Chapter 18 Acid-Base I

Acids and Bases Chapter 18 Stuff

I need the Exam 1 in my box by Friday. Those who don’t turn it in will use the Make-Up Exam for Exam 1.

March 10 Exam 3 (Chapter 18/19)

March 17 Make-Up Exam (Comprehensive + 20)

Final Exam (Comprehensive) March 24 The effects of acid rain on a statue of George Washington taken in 7:30-9:30AM 1935 (left) and 2001 (right) marble.

CaCO3(s) + 2HCl(aq) ==> H2O(l) + CO2(g) + CaCl2(aq)

Acids and bases have distinct properties. Acid-Base Equilibria Acids: • Acrid sour taste Acid 18.1 Acids and Bases in Water • React with metals (Group I,II) to yield H2 gas 18.2 Autoionization of Water and the pH Scale • Changes plant dye litmus from blue to red • React with carbonates and bicarbonates to 18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition produce CO2 gas 200 Million MT H2SO4 18.4 Solving Problems Involving Weak-Acid Equilibria

18.5 Weak Bases and Their Relations to Weak Acids Base 18.6 Molecular Properties and Acid Strength

18.7 Acid-Base Properties of Salt Solutions Bases: 18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect • Bitter taste 50 Million MT NaOH/yr 3 million containers • 18.9 Electron-Pair Donation and the Lewis Acid-Base Definition Slippery feel • Changes plant dye litmus from red to blue • React and neutralizes the effects of acids

Acids and bases are everywhere.

Common Acids have three definitions

, soft drinks

Most anti-perspirants, water treatment plants, paper

Common Bases Chemistry uses three definitions of acids and Arrhenius Acid/Base definitions arise from the reaction with water. bases: 1) Arrehenius 2) Bronsted-Lowry 3) Lewis Acid: a substance that has a covalent H atom in its formula, and releases a proton H+ when dissolved in water. Name Acid Definition Base Definition HCl(g), HBr(g), HI (g), HNO3(l), H2SO4(l), HClO4(l)

Substance that Substances that HCl(aq), HBr(aq), HI (aq), HNO3(aq), H2SO4(aq), HClO4(aq) Arrhenius increases H+ increase OH- H2O HA(g) <==> H+(aq) + A-(aq) Brønsted- Substances that Substances that Base: a substance that contains OH in its formula, and Lowry donate H+ accept H+ releases hydroxide ions (OH-) when dissolved in water.

Electron-pair Electron-pair NaOH(s), KOH (s), LiOH (s), Mg(OH)2(s), Ca(OH)2(s) Lewis acceptor donar NaOH(aq), KOH (aq), LiOH (aq), Mg(OH)2(aq), Ca(OH)2(aq)

H2O MOH(s) <==> OH- + M+

H+ has never been isolated in the lab, but chemists Acid and base formation is represented symbolically use it symbolically anyway. H+ reacts with water to in two ways semi-confusing ways. Learn both & + form the hydronium ion, H3O don’t let it distract you.

H2O The lazy but 1. HA(g) <==> H+ + A- convenient way. H2O written over arrow H2O with H+ as the acid. MOH(s) <==> OH- + M+

+ + H and H3O are equivalent in chemistry. Reacting + - 2. HA(g) + H2O(l) <==> H3O (aq) + A (aq) with H2O in the equation with H3O - + MOH(s) + H2O(l) <==> OH (aq) + M (aq) + H2O as acid.

An equilibrium constant called “Acid-Dissociation Chapter 17 & 18 Homework Problems Constant, Ka quantifies the extent of acid dissociation (i.e. acid strength). Problems have been updated for the Principles of Chemistry, Silberberg, 2nd Edition Textbook.

H2O [H+][A-] HA(g) <==> H+ + A- K = a [HA] Exam 2 February 12/13/14?

+ - + - [H3O ][A ] HA(g) + H2O(l) <==> H3O (aq) + A (aq) K = a [HA] Chapter 17: 1,2,3,9,13,17,19,23,25,28,30,32,35,39,42,44,46,50 ,51,54,59,64,66,69,72,80,81,86,89,90. + stronger acid => higher [H3O ] => large Ka Chapter 18: 5,6,7,9,11,15,16,18,22,26,27,30,32,34,38,44,45,47 + weaker acid => lower [H3O ] => small Ka ,49,51,60,64,66,74,81,83,90,92,104,120 Strong acids and strong bases completely Because of complete dissociation, strong acids dissociate or completely ionize in water: Ka >> 1. and strong bases conduct electricity and are strong electrolytes. - + [A ][H3O ] + - HA(g) + H2O(l) H3O (aq) + A (aq) Ka = Uni-directional arrow used [HA] to heavily product-favored product-favored HCl(aq) H+(aq) + Cl-(aq) Strong Strong + Before After H3O dissociation dissociation + - HNO3(aq) H (aq) + NO3 (aq)

+ - H2SO4(aq) H (aq) + HSO4 (aq)

NaOH(aq) Na+(aq) + OH-(aq)

2+ - + - Ca(OH)2(aq) Ca (aq) + 2OH (aq) HA HA H3O A Ka >>1

Weak acids and weak bases dissociate only to a Weak acids and weak bases do not dissociate or slight extent in water Ka << 1. ionize to a large extent in solution, and so are weak electrolytes (small value of Ka). - + [A ][H3O ] + - - + HA(aq) + H2O(l) H3O (aq) + A (aq) Ka = [A ][H3O ] [HA] Ka = = << 1 All not strong is weak! reactant-favored [HA]

Before After - + HNO2 (aq) <=> NO2 (aq) + H (aq) dissociation dissociation - + HA CH3CO2H(aq) <=> CH3CO2 (aq) + H

HA - + HA H2SO3 (aq) <=> HSO3 (aq) + H (aq) HA

HA HA

+ - HA HA H3O A

Ka Values For A Few Monoprotic Acids at 25C Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base.

(a) H2SeO4 (b) (CH3)2CHCOOH (c) KOH (d) (CH3)2CHNH2

Pay attention to the text definitions of acids and bases. Look at O for acids as well as the -COOH group; watch for amine groups and cations in bases. Classify each of the following compounds as a strong Acids can have one, two or three acidic protons acid, weak acid, strong base, or weak base. depending on their structure (and Ka1 Ka2 Ka3).

Monoprotic acids--only one H+ available (a) H2SeO4 (b) (CH3)2CHCOOH (c) KOH (d) (CH3)2CHNH2 HCl H+ + Cl- Strong electrolyte, strong acid Pay attention to the text definitions of acids and bases. Look at O for acids + - HNO3 H + NO3 Strong electrolyte, strong acid as well as the -COOH group; watch for amine groups and cations in bases. + - CH3COOH H + CH3COO Weak electrolyte, weak acid

Diprotic acids--two acidic H+ available for reaction (a) Strong acid - H2SeO4 - the number of O atoms + - exceeds the number of ionizable protons by 2. H2SO4 H + HSO4 Strong electrolyte, strong acid - + 2- Weak electrolyte, weak acid (b) Weak acid - (CH3)2CHCOOH is an organic acid HSO4 H + SO4 having a -COOH group. + (c) Strong base - KOH is a Group 1A(1) hydroxide. Triprotic acids--three acidic H H PO H+ + H PO - (d) Weak base - (CH ) CHNH has a lone pair of 3 4 2 4 3 2 2 H PO - H+ + HPO 2- Weak electrolyte, weak acid electrons on the N and is an amine. 2 4 4 2- + 3- HPO4 H + PO4

Acid and bases react with one another in a The Arrhenius acid-base definition fails to cover neutralization reaction forming a salt and water. all cases..so chemists invented the Bronsted- Lowry definition. + - NaOH(aq) + HCl(aq) <==> H2O + Na + Cl How Arrhenius Defintion Fails base + acid <==> Water + salt 1. NH3 or amines (like R-NH2) don’t have OH group but are + - Net Ionic Equation: H (aq) + OH (aq) => H2O (aq) bases.

2. Many transition metals Cr, Al, Fe, Cu are acidic but have no H+. + - CH3COOH(aq) + NaOH (aq) <==> H2O + Na + CH3COO 3. Arrhenius definition requires Weak acid + strong base <==> Water + salt water...yet some acid base reactions occur without water HCl(g) + NH3(g) => NH4Cl(s) as solvent. gaseous without water

Acid-BaseThe Bronsted-Lowry Reaction concept = Proton views acids Transfer as proton An Bronsted acid is a substance that “donates” or donars and bases as proton acceptors. a proton to a Bronsted base. A Bronsted base is a substance that “accepts” a proton from another a Bronsted acid.

Conjugate Acid Conjugate Base

Conjugate Acid Conjugate Base • For a molecule or ion to be a Bronsted base, it Generalized Bronsted Acid and Base must have at least one unshared pair of electrons for accepting a proton. HA + B <------> A-- + BH+

All of these • H2O have a • H2O Acid Base, Conjugate Conjugate lone-pair • HF a H+ donor a H+ acceptor Base Acid of e- and • F- All can are bases! donate H+ • HCl • Cl- Conjugate pair Conjugate pair and are • HNO acids! 3 • NH3 The Lone pairs • HBr A- is called the conjugate base of the acid HA. Has a free are there if you • OH- pair of electrons to “accept a proton”. draw the Lewis • HClO4 structure! BH+ is called the conjugate acid of the base B Both are called conjugate acid-base pairs!

Amphoteric substances can act an Bronsted acid Water is amphoteric, it can act as a base or acid (proton donar) or a base (a proton acceptor). and depending on what it reacts with.

Acid Base--H+ acceptor Conjugate Acid Conjugate Base

Solvent

proton donor proton acceptor

Base Acid--H+ donor Base Acid

Identifying the following conjugate acid-base pairs Write the equilibrium expression nitrous acid + (H donars and H+ acceptors). (HNO2) and explain its acidity in terms of a) the Arrhenius definition, and b) the Brønsted-Lowry conjugate take away H+ acid base theory, c) identify the acid base conjugate pairs. + + acid1 base2 base1 acid2 add H+

reaction 1 HF + H O - + + 2 F H3O

reaction 2 HCOOH + CN- HCOO- + HCN

reaction 3 + + 2- NH + - NH4 CO3 3 HCO3

reaction 4 - + - 2- + H O H2PO4 OH HPO4 2 reaction 5 H SO + + - + N H 2+ 2 4 N2H5 HSO4 2 6

reaction 6 2- + 2- 3- + - HPO4 SO3 PO4 HSO3 a) HNO2 is an Arrhenius acid because it dissociates The stronger the acid (base) is, the weaker is in water to produce H+ ions: its conjugate base (acid).

+ - HNO2(aq) <===> H (aq) + NO2 (aq) - + HCl(g) + H2O(aq) <===> Cl (aq) + H3O (aq)

• b) Nitrous acid is a Brønsted-Lowry acid because it acts very weak base as a proton donor, transferring a proton to water to give + - NaOH(s) + H2O(l) <===> Na + H2O + OH hydronium ion very weak acid Weak acids (bases) give strong conjugate bases (acids). - + HCOOH + H2O <===> HCOO + H3O

basic in water - + - • The conjugate base of HNO2 is NO2 , the species that NH3(aq) + H2O(aq) <===> NH4 (aq) + OH (aq)

remains after HNO2 has lost a proton acidic in water

The weaker the acid--the stronger its conjugate base. An acid-base reaction will go in that direction that The stronger the acid--the weaker its conjugate base. forms the weaker acid-base pair. Relative Strengths of Conjugate Acid-Base Pairs Acid (HA) Conjugate Base, A- -4 Given that Ka for HF = 6.8 X 10 and -8 Ka for H2S = 9.0 X 10

In which direction will this reaction proceed?

- - HS (aq) + HF(aq) F (aq) + H2S(aq)

In which direction will these reactions proceed? In which direction will these reactions proceed?

+ - + - H2SO4(aq) + NH3(aq) NH4 (aq) + HSO4 (aq) H2SO4(aq) + NH3(aq) NH4 (aq) + HSO4 (aq)

- 2- - 2- - 2- - 2- HCO3 (aq) + SO4 (aq) HSO4 (aq) + CO3 (aq) HCO3 (aq) + SO4 (aq) HSO4 (aq) + CO3 (aq) In which direction will these reactions proceed? Water is usually a “reactive-solvent” with all acids and bases. + - H2SO4 and NH4 are acids, NH3 and HSO4 are bases. H2SO4 is + - stronger acid than NH4 , and NH3 stronger base than HSO4 , + - HClO4(l) + H2O(l) <==> H3O (aq) + ClO4 (aq) therefore NH3 gets the proton - + NaOH(s) + H2O(l) <==> OH (aq) + Na (aq) + H2O(l) + - H2SO4(aq) + NH3(aq) ! NH4 (aq) + HSO4 (aq) stronger acid stronger base weaker acid weaker base Water’s reaction can be viewed as acid-base - - 2- 2- reaction. HCO3 and HSO4 are acids, SO4 and CO3 are bases. - 2- HSO4 is the stronger acid, and CO3 is the stronger base. + - HClO4(l) + H2O(l) <==> H3O (aq) + ClO4 (aq) 2- Therefore, CO3 gets the proton acid + base <==> conjugate acid + base - 2- - 2- HCO3 (aq) + SO4 (aq) " HSO4 (aq) + CO3 (aq) weaker acid weaker base stronger acid stronger base + - Net Ionic Equation: H (aq) + OH (aq) => H2O (aq)

+ - The hydronium ion, H3O , and the OH are the Acid Strength Is Summarized in Tables of Ka and pKa strongest acid and base that can exist in the presence Acid Ionization Equilibrium Ka pKa of water. + - Hydroiodic Acid HI + H2O <==> H3O + I 3 X 109 -9 + - We can not Hydrochloric Acid HCl + H2O <==> H3O + Cl 1.3 X 106 -6 + - + - + H2SO4 + H2O <==> H3O + HSO4 H2O(l) + HCl(aq) ==> H3O + Cl distinguish H3O Sulfuric Acid 1000 -3 + - H3O + H2O <==> H3O + IO3 + - in these cases. Hydronium Ion 55 -1.7 H2O(l) + HNO3(aq) ==> H3O + Cl + - All proton Nitric Acid HNO3 + H2O <==> H3O + NO3 28 -1.4 + - + - -1 H2O(l) + H2SO4(aq) ==> H3O + HSO4 acceptors give Iodic acid HIO3 + H2O <==> H3O + IO3 1.61 X 10 0.80 - OH in aqueous + - Chlorous Acid HClO2 + H2O <==> H3O + ClO2 1.1 X 10-2 1.96 + solutions. + - Strong acid ===> produce H3O Phosphoric Acid H3PO4 + H2O <==> H3O + H2PO4 7.6 X 10-3 2.12 + - Nitrous Acid HNO2 + H2O <==> H3O + NO2 7.2 X 10-4 3.14 + - We can not Hydrofluoric Acid HF + H2O <==> H3O + F 6.61 X 10-4 3.18 - - - O2 (l) + H2O(l) <==> 2OH distinguish OH in + - Formic Acid HCOOH + H2O <==> H3O + HCO2 1.8 X 10-4 3.74 these cases. All + - - - CH3CO2H + H2O <==> H3O + CH3CO2 -5 NH2 (aq) + H2O(l) <==> NH3(aq) + OH Acetic Acid 1.8 X 10 4.74 proton acceptors + - H2S + H2O <==> H3O + HS -7 give OH- in Hydrosulfuric Acid 1.32 X 10 6.88 + - aqueous Hypochlorous Acid HClO + H2O <==> H3O + ClO 2.88 X 10-8 7.54 + - solutions. Hydrocyanic Acid HCN + H2O <==> H3O + CN 6.2 X 10-10 9.21

+ - -14 The product of [H3O ][OH ] = 1 X 10 a constant. Kw is the equilibrium constant that describes Once we know one...... we know the other! the auto or self-ionization of water. + - [H3O ] [OH ] Kw

+ – + - -14 -0 -14 H2O(l) H (aq) + OH (aq) Kw = [H ][OH ] 1 X 10 1 X 10 1 X 10 -13 -1 -14 + – + - 1 X 10 1 X 10 1 X 10 H2O(l) + H2O(l) H3O (aq) + OH (aq) Kw = [H3O ][OH ] -12 -2 -14 Base Acid Acid Base 1 X 10 1 X 10 1 X 10 1 X 10-11 1 X 10-3 1 X 10-14 1 X 10-10 1 X 10-4 1 X 10-14 + - -14 1 X 10-9 1 X 10-5 1 X 10-14 Kw = [H3O ][OH ] = 1 X 10 1 X 10-8 1 X 10-6 1 X 10-14 1 X 10-7 1 X 10-7 1 X 10-14 1 X 10-6 1 X 10-8 1 X 10-14 This equilibrium constant for water is called the 1 X 10-5 1 X 10-9 1 X 10-14 “Ion-Product of Water”. Memorize the value. 1 X 10-4 1 X 10-10 1 X 10-14 [OH-] and [H+] can vary over 14 orders of magnitude pH is a defined logarithmic scale for measuring acidity. It is directly related to [H+] If we plot a few concentration in solution. points of [OH-] vs [H+] then we see that we can not + + plot all the various pH = -log10 [H ] = -log10 [H3O ] logarithmic form powers of 10 that - + Basic OH > H3O can exist in ] - solution. p = -log10

[OH In 1909, [H+] = 10-pH exponential form + - Sorenson Neutral H3O = OH suggests using a logarithmic plot + - acidic H3O > OH and this makes things easy to see 1 pH unit represents a power of 10X in concentration and to plot. It (or an order of magnitude as it is called). + [H3O ] defines pH!

We can compress all 14 orders of magnitude using Acidic, basic and neutral solutions are identified a logarithmic scale--called the pH scale! by their pH values + - [H3O ] pH [OH ]

+ pH = -log [H3O ] Basic (pH > 7, pOH < 7) ] - 14 orders of connect these dots magnitude can (pH = 7, pOH = 7) be seen in a Neutral + - -14 single graph! Kw = [H3O ][OH ] = 10

pH = -log [OH Acid (pH < 7, pOH > 7) The ion-product of water is a constant!

+ pH = -log [H3O ]

+ + - A solution with a pH of 4 is 10X lower in [H ] than The amounts of H3O and OH in aqueous solution one with pH=3, 100X lower than pH=2 and 1000X determines whether a solution is acidic, neutral or lower than pH = 1. basic.

WHEN [H+] -log[H+] = pH + - + - + - [H3O ] > [OH ] [H3O ] = [OH ] [H3O ] < [OH ] 0.5M 0.301 acidic neutral basic 0.05M 1.301 solution solution solution pH < 7 pH = 7 pH > 7 0.005M 2.301 0.0005M 3.301 Litmus Paper Color

When concentration changes by factor of 10, pH changes by +/- 1 A Logarithm of a number is an Exponent Logs are easy if you see the riff.

Let y be any number. Let b also be a number and let’s What is the log10 of the following numbers? call it “the base”, b. Lastly let’s call the power we raise the base to an “exponent” and label it as “x”. In 1000, 100, 10, 1, 0, 0.1, 0.01, 0.001 equation form:

means X y = bx b = base, x = exponent and y = a number log10 1000 = log 1000 = x 10 = 1000 means X 100 = 10x b = 10 and y =100 and x = exponent log10 100 = log 100 = x 10 = 100 log 10 = x means 10X = 10 The logarithm of 100 = log (100) is the value of the X exponent, x, needed to raise 10 to in order to obtain log 1 = x 10 = 1 100. log 0 = x 10X = 0 The logarithm of 100 is the value of the exponent in the log 0.1 = x 10X = 0.1 equation: 10X = 100 log10 (100) = x = 2 log 0.01 = x 10X = 0.01

Two very common logarithm functions are log to Just as there are distributive, associative the base 10 and the natural log base e (2.7183). properties with equations there are “properties of logs that should be memorized.

Properties of Logs

log = log10 “log to the base 10” log (A B) = log A + log B × loge = log2.7183 = ln the “natural log” A log = log A log B B − log An = n log A

antilog(log x) = 10log x = x

Antilog of a number means use the LogKnowing and the Exponential math can help youForms go fast number as an exponent.

Sometimes we know the base and the exponent, but we need Logarithmic Form Exponential Form to compute the number. We can write it like this: + -pH + pH = -log [H3O ] 10 = [H3O ] log (?) = 2 - -pOH - In words we would say: “The log10 of what number is equal to 2”? pOH = -log [OH ] 10 = [OH ] Finding this number is called taking the “antilogarithm” -pKa pKa = -log Ka 10 = pKa Antilog [log (?)] = Antilog (2)

+ - ? = 102 = 100 When H3O , OH , Ka are large values----the pK’s are small values. Strong acids have large H3O and Ka

The antilog10 is the same as the base 10 operator! therefore small pH and small pKa Figure 18.7 Methods for measuring the pH of an aqueous solution

Kw is related to pH via logarithm. pH is measured with dyed paper or with a commercial instrument. + - -14 KW = [H3O ][OH ]= 1.0 X 10 pH meter take -log of both sides

+ - -14 -log Kw = -log ([H3O ][OH ]) = -log(1.0 X 10 )

+ - -14 -log KW = - log[H3O ] + -log[OH ] = -log (10 ) pH (indicator) paper pKW = pH + pOH = -(-14)

pKW = pH + pOH = 14

+ - -14 KW = [H3O ][OH ] = 10

What is the concentration of OH- ions in What is the concentration of OH- ions in a HCl solution whose hydrogen ion a HCl solution whose hydrogen ion concentration is 1.3 M? concentration is 1.3 M? Key: Strong acids dissociate completely-- stoichiometry dictates [H+]. Memorize all strong acids (HCl, HBr, HI, HNO3, H2SO4, HClO4)

[H+] = 1.3 M

+ - -14 Kw = [H ][OH ] = 1.0 x 10

-14 Kw 1 x 10 [OH-] = = = 7.7 x 10-15 M [H+] 1.3

A research chemist adds a measured amount of A research chemist adds a measured amount of HCl gas to pure water at 25oC and obtains a HCl gas to pure water at 25 oC and obtains a + -4 - + -4 - solution with [H3O ] = 3.0 x 10 M. Calculate [OH ]. solution with [H3O ] = 3.0 x 10 M. Calculate [OH ]. Is the solution neutral, acidic or basic? Is the solution neutral, acidic or basic?

Again, strong acids dissociate completely-- stoichiometry dictates [H+]

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

- + -14 -4 -11 [OH ] = Kw/ [H3O ] = 1.0 x 10 /3.0 x 10 = 3.3 x 10 M

+ - [H3O ] > [OH ]; the solution is acidic. Significant figures in logarithm calculations follow different rules than we are used to. The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. 1. When taking the log of a measured number, retain in the What is the H+ ion concentration of the rainwater? mantissa (the numbers to the right of the decimal point in the answer) the same number of significant figures as there are in the number whose logarithm you are taking. log(3.000 x 104) = ? = 4.4771 4 SF’s 4 SF’s after decimal pt log(3. x 104) = ? = 4.5 The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood? 1 SF’s 1 SF’s after decimal pt 2. When taking an antilog, the number of SF’s in the answer is the same as the number of digits after the decimal point. antilog(4.4711) = 104.4711 = 3.000 x 104 4 SF’s after decimal pt 4 SF’s

+ - The pH of rainwater collected in a certain region of the Problem: Calculating [H3O ], pH, [OH ], and pOH northeastern United States on a particular day was 4.82. In a restoration project, a conservator prepares What is the H+ ion concentration of the rainwater? copper-plate etching solutions by diluting + pH = -log [H ] concentrated nitric acid, HNO3, to 2.0 M, 0.30 M, + + - antilog(-pH) = antilog (log [H ]) and 0.0063 M HNO3. Calculate [H3O ], pH, [OH ], and pOH of the three solutions at 25 oC. [H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

- -7 + The OH ion concentration of a blood sample is 2.5 x 10 M. PLAN: HNO3 is a strong acid so [H3O ] = [HNO3]. Use Kw to What is the pH of the blood? find the [OH-] and then convert to pH and pOH. pH + pOH = 14.00 pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

+ - Problem: Calculating [H3O ], pH, [OH ], and pOH Acid-Base Equilibrium Problems In a restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated nitric acid, HNO3, 1 2 + to 2.0 M, 0.30 M, and 0.0063 M HNO3. Calculate [H3O ], pH, Strong Acids/Bases Weak Acid Weak Bases [OH-], and pOH of the three solutions at 25 oC. For 2.0 M 100% dissociation + + 1. Find Ka /Kb given equilibrium values HNO3, [H3O ] = 2.0 M, and pH = -log [H3O ] = -0.30 = pH [OH-] = K / [H O+] = 1.0 x 10-14/2.0 = 5.0 x 10-15 M; pOH = 14.30 2. Find equilibrium values given Ka w 3 No need for ICE table. + [H3O ] = [HA]. For 0.3 M HNO3, + + [H3O ] = 0.30 M and -log [H3O ] = 0.52 = pH Identify and write the correct equation - + -14 -14 [OH ] = Kw/[H3O ] = 1.0 x 10 /0.30 = 3.3 x 10 M; pOH = 13.48

For 0.0063 M HNO3, 1. Simplify: If 100 x Ka/b < [A]0 then [HA] - x = [HA] + + [H3O ] = 0.0063 M and -log [H3O ] = 2.20 = pH 2. Look for perfect square - + -14 -3 -12 3. Quadratic equation [OH ] = Kw/ [H3O ] = 1.0 x 10 /6.3 x 10 = 1.6 x 10 M; pOH = 11.80 Strong acids and bases dissociate completely while Know all strong acids and bases and you will weak acids (bases) are equilibrium dependent. recognize be able to recognize all weak acids/bases.

+ HCl(aq) + H O H O (aq) + Cl−(aq) 2 −−→ 3

0.004% at 99.996% at STRONG ACID equilibrium equilibrium hydrohalic Group I acids + hydroxides 1.0M HCl 1.0M H3O perchloric + nitric CH3CO2H(aq) + H2O H3O (aq) + CH3CO2−(aq) ←−→ sulfuric Group II WEAK hydroxides 98.7% at 1.3% at ACID equilibrium equilibrium

+ -2 % ionization of an acid is the ratio of [H ] at What is the pH of a 1.8 x 10 M Ba(OH)2 solution? equilibrium to [HA]initial X 100. Ba(OH)2 is a strong base – 100% dissociation. - 1 mol Ba(OH)2 = 2 mol OH + Ba(OH) (s) Ba2+ (aq) + 2OH- (aq) [H ]equilbrium 2 % ionization = x 100% [HA] Initial 0.018 M 0.0 M 0.0 M inital Change -0.018 M 0.018 M 0.036 M Equilibrium 0.0 M 0.018 M 0.036 M

- [OH ]equilbrium pKw = pH + pOH = 14.00 % ionizationb = x 100% [OH-] pH = 14.00 – pOH inital pH = 14.00 - (- log(0.036)) = 12.56

As the concentration of a weak acid (base) decreases, the Strong Acid Equilibrium Sample Problem % ionization of the acid (base) increases!

-3 What is the pH of a 2.00 x 10 M HNO3 solution? Strong Acid Rationale: larger solution volume accommodates more ions (analogous to pressure effects on gas equilibria when 1. What does the problem ask--write it down make a picture? #ngas is non-zero) 2. Is the acid or base--strong or weak? 3. Can we make an ICE table? % % Ionization + - [H3O ][A ] Weak Acid Ka = Weak Acid [HA] n n H O+ A- 1 K = 3 Acid Molarity a n HA V What is the pH of a 2 x 10-3 M HNO solution? There are 2 key simplifications that can help us 3 solve equilibrium problems rapidly.

HNO3 is a strong acid it is100% dissociated. + 1. [H3O ] from the auto-ionization of water is negligible in comparison to H+ from an acid, HA in solution. + - HNO3 (aq) + H2O (l) H3O (aq) + NO3 (aq)

2. If 100 x Ka < [HA]initial then [HA]initial - x = [HA] Initial 0.002 M 0.0 M 0.0 M +0.002 M 0.002 M Change -0.002 M Why Does it Work? Equilibrium 0.0 M +0.002 M +0.002 M Because Ka values are known to about ± 5% accuracy.

+ [H ]equilbrium Check using the 5% Rule x 100 < 5% [HA] + inital pH = -log [H3O ] = -log(0.002) = 2.7 3. If #2 fails then you must use the quadratic equation

Find the Ka of a weak acid from the pH of its solution Find the Ka of a weak acid from the pH of its solution

What is the Ka of phenylacetic acid (HPAc)? Phenylacetic acid (C6H5CH2COOH, denoted as HPAc) HPAc( ) + H O( ) + - builds up in the blood of people afflicted with aq 2 l H3O (aq) + PAc (aq) phenylketonuria, an inherited genetic disorder that, if initial 0.12 - 1 x 10-7 0 left untreated, causes mental retardation and death. A change -x - +x +x study of the acid shows that the pH of a 0.12 M solution equilibrium 0.12 - x - x +(1 x 10-7) x of HPAc is 2.60. What is the Ka of phenylacetic acid? + - -7 2 [H3O ][PAc ] [x + (10 ) ][x] [x ] PLAN: Same as last chapter K = = = a [HPAc] [0.12 - x] [0.12 - x] 1. Write a balanced equation for acid in water. + Solution pH = 2.6 = -log[H3O ] -7 + 2. Write the equation for Ka. >> 10 (the [H3O ] + -pH -2.6 -3 contribution from 3. Set-up and fill in the ICE Table [H3O ] = 10 = 10 = 2.5 x 10 M 4. Use pH, initial solution concentration and ICE water) + - -3 table to find Ka. [H3O ] $ [PAc ] = x = 2.5 x 10 M -3 [HPAc]eq = 0.12 - x = 0.12 - 2.5 x 10 ! 0.117 M

Determining concentrations and pH from Ka and initial [HA]

+ - [H3O ][PAc ] [x ]2 The pH of a solution containing 0.250 M HF Ka = = [HPAc] [0.12 - x] solution is 2.036. What is the value of Ka for HF at equilibrium?

(2.5 x 10-3)2 K = = 5.2 x 10-5 a 0.117 Determining concentrations and pH from Ka and initial [HA] Determining concentrations from Ka and initial [HA] The pH of a solution originally containing 0.250 M HF Propanoic acid (CH3CH2COOH, simplified as HPr) is solution is 2.036. What is the value of Ka for HF at an organic acid whose salts are used to retard mold + equilibrium? growth in foods. What is the [H3O ] of a 0.10 M + - -5 HF + H2O H3O + F aqueous solution of HPr (Ka of HPr = 1.3 x 10 )? Initial .250 M ? M ? M Change -x M +x M +x M PLAN: 1. Write a balanced equation for acid in water. Equilibrium 0.250-x x x 2. Write the equation for Ka. 3. Set-up and fill in the ICE Table + 4. Use pH, initial solution concentration and ICE [H3O ][F−] pH = 2.036 = -log [H O+] K = table to find K . 3 a [HF] a -pH + -2.036 -3 10 = [H3O ] = 10 = 9.20 x 10 M [HF] = 0.25 – 0.00920 = 0.241 M

-3 2 -4 Ka = (9.2 x 10 ) /0.241 = 3.52 x 10

+ - HPr(aq) + H O(l) H O (aq) + Pr (aq) + - 2 3 [H3O ][Pr ] x2 K = = 1.3 x 10-5 = initial 0.10 - 0 0 a [HPr] 0.10 change -x - +x +x equilibrium 0.10 - x - x x solving for x + - [H3O ][Pr ] x2 K = = 1.3 x 10-5 = a 0.10 - x -3 + [HPr] = 1.1 x 10 M = [H3O ]

Can We Simplify?

100 x Ka < [HA]initial then + Checking assumption with 5% Rule: If [H ]equilibrium If No [HA]initial - x = [HA] -b ± ! b2 – 4ac divided by [HA]initial (x 100) is < 5% then assumption is valid. x = 2a 100 x 1.3 x 10-5 < [.10]? 3 Yes, it is: .10 - x = 0.10 1.1 10− M %[HPr]diss = × = 1.1% << 5% + - 0.10M [H3O ][Pr ] x2 K = = 1.3 x 10-5 = a [HPr] 0.10 our assumption is valid

Using Molecular Scenes to Determine the Extent of HA Using Molecular Scenes to Determine the Extent of HA Dissociation Dissociation A 0.15 M solution of acid HA (blue and green) is 33% A 0.15 M solution of acid HA (blue and green) is 33% dissociated. Which scene best represents a sample of the dissociated. Which scene best represents a sample of the solution after it is diluted with water? solution after it is diluted with water?

+ [H3O ] % dissociation = + x 100% [HA] + [H3O ] Percent dissociation increase as solution is diluted. Calculate the percent dissociation of each sample. SOLUTION: Solution 1. % dissociated = 4/(5 + 4) x 100 = 44% Solution 2. % dissociated = 2/(7 + 2) x 100 = 22% + Solution 3. % dissociated = 3/(6 + 3) x 100 = 33% [H3O ] % dissociation = + x 100% Scene 1 represents the diluted solution. [HA] + [H3O ] Polyprotic acids are acids that have more than one ionizable H+ when reacted with water

Phosphorous Acid Each H has its (H3PO3) own Ka

Sulfurous Acid (H2SO3) The electronegativity Polyprotic Acids Phosphoric Acid of the bonded atoms influence the ionizability Arsenic Acid of an acidic (H3AsO4) hydrogen in a substance.

Citric Acid (H3C6H5O7)

Phosphoric acid, H3PO4 Ka quantifies and communicates the degree of acid dissociation = an acids strength + - - + [H3O ] [H2PO4 ] [H O+][A-] H3PO4(aq) + H2O(l) H2PO4 (aq) + H3O (aq) Ka1 = + - 3 HA(aq) + H2O(l) H3O (aq) + A (aq) K = [H3PO4] a [HA] = 7.2 x 10-3 strong acid = Ka is large, weak acid = Ka is small,

+ 2- pKa is small pKa is large - 2- + [H3O ] [HPO4 ] H2PO4 (aq) + H2O(l) HPO4 (aq) + H3O (aq) Ka2 = - [H2PO4 ] HA = 6.3 x 10-8 HA HA HA + 3- + 2- 3- + [H3O ] [PO4 ] H HPO4 (aq) + H2O(l) PO4 (aq) + H3O (aq) Ka3 = [HPO 2-] 4 H+ H+ HA HA = 4.2 x 10-13 H+ Ka1 > Ka2 > Ka3

Kb is the equilbrium constant that quantifies a KEY DOT CONNECTION: The Ka of an acid and the degree OH dissociation, a base’s strength. Kb of the acid’s conjugate base are related through the ion-product constant of water. + - [BH+][OH-] B + H2O(l) BH (aq) + OH (aq) Kb = + [B] [H ][A−] HA (aq) H+ (aq) + A- (aq) Ka = [HA] ---Strong bases dissociate completely [OH ][HA] A- (aq) + H O (l) OH- (aq) + HA (aq) K = − 2 b [A ] 2+ - − Ca(OH)2(s) + H2O(l) ==> Ca (aq) + 2OH (aq) + H2O(l) + - + H2O (l) H (aq) + OH (aq) Ka K = [OH−][H ] H2O b NaOH(s) ====> Na+ (aq) + Cl-(aq)

If you know Ka for an acid you can determine Kb for ---Weak bases dissociate partially its conjugate base! NH (aq) + H O (l) NH + (aq) + OH- (aq) K = 1.76 x 10-5 3 2 4 b -14 Kw = Ka Kb = 10 + - (CH3)2NH2 (aq) + H2O (l) (CH3)2NH (aq) + OH (aq) -4 pKw = pK pK = 14 Kb = 5.4 x 10 a + b A table of bases and their conjugate acids, BH+. Ka and Kb of Some Common Weak Acids and Base 14 Acid Formula Ka Conjugate Kb Notice! Kb x Ka = Kw = 1 x 10 Iodic acid HI 1.61 X 10-1 I- 6.1 X 10-14 -2 - -13 Chlorous Acid HClO2 1.1 X 10 ClO2 9.1 X 10 -3 - -12 Phosphoric Acid H3PO4 7.6 X 10 H2PO4 1.3 X 10 -4 - -11 Nitrous Acid HNO2 7.2 X 10 NO2 1.4 X 10 Hydrofluoric Acid HF 6.6 X 10-4 F- 1.5 X 10-11 -4 - -11 Formic Acid HCOOH 1.8 X 10 HCO2 5.6 X 10

-5 - -10 Acetic Acid CH3CO2H 1.8 X 10 CH3CO2 5.6 X 10 -7 - -8 Hydrosulfuric Acid H2S 1.32 X 10 HS 7.6 X 10 Hypochlorous Acid HClO 3.5 X 10-8 ClO- 2.9 X 10-7 -8 - -7 Hydrazine N2H4 1.1 X 10 OH 8.9 X 10 -7 - -7 Water H2O 1.0 X 10 OH 1.0 X 10 -10 + -5 Ammonia NH3 5.6 X 10 NH4 1.8 X 10 Hydrocyanic Acid HCN 4.0 X 10-10 CN- 2.5 X 10-5 -11 + -4 Dimethyl Amine (CH3)2NH 1.9 X 10 (CH3)2NH2 5.4 X 10

We solve weak base problems just like weak acid Determining pH from Kb and initial [B] + ones, except we solve for [OH-] instead of [H ]. Dimethylamine, (CH3)2NH, a key intermediate in -4 detergent manufacture, has a Kb = 5.9 x 10 . What + - is the pH of a 1.5 M aqueous solution of (CH ) NH? + - [NH4 ][OH ] 3 2 NH3 (aq) + H2O (l) NH4 (aq) + OH (aq) Kb = [NH3]

-5 For ammonia, Kb = 1.76 x 10 what is the conjugate acid and what is it’s pKa? How would you write the reaction of + NH4 with water?

Generalized Base Equilibria [AH+][OH-] A- + H O(l) AH+ (aq) + OH- (aq) K = 2 b [A-] Perhaps the hardest part for students is writing the base reaction in water...just think proton donar and acceptor.

Determining pH from K and initial [B] b Check [3.0 x 10-2 M / 1.5 M] x 100 = 2% (error is < 5%; assumption: Dimethylamine, (CH3)2NH, a key intermediate in thus, assumption is justified) -4 detergent manufacture, has a Kb = 5.9 x 10 . What is + - -14 -2 -13 the pH of a 1.5 M aqueous solution of (CH3)2NH? [H3O ] = Kw/[OH ] = 1.0 x 10 /3.0 x 10 = 3.3 x 10 M + - concentration (CH3)2NH(aq) + H2O(l) (CH3)2NH2 (aq) + OH (aq) -13 initial 1.50 - 0 0 pH = -log (3.3 x 10 ) = 12.5 change - x - + x + x equilibrium 1.50 - x - x x

[(CH ) NH +][OH-] 2 -4 3 2 2 x Kb = 5.9 x 10 = = [(CH3)2NH] 1.5 - x [B ] [1.5] K = initial > 100 Simplify?b Kb = 4 = 2542 > 100 Kb 5.9 10 × − x2 K = 5.9 x 10-4 = x = 3.0 x 10-2 M = [OH-] b 1.5 - Determining the pH of a solution of A Sodium acetate (CH3COONa, abbreviated NaAc) has Sodium acetate (CH COONa, abbreviated NaAc) has applications in photographic development and textile 3 dyeing. What is the pH of a 0.25 M aqueous solution of applications in photographic development and textile NaAc? The K of acetic acid (HAc) (conjugate acid of Ac) dyeing. What is the pH of a 0.25 M aqueous solution of a is 1.8 x 10-5. NaAc? The Ka of acetic acid (HAc) (conjugate acid of Ac) Ac-(aq) + H O(l) HAc(aq) + OH-(aq) is 1.8 x 10-5. concentration 2 initial 0.25 - 0 0 1. Write balanced equation knowing that sodium salts are change -x - +x +x completely soluble in water and acetate will be basic [Ac-] = 0.25 M. equilibrium 0.25 - x - x x

2. Write the equilibrium expression--looking for pH. - [HAc][OH ] Kw Kb = = [Ac-] K 3. Be organized set up ICE table and fill in the unknowns a K -14 w 1.0 x 10 - = Kb = = 5.6 x 10 10 4. Use Ka to find Kb and recall pH + pOH = 14. -5 Ka 1.8 x 10

- [Ac ] = 0.25 M - x ! 0.25 M (since Kb is small) The acid strength of non-metal hydrides increase across a period and down a group. - -10 Check assumption: [Ac ]in / Kb > 100 = .25/ 5.6 x 10 > 100 Electronegativity increases - [HAc][OH ] Acid strength increases Kb = - Use Weak HF [Ac ] to remember 5.6 x 10-10 $ x2/0.25 M

- -5 [OH ] = x $ 1.2 x 10 M Bond H2S strength + - [H3O ] = Kw/[OH ] decreases, H2Se acid + - -14 -5 -10 [H3O ] = Kw/[OH ] = 1.0 x 10 /1.2 x 10 = 8.3 x 10 M strength H2Te increases pH = -log (8.3 x 10-10 M) = 9.1

Oxo-acids are compounds with one or more OXO-ACIDS: The greater electronegativity and number acidic hydrogens bonded to oxygen atom which of internal atoms the more acidic the compound. is further bound to another central atom, Z. electronegative central atom stable anion O - + and acidic " " - + O " " O Z H Z O H Z O- + H+ O electronegative central atom Less More Electro- acidic acidic negativity increases IIA IVA VA VIA VIIA acidity increases Sulfuric Acid H3BO3 H2CO3 HNO3 More H2SiO4 H3PO4 H2SO4 HClO3 acidic .. H3AsO4 H2SeO HBrO3 4 Less Sulfurous acid H6TeO6 HIO3 Phosphorous Acid acidic The acidity of oxo-acids is determined by Indicate whether each is a strong or weak acid, name electronegativity (electron withdrawing effects). each acid and its sodium salt, and predict the relative strengths of the following groups of Acid Strength Increases oxoacids:

a) HClO2, HBrO2, and HIO2. Electronegativity Increases

b) HIO HIO2, HIO3

c) HNO3 and HNO2.

Very Weak Acid Strong Acid d) H3PO3 and H3PO4.

HClO < HClO2 < HClO3 < HClO4 More oxygens, more electron-withdrawing, increase polarity e) H2SO4 and H2SO3 and acidity

Indicate whether each is a strong or weak acid, Transition metal ions (Al3+, Fe3+, Cr3+) react with predict the relative strengths of the following groups water and are acids in aqueous solution. of oxoacids and the strengths of the conjugate base 3+ 2+ + salts. Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H (aq) Acid Hydrolysis of Al3+ a) HClO2, HBrO2, and HIO2. HClO2 > HBrO2, > HIO2. electron density drawn solvent H2O acts toward Al3+ as base b) HIO HIO2, HIO3 HIO < HIO2 < HIO3

c) HNO3 and HNO2. HNO3 > HNO2. H O + 2 H3O d) H PO and H PO . 3 3 3 4 H3PO3 < H3PO4.

e) H2SO4 and H2SO3 H2SO4 > H2SO3 H+ lost

Aqueous solutions of many metal ions are acidic because the hydrated metal ions transfers an H+ ion to water.

n+ - M(NO3)n(s) + xH2O(l) M(H2O)x (aq) + nNO3 (aq) n+ (n-1)+ + M(H2O)x (aq) + H2O(l) M(H2O)x-1OH (aq) + H3O (aq) Hydrolysis of Ionic Salts Ionic Salt + Water ==> ? pH

high charge and small size enhance acidity Acid and bases react with one another in a Salts are the product of an acid-base reaction. neutralization reaction. Salts that have either an anion or cation derived from a weak acid (anion) or weak base (cation) + - NaOH(aq) + HCl(aq) <==> H2O + Na + Cl will react with water (hydrolysis) in an acid base reaction called hydrolysis. base + acid <==> Water + salt 1. look at the cation and anion 2. If its from a Hydrolysis yields Most salts can be viewed and ask is it weak parent...it opposite pH as as coming from a “parent NaCl(s) strong or weak will hydrolyze the parent. acid and parent base”. If parent acid/base? the parents are “strong” the Ca(NO3)(s) cation or anion will not react with water, if the NH4Br(s) NH4Cl parents are “weak” the ion CH3CO2K will react with water in a CH3COOK(s) process called hydrolysis. Ca(NO3)2

There are 4 acid-base neutralization reactions 1. Salts formed from a strong acid and a strong yielding salts that can react with water. base are neutral when dissolved in H2O.

1. Strong acid + Strong base ==> Ionic salt + H2O + - parent is NaOH a strong base HCl (aq) + NaOH (aq) ==> Na + Cl + H2O Neutral salt neither Na+ or Cl- react parent is HCl a strong acid with water. Neutral!

2. Strong acid + Weak base ==> Ionic salt + H2O + - + - NaCl(s) + H2O(l) <=> Na (aq) + Cl (aq) + H2O pH = 7 HBr (aq) + NH3 (aq) ==> NH4 + Br + H2O Acidic salt

parent is KOH a strong base + - 3. Weak acid + Strong base ==> Ionic salt + H2O neither K or SO4 react +2 - parent is H2SO4 a strong acid 2HCOOH (aq) + Ca(OH)2 (aq) ==> Ca + 2HCOO + 2H2O with water. Neutral!

Basic Salt + - K2SO4(s) + H2O(l) <=> K (aq) + SO4 (aq) + H2O pH = 7

4. Weak acid + Weak base ==> Ionic salt + H2O - -, - - Why? The anion of a strong acid (Cl , I Br , NO3 ) is a much + - CH3COOH (aq) + NH3 (aq) ==> NH4 + CH3COO + H2O weaker base than water, and the cation of from the strong acid Depends on pKa vs pkb is a much weaker acid than water!

2. Solutions of salts formed from a strong acid 3. Solutions of salts formed from a weak acid and and a weak base are acidic when dissolved in a strong base are basic when dissolved in H2O. H2O. cation will parent is parent is hydrolyze or anion is from parent acid of the anion is CH3CO2H = weak acid NH4OH = HBr = react with a strong acid anion hydrolyzible weak base strong acid water and does not parent of cation is KOH hydrolyze. = strong base H2O + - CH3COOK (s) K (aq) + CH3COO (aq)

+ - - - NH4Br(s) NH4 (aq) + Br (aq) CH3COO (aq) + H2O (l) CH3COOH (aq) + OH (aq)

hydrolysis of salt of weak acid gives a basic solution NH +(aq) + H O(l) NH (aq) + H O+ (aq) + - 4 2 3 3 NaH2PO4(s) Na (aq) + H2PO4 (aq) H PO -(aq) + H O(l) HPO 2-(aq) + H O+(aq) hydrolysis gives acidic solution 2 4 2 4 3 4. Solutions of salts formed from a weak acid and Predict whether aqueous solutions of the following a weak base depend on the Ka and Kb of the salt. compounds are acidic, basic, or neutral (write an equation for the reaction of the appropriate ion with Both are weak and thus cation and anion will hydrolyze. water to explain pH effect). Which one wins depends on strength (Ka vs Kb) or the ions.

Example: NH4HS (a) KClO4 (b) sodium benzoate, C6H5COONa + + NH (aq) + H O NH (aq) + H O (aq) (c) CrCl3 (d) NaHSO4 4 2 ←−→ 3 3 HS −(aq) + H O H S(aq) + OH −(aq) 2 ←−→ 2 PLAN: Consider the acid-base nature of the anions and + -10 - -7 Ka(NH4 ) = 5.7 x 10 vs Kb(HS ) = 1.0 x 10 cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic. Since Kb > Ka, the solution is basic.

You are given 0.1M solutions of the following. What (a) KClO4 (b) sodium benzoate, C H COONa 6 5 is the nature of the solutions acid/ base/ neutral or (c) CrCl3 (d) NaHSO4 amphoteric?

• KClO2 + - (a) The ions are K and ClO4 , which come from a strong base • CsI (KOH) and a strong acid (HClO4). The solution will be neutral. • NaHSO4 • HCO2K (b) Na+ comes from the strong base NaOH while - • Al(H2O)6 C6H5COO is the anion of a weak organic acid. The salt solution will be basic. • NaNO3 (c) Cr3+ is a small cation with a large + charge, so its • Na3PO4 + - hydrated form will react with water to produce H3O . Cl • Na2HPO4 comes from the strong acid HCl. The salt solution will be acidic. • FeCl3 + - • KClO4 (d) Na comes from a strong base. HSO4 can react with + water to form H3O . The salt solution will be acidic. • CH3COONH4

You are given 0.1M solutions of the following. What Predicting the relative acidity of weak-weak salt solutions is the nature of the solutions acid/ base/ neutral or Determine whether an aqueous solution of zinc formate, amphoteric? Zn(HCOO)2, is acidic, basic, or neutral. The Ka for Zn 2+ -9 -4. (H2O)6 = 10 and Ka for formic acid is 1.8 x 10 Basic • KClO2 Both Zn2+ and HCOO- come from weak conjugates. In order • CsI Neutral to find the relatively acidity, write the dissociation reactions and use the information in Tables 18.2 and 18.7. Acidic • NaHSO4 Basic • HCO2K Acidic • Al(H2O)6 Neutral • NaNO3 Basic • Na3PO4 Depends on Ka vs Kb • CH3COONH4 Acidic • FeCl3 Predicting the relative acidity of weak-weak salt solutions A Lewis Acid is any substance that is an electron- Determine whether an aqueous solution of zinc formate, pair acceptor. Zn(HCOO)2, is acidic, basic, or neutral. The Ka for Zn 2+ -9 -4. (H2O)6 = 10 and Ka for formic acid is 1.8 x 10 + BF Neutral Proton Donors H 3 Lewis acids Both Zn2+ and HCOO- come from weak conjugates. In order lack to find the relatively acidity, write the dissociation reactions and electrons use the information in Tables 18.2 and 18.7. and have an SOLUTION: orbital.

2+ + + Some Zn(H2O)6 (aq) + H2O(l) Zn(H2O)5OH (aq) + H3O (aq) Lewis - - HCOO (aq) + H2O(l) HCOOH(aq) + OH (aq) Acids Cations 2+ -9 Ka Zn(H2O)6 = 1 x 10

K HCOOH = 1.8 x 10-4 ; K = K /K = 1.0 x 10-14/1.8 x 10-4 = 5.6 x 10-11 a b w a Transition Metal Cations Al3+, Ti4+, 3+ 2+ 2+ - Fe , Zn , Ka for Zn(H2O)6 > Kb HCOO ; the solution is acidic.

A Lewis Base is any substance that is an electron- Neutralization (in the Lewis acid context) occurs when pair donor (nucleophile). a Lewis acid reacts and forms a covalent bond with a Lewis base. The bond is called a coordinate bond. 2–2- H2O NH3 O The adduct Atoms contains a having new covalent lone pairs of bond. electrons acid base adduct is the distinguish ing feature adduct

M2+ 2+ H2O(l) M(H2O)4 (aq)

Identify the Lewis acids and Lewis bases in Identify the Lewis acids and Lewis bases in the following reactions: the following reactions:

+ - + - (a) H + OH H2O (a) H + OH H2O - - - - (b) Cl + BCl3 BCl4 (b) Cl + BCl3 BCl4 + + + + (c) K + 6H2O K(H2O)6 (c) K + 6H2O K(H2O)6

Look for electron pair acceptors (acids) and donors (bases).

acceptor donor + - (b) Cl- + BCl BCl - (a) H + OH H2O 3 4 donor acceptor acceptor + + (c) K + 6H2O K(H2O)6 donor