CHAPTER 14

ACID/ BASE DEFINITIONS ARRHENIUS BRONSTED-LOWRY LEWIS

1. ARRHENIUS  Most limited definition of acids and bases.  Acids supply H + in aqueous solution.  Bases supply OH -1 in aqueous solution.  Limited since many bases do not contain OH -1. Ex. NH 3

2. BRONSTED-LOWRY  More general definition.  Acids are proton (H + ) donors. o When an acid loses a proton, it becomes a conjugate base.  Bases are proton acceptors. o Once the base accepts a proton, it becomes a conjugate acid.  General form of a BL acid base reaction: HA + H 2 O  A -1 + H 3 O +

EXAMPLES: Formic acid dissociates (HCOOH) Perchloric acid dissociates Acetic acid dissociates

 Relationship between acids and their conjugates: The stronger the acid, the weaker its conjugate base.

3. LEWIS  Most general definition.  Acids are electron pair acceptors. Ex. BF 3  Bases are electron pair donors. Ex. NH 3

ACID DISSOCIATION EQUILIBRIUM  Ka expressions can be written for an acid dissociation reaction.  Ka = [ products] power [reactants] power Write the Ka expressions for the sample BL reactions:

 Higher Ka values, more dissociation, stronger acids.  Strong acids do not have a Ka value s(extremely large) since the equilibrium lies so far to the right due to complete dissociation of a strong acid.  The strong acids are: Sulfuric Nitric Perchloric Hydrochloric Hydrobromic Hydroiodic  Other Ka values are given in the appendix. Take note of acetic acid’s Ka value and memorize it for the AP exam.

WATER Water’s Ka value is 1.0 x How? Consider two waters reacting…this is called auto-ionization of water. What is the concentration of H + and OH -­ in a sample of water.

MEASURING ACID and BASE STRENGTH pH scale ranges from 0-14 pH = - log [H + ] pOH = -log [OH -1 ] pH + pOH = 14

For 25 °C pH = 7 [H+] = 10 -pH pOH = 7 [OH -1 ] = 10 -pH Kw = [H + ][OH -1 ] and pKw = pH + pOH = 14

SolutionpHpOH[H +1 ][OH -1 ]Acid/base/ neutral A6.88 B8.4 E -14 C3.11 D1.0 E -7

ACID STRENGTH and CHEMICAL STRUCTURE When a substance is dissolved in water, it may behave as an ACID, a BASE, or produce a NEUTRAL solution. How does the chemical structure of a substance determine such behaviors?

FACTORS AFFECTING ACID STRENGTH  Strength measured by: o Ka value Recall Ka meaning. o pH pH = -log [H + ] measures how much an acid dissociates

1. BOND POLARITY a proton is transferred only when the H is the positive pole of the compound H +  X o in hydrides, ex. NaH o the H is negatively charged, so no H + could result. o in non-polar molecule, ex. CH 4 the electrons are not being pulled from the H, so no H + could result.

2. BOND STRENGTH also affects acid strength o strong bond – less likely to dissociate, weaker acid o weak bond – stronger acid

3. CONJUGATE STRENGTH o stable conjugates – come from a strong acid

TYPES of ACIDS and STRENGTH TRENDS  Acids that can donate more than one proton are called POLYPROTIC acids o 2 protons, ex. H 2 SO 4 = diprotic o 3 protons, ex. H 3 PO 4 = triprotic

1. BINARY ACIDS made of H and one other element. General form H-X.  H-X bond strength determines strength of the acid.  Bond strength DECREASES as the size of X increases.  Changes more drastically down a group.

Going across a period, look at ELECTRONEGATIVITY. IVVVIVII 2CH 4 NH 3 H2OH2OHF 3SiH 4 PH 3 H2SH2SHCl

2. OXYACIDS  contain OH groups bound to a central atom.  OH groups acting as bases:  When OH is bound to a group with extremely low ELECTRONEGATIVITY. Ex. metals Ca(OH) 2 KOH

OH groups acting as acids: Bound to a NON-METAL.  Does not readily lose OH.  As the ELECTRONEGATIVITY of Y increases, so will the acidity.  The O-H bond becomes more POLAR and loss of H + is favored.  When an additional OXYGEN is bound to the central Y, further increases the POLARITY of the OH bond favoring loss of the H +.

Rules for comparing oxyacid strength: 1. for oxyacids that have the same number of OH groups and the same number of O atoms, acid strength increases with increasing electronegativity of the central atom 2. for oxyacids with the same central atom, acid strength increases as the number of oxygens attached to Y increases.

Example: Place the following oxyacids in order of increasing strength: HClO HClO 2 HClO 4 HClO 3 HBrO

3. CARBOXYLLIC ACIDS  contain a carboxyl group  additional oxygens attached to the carboxyl group – draws electron density from the OH group, increasing its polarity.  strength of the carboxylic acid increases when the number of electronegative atoms increases.

Example: CH 3 OH – is this an acid? CH 3 COOH – what acid is this? How does its strength compare to CH 3 OH?

MORE PRACTICE Arrange the compounds in each of the following series in order of increasing acid strength: AsH 3 HI NaH H 2 O

Arrange the compounds in each of the following series in order of increasing acid strength: a. H 2 SeO 3 b. H 2 SeO 4 c. H 2 O

Explain why a. HCl is stronger than H 2 S as an acid. b. benzoic acid (C 6 H 5 COOH) is a stronger acid than phenol (C 6 H 5 OH) c. H 2 SO 4 is stronger than HSO 4 -1

Calculating pH of strong acids Large Ka, so products are favored. Assume the [acid] 0 = [ H + ] due to complete dissociation. What else contributes [ H + ] in solution? Auto- ionization of water Assumed to be negligible for strong acids (unless they are dilute >10 -6 M) So, pH = - log [acid]o

pH example (strong) Calculate the pH and [OH -1 ] of a 5 x 10 _3 M solution of HClO 4.

pH of weak acid solutions Ka is small Consider all sources of H+ in solution ICE table Ka expression Solve, approximate whenever possible (check <5%)

pH example (weak) Calculate the pH of a M solution of formic acid HCOOH (Ka = 1.77 x ).

Percent dissociation Shows how much of the initial acid is turned into H+ in solution. % dissociation = amount dissociated x 100 initial concentration Calculate by comparing H+ at equilibrium to [acid]0. Ex. What is the percent dissociation in the formic acid problem?

pH of a weak acid mixture Consider all sources of H+ and determine the one that is most significant. Assume the others to be negligible. LeChatlier’s principle predicts that acids with lower Ka values will not be able to dissociate due to an abundance of H+ from the stronger acid’s dissociation, so the weaker acid’s dissociation is suppressed.

pH example (mixture of acids) Calculate the pH of a mixture of 2.00 M formic acid and 1.50 molar hypobromous acid (Ka = 2.06 x ). 1. Write dissociation reactions for all species in solution. HCOOH HOBR H 2 O

2. Determine the source of H+ in solution 3. ICE HCOOHH+COOH-

4. Ka expression 5. H+ at equilibrium, pH

Ka from percent dissociation A M solution of uric acid is 1.6% dissociated. Calculate the value of Ka for uric acid. HAH+A-

BASES Basically, the problems are the same as acids. Focus on [OH-] at equilibrium. Consider all sources of [OH-] including auto-ionization of water. pOH = - log [OH-]

pH example (strong base) Calculate the pH of a solution made by putting 4.63 grams of LiOH into water and diluting to a total volume of 400 ml.

pH example (weak base) Calculate the pH of a M solution of methylamine, CH3NH2 (Kb = 4.38 x )

Polyprotic acids Supply more than one proton to the solution. Each step has its own Ka value. Ka1 >> Ka2 > Ka3, so many times it is possible to ignore the contribution of H+ from the second (and third)dissociation. Sulfuric acid is the exception to this rule!!!

pH example (polyprotic acid) Calculate the pH, [PO4 -3 ] and [OH-] of a 6.0 M phosphoric acid solution. 1. Write all relevant dissociation reactions and find Ka values for each.

2. ICE, assumptions, Ka1 3. Ka2 to find [HPO4 -2 ]

4. Ka3 to find [PO4 -3 ] 5. [OH-], pOH, pH

ACID/ BASE properties of SALTS Ionic compounds dissociate in water and the resulting ions can make the solution acidic, basic, or neutral. Na+ and other alkali and alkaline earth metals do NOT exhibit acid or base properties. What about Na (s)? Conjugate of strong acids or bases do NOT exhibit acid or base properties. Kw = Ka x Kb = 1x If Ka> Kb, acidic If Kb> Ka, basic

Salt examples Predict whether each of the following will create an acid, base, or neutral aqueous solution. Na3PO4 KI NH4F

pH example (of a salt) Calculate the pH of a M NaNO 2 solution (Ka = 4.0 x ) Reaction Kb ICE

ACID-BASE PROPERTIES of SALT SOLUTIONS When most salts are dissolved in water they will form acidic, basic, or neutral solutions. When both the cation and the anion have an effect, the ion with the weakest conjugate acid or base will have the greatest influence on pH.

ANIONS Reaction of anion in water: A -1 + H 2 O  HA + OH -1 If a strong acid is predicted as a product, the equilibrium will lie to the left. No considerable acidic/ basic effects will be noted.  If a weak acid is predicted as a product, the equilibrium will lie to the right. Due to the production of hydroxide ion, the pH of the solution will increase (become more basic).

CATIONS Polyatomic cations are considered to be conjugates of weak bases. Ex. NH H 2 O  hydroxide ions are formed and the pH is increased. Most metal ions can also react with water to decrease the pH of an aqueous solution. Some exceptions include the alkali metals and some alkaline earth metals (usually produce bases in H 2 0). Metal ions are considered Lewis acids.

LEWIS DEFINITION of ACIDS and BASES  Lewis base – electron pair donor Ex. NH 3  Lewis acid – electron pair acceptor

Metal ions Positively charged metal ions attract the oxygens from the water molecules. This reaction is known as hydration. usually coordinate with # waters = 2 x metal ion’s charge Example: Fe +3 in water: (coordinates with 6 water molecules) Acid dissociation constants for hydrolysis reactions incerase with increasing charge and decreasing radius on the metal ion. Ex. Which would form a more acidic solution Cu +2 or Fe +3 ?

COMMON ION EFFECT (add on) Adding a salt to an acid base equilibrium that contributes to the initial conjugate concentration. Ex. What is the pH of a solution that is 0.5M in acetic acid and 2.5M in sodium acetate, NaCH 3 COO? Consider both dissociation rxns. What ions does the salt contribute to the acid’s equlibrium? (common ion) What does LeChatelier predict? Would the acid pH increase or decrease due to the common ion?

BUFFERS  Solutions that contain weak conjugate acid-base pairs.  Resist changes in pH when small amounts of strong acid or base are added.  Examples of common buffer systems:  Blood  Hydrogen carbonate/ carbonic acid  Contain an acid species to neutralize hydroxide and a basic species to neutralize hydrogen ions.

WEAK ACID BUFFER HA  H+ + A- With equilibrium expression: Ka = [H+] [A-] / [HA] Adding OH -1 to the solution: OH -1 + HA  H2O + A- HA = weak acid in the buffer – used to absorb excess base Adding H + to the solution: H + +A -  HA A- = conjugate base of the weak acid in the buffer – used to absorb excess acid

BUFFER CAPACITY and PH  Buffer capacity – the amount of acid or base a buffer can neutralize before pH changes to a considerable degree.  PH depends on the Ka for the acid and the relative concentrations of the acid and base that comprise the buffer. Calcualting pH and relating to Ka  Uses the Henderson - Hasselbalch equation. pH = pKa + log ([A-] / [HA] )

Henderson Hasselbalch (*hoff)  Because weak acids and bases only slightly ionize, it is possible to use the initial amounts of acid and base when using the Henderson – Hasselbalch relationship. pH = pKa + log ([A-] / [ HA]) Example: Calculate the pH of a buffer composed of 0.12 M lactic acid (HC 3 H 3 O 3 ) and 0.10 M sodium lactate. Ka for lactic acid = 1.4 x

Example: What is the ratio of HCO 3 -1 to H 2 CO 3 in blood of pH 7.4?

Weak base/ conjugate acid buffer pOH = pKb + log ([BH+]/[B])

ADDITION of A STRONG ACID or BASE to BUFFER SYSTEM  Reactions between STRONG ACIDS and WEAK BASES or a STRONG BASE and WEAK ACID occur essentially to completion. So the buffer should completely consume the additional acid or base. Calculating pH after the addition of an acid or a base: 1. consider the acid-base neutralization reaction and determine its effect on [HA] and [A - ]. (Use stoichiometry). 2. Use Ka and the new concentration of [HA] and [A - ] to calculate [H + ] using an ICE table or the Henderson – Hasselbalch equation.

H+ + A-  HA OH- + HA  H2O + A- Stoichiometry determines new concentrations of the conjugates adding H+ or OH- to HA, A- at equilibrium pKa + log of the ratio of the [conjugate] [original acid or base] Neutralization of excess acid Neutralization of excess base

Example: A buffer is made by adding mole acetic acid to mole sodium acetate to make a 1.00L solution. The pH of the buffer is calculate the pH of the solution after moles of NaOH is added

calculate the pH of the solution after.020 moles of HCl is added.