 L )NaClO Na+ + ClO-1 mole1 mole 1 moleAnswer

ClO-(aq) + H2O(l) HClO(aq) + OH-(aq) Let say we start with 1.0 M NaClO : conc. of ClO- is 1.0 M. Then,ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)1 - xxxx; concentration of OH-* If we dont know the ClO-. Let say ClO-: zBut we know the [OH-] x . Then, z - xxx

Answer

Given pH = 10.50from : pH + pOH = 14.0pOH = 14.0 - 10.50 = 3.50pOH = - log [OH-][OH-] = 10 -3.50= 3.2 x 10-4 M x = 3.2 x 10-4 MAnswer

Tabulate in equilibrium table: ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)

ClO-(aq)H2O(l)HClO(aq)OH-(aq)Initialz-00Change-3.2 10-4-3.2 10-43.2 10-4finalz - 3.2 10-4-3.2 10-43.2 10-4

Z = [ClO- ] = 0.31 M NaClO = 0.31 M

Mol of NaClO = 0.31 (mol/ L) x (2 L)= 0.62 mol

Answer

Relationship between Ka and KbWe need to quantify the relationship between strength of acid and conjugate base.NH4+(aq) NH3(aq) + H+(aq)NH3(aq) +H2O(l) NH4+(aq) + OH-(aq)

NH4+(aq) NH3(aq) + H+(aq)NH3(aq) +H2O(l) NH4+(aq) + OH-(aq) H2O(l) H+(aq) + OH-(aq)

When two reactions are added to give a third, the equilibrium constant for the third reaction is the product of the equilibrium constant for the two added reactions:reaction 1+ reaction 2 = reaction 3 K1 x K2 = K3Relationship between Ka and Kb

For a conjugate acid -base pair Therefore , the larger the Ka the smaller the Kb. That is, the stronger the acid, the weaker the conjugate base. Taking negative logarithms:pKa = -log (Ka) and pKb = -log(Kb)then pKa + pKb = pKw

END of CHAPTER 13