Why study chemistry -...

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253 Acids/Bases

Transcript of Why study chemistry -...

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Acids/Bases

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Test Content from AP Chemistry Course Description

I. Structure of Matter (20%)B. Chemical bonding

3. Geometry of molecules and ions, structural isomerism of simple organic molecules and coordination complexes; dipole moments of molecules; relation of properties to structure

III. Reactions (35-40%)A. Reaction types

1. Acid-base reactions; concepts of Arrhenius, Brønsted-Lowry, and Lewis; coordination complexes; amphoterism

C. Equilibrium1. Concept of dynamic equilibrium, physical and chemical; Le Châtelier's

principle; equilibrium constants2. Quantitative treatment

a. Equilibrium constants for gaseous reactions: Kp, Kc

b. Equilibrium constants for reactions in solution(1) Constants for acids and bases; pK; pH

IV. Descriptive Chemistry (10-15%)Knowledge of specific facts of chemistry is essential for an understanding of principles and concepts. These descriptive facts, including the chemistry involved in environmental and societal issues, should not be isolated from the principles being studied but should be taught throughout the course to illustrate and illuminate the principles. The following areas should be covered:

1. Chemical reactivity and products of chemical reactions

Types of Calculations found on Exam adapted from AP Chemistry Course Description

Summary of types of problems either explicitly or implicitly included in the preceding material.

8. Equilibrium constants and their applications, including their use for simultaneous equilibria

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ACIDS, BASES AND SALTSWater (H2O) is made of two ions: H+ (aq) hydrogen ion, OH- (aq) hydroxide ion

Acids – substances that increase H+ (aq) conc.- definition of Arrhenius acid

Strong acids- strong electrolytes, i. e., completely dissociate- memorize list

HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4

Weak acids- weak electrolytes

HF (aq) H+ (aq) + F- (aq)H3PO4 (aq) H+ (aq) + H2PO4

- (aq)

Bases – substances that increase OH- concentration- definition of Arrhenius base

Strong bases- strong electrolytes- all soluble hydroxides LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2,

Sr(OH)2, Ba(OH)2

Weak bases- weak electrolytes- usually increase OH- (aq) conc. “indirectly” by decreasing H+ (aq) conc.

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

Neutralization reactionAcid + Base Water + SaltSalt

- ionic compound- cations and anions remaining after water is made

HNO3 (aq) + KOH (aq) H2O (l) + KNO3 (aq)

Learning objective 3.7 The student is able to identify compounds as Bronsted-Lowry acids, bases, and/or conjugate acid-base pairs, using proton-transfer reactions to justify the identification. (See SP 6.1; Essential Knowledge 3.B.2

BRØNSTED – LOWRY ACID/BASE THEORYAcid/base reactions are simply transfer of proton (hydronium ion) from acid to base.

acid base water salt

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Brønsted – Lowry Acid- proton donor- note HCl, H2SO4, HC2H3O2 all donate protons

Brønsted – Lowry Base- proton acceptor- note OH- and NH3 accept protons

In Brønsted – Lowry theory, water is an acid or base depending on the circumstances.

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)- water accepts proton; therefore, it is a base.

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

- water donates proton; therefore, it is an acid.

Conjugate Acids and BasesConsider a typical acid/base reaction.

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

- H2O has donated proton, it is an acid.- NH3 accepted proton, it is a base.

Consider the reverse reaction.NH4

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

+ has donated proton, it is an acid.- OH- accepted proton, it is a base.

- An acid is changed into a base and a base is changed into an acid.- These pairs of acids and bases are called conjugate acid-base pairs.- A conjugate acid-base pair differ from each other only by a proton.

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

Another Example:

H2SO4 (aq) + H2O (l) HSO4- (aq) + H3O+ (aq)

Another Example:

HC2H3O2 + NaOH H2O + NaC2H3O2

base acid conjugate base

conjugate acid

acid base conjugate acid

conjugate base

acid base conjugate base

conjugate acid

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LEWIS ACIDS AND BASESDefinitions

Lewis acid – electron pair acceptorLewis base – electron pair donor

Lewis acid/base theory is more general than Brønsted – Lowry acid/base theory.

- Brønsted – Lowry acids are Lewis acids, though not vice versa.- Brønsted – Lowry bases are Lewis bases, though not vice versa.

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

- NH3 is electron pair donor- H2O is electron pair acceptor

AlCl3 + CH3Cl AlCl4- + CH3

+ other products

This is the first step in the “Friedel-Crafts alkylation” that adds a methyl group to an organic molecule. More next year in organic chemistry!

EXCLUSION STATEMENT Essential Knowledge 3.B.2

Lewis acid-base concepts are beyond the scope of this course and the AP Exam.

Learning objective 6.17 The students can, given an arbitrary mixture of weak and strong acids and bases (including polyprotic systems), determine which species will react strongly with one another (i.e., with K > 1) and what species will be present in large concentrations at equilibrium (See SP 6.4; Essential Knowledge 6.C.1)

Lewis acid – accepts e- pair from Cl-

Lewis base – donates e- pair via Cl- ion

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Relative strengths of acids and basesThe strength of a conjugate base is related to the strength of its acid.The strength of a conjugate acid is related to the strength of its base.

Consider nitric acid:HNO3 (aq) – acidNO3

- (aq) – conjugate base

HNO3 is a strong acid, meaning that it fully dissociates.HNO3 (aq) H+ (aq) + NO3

- (aq)- This means that the NO3

- ion has no desire to accept a proton.- Therefore the nitrate ion is a very weak base.

Consider aqueous sodium hydroxide:OH- – base H2O – acid

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

- OH- ion reacts fully (almost) with H+ ions.- If any additional H+ is added, OH- will grab it to form water.- OH- is a very strong base.- The conjugate acid, H2O is a very weak acid.

As the strength of acid increases, the strength of its conjugate base decreases.As the strength of base increases, the strength of its conjugate acid decreases.

Example: HC2H3O2 is a weaker acid than HF. Which ion will more readily accept protons in aqueous solution, C2H3O2

- or F-?

HC2H3O2 is weaker acid; therefore, C2H3O2- is stronger base.

Thus C2H3O2- will accept a proton more readily.

Example: Pyridine, C5H5N, is a weaker base than ammonia, NH3. Which ion is the stronger acid, NH4

+, C5H5NH+?

C5H5N is weaker base; therefore, C5H5NH+ is stronger acid.Therefore, C5H5NH+ will donate its proton easier that NH4

+.

Pyridine is a base used in methods that determine the moisture content of foods.

Learning objective 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. (See SP 7.2, connects to Big Idea 5, Big Idea 6; Essential Knowledge 2.A – 2.D)

CHEMICAL STRUCTURE OF ACIDS AND BASES

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Basic Principles of Acidity

1. Acidity is related to bond strength.- H2Se is more acidic than H2S because the H – Se bond is weaker than the

H – S bond (Se is larger than S; therefore the bond overlap for the H – Se bond is weaker than for the H – S bond).

2. Acidic hydrogens ionize from a polar bond.

3. Polarity of the acidic hydrogen bond can be affected to a lesser extent by other atoms.

4. Strong attraction of the water (solvent) molecules is the driving force.

Binary Acids- Bonds between water molecules and ions are stronger than bonds between

acid molecules.

- Bond strength is very important.- HI is more acidic than HBr which is more acidic HCl.

Hydrofluoric acid’s acidity is peculiar.- Bonds between HF molecules are very strong.

HF (aq) + HF (aq) [F – H – F]- (aq) + H+ (aq)

- Even though HF is very polar, it is a weak acid since it forms strong intermolecular bonds with itself.

Oxyacids- Oxyacids are essentially molecular compounds.- The H – O bond is strongly polar and allows H+ to disassociate.

H O

H

H O

H

H Cl H Cl

H O

H

H+ H O

H

Cl-

These bonds are stronger than these bonds.

polar bond

Cl

O

O O

O

H

H O

H

H+

H O

H

F-

These bonds are stronger than these bonds.

HF etches glass (and bone!). It cannot be stored in a glass bottle.

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- Electronegativities of secondary and tertiary atoms are very important.- Electron withdrawing of other oxygen atoms is the difference between

HClO4 being a strong acid rather than a weak acid.

Compare to HClO2.

Example: Which is the stronger acid, acetic acid or trichloroacetic acid?

Chlorine atoms are much more electronegative that hydrogen atoms; therefore, the chlorine atoms are making the O – H bond more polar thus increasing the acidity.

Chlorine atoms are much more electronegative that hydrogen atoms; therefore, the chlorine atoms are making the O – H bond more polar thus increasing the acidity.

Trichloroacetic acid is the stronger acid.

Trichloroacetic acid is used in some tattoo removal processes.

HC2H3O2(aq) + CN-(aq) HCN(aq) + C2H3O2-(aq)

MC Question: The reaction represented above has an equilibrium constant equal to 3.7 104. Which is the following can be concluded from this information?

(A) CN-(aq) is a stronger base than C2H3O2-(aq)

(B) HCN(aq) is a stronger acid than HC2H3O2(aq) (C) The conjugate base of CN-(aq) is C2H3O2

-(aq) (D) The equilibrium constant will increase with an increase in temperature. (E) The pH of a solution containing equimolar amounts of CN-(aq) and

HC2H3O2 is 7.0.

THE AUTOIONIZATION OF WATER- Water is not a nonelectrolyte, actually it is a very weak electrolyte.- One out of every 10 million water molecules dissociates into an H+ ion and an OH- ion.

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

bond is less polar since fewer oxygen atoms are drawing away electron density

ClO O H

acetic acid trichloroacetic acid

C

H

H C

H

O

O H C

Cl

Cl C

Cl

O

O H

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Learning objective 6.14 The student can, based on the dependence of Kw on temperature, reason that neutrality requires [H+] = [OH-] as opposed to requiring pH = 7, including especially the applications to biological systems. (See SP 2.2, 6.2; Essential Knowledge 6.C.1)

Ion product of water

Write equilibrium expression

- Since H2O is solvent, we will neglect it in equilibrium expression.

No matter what conditions exist in an aqueous solution, the following expression is true. (at 25 C)

Ion product is also known as Kw.

Aside: Other solvents autoionize as well.

NH3 + NH3 NH4+ + NH2

- (Ks 10-33)

CH3OH + CH3OH CH3OH2+ + CH3O- (Ks 2 10-17)

Since is always true- if we know [H+], we can calculate [OH-]- if we know [OH-], we can calculate [H+]

Note: When [H+] = [OH-], the amount of acid and base are in equal amounts. The solution is called neutral.

THE HYDRONIUM IONH+ is simply a proton.

- A proton will hydrogen bond with water

H+

H H

OH H

O

H

+

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H3O+ is a hydronium ion.

Current theories of water treat proton as surrounded by 4 or 6 water molecules. Therefore, H+ may be truly H9O4

+ or H13O6+.

H3O+ is a concept, not an actual structure.

Reconsider autoionization of water asH2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

pH SCALEDefinition: pH = - log [H + ] = -log[H 3O + ]

pH scale is used strictly for convenience- convenient to avoid scientific notation.

[H+] for pure water equals 10-7; therefore, pH = - log(10-7) = - (-7) = 7

O

H H

O

H H

O

H

H

O

H

H

H+

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All neutral solutions have pH = 7.If solution is acidic, pH < 7.If solution is basic, pH > 7.pH scale of common substances

gastric acid

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1234567

8910111213

very acidic

slightly acidic

neutral

coke

coffee

lemon juice

blood

milk of magnesia

household ammoniahousehold bleach

very basic

slightly basic

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**In a number that is a logarithm such pH, only the digits to the right of decimal point are significant.**

[H+] = 5.38 10-2 pH = 1.269[H+] = 5.38 10-5 pH = 4.269[H+] = 5.38 10-8 pH = 7.269

[H+] = 5.38 10-12 pH = 11.269

Other definitions: pOH = - log [OH-] pKw = - log Kw = 14.00

Autoionization of water in terms of pH- log([H+][OH-]) = -log Kw pH + pOH = pKw = 14.00

Example: The pH of milk of magnesia is 10.0, what is the pOH?

pOH = pKw – pH = 14.00 – 10.0 = 4.0

Example: A sodium cyanide solution used to extract gold from low-grade ore has a pH of 10.67, what is the hydronium ion concentration of the solution?

4 Au (s) + 4 CN- (aq) + O2 (aq) + H2O (l) 4 Au(CN)2- + 4 OH- (aq)

Example: A basic solution of trisodium phosphate (TSP) has a pOH of 2.14, what is its hydroxide ion concentration?

You know that the proper name for TSP is sodium phosphate!

Learning objective 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. (See SP 1.1, 1.4, 2.3; Essential Knowledge 6.C.1) (Missing from notes)

Learning objective 6.12 The student can reason about the distinction between strong and weak acid solutions with similar values of pH, including the percent ionization of the acids, the concentrations needed to achieve the same pH, and the amount of base needed to reach the equivalence point in a titration. (See SP 1.4; Essential Knowledge 6.C.1, connects to 1.E.2)

STRONG ACIDS AND BASESRemember lists

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Strong Acids: HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4

Strong Bases: Alkali metal hydroxides, Ba(OH)2, Sr(OH)2, Ca(OH)2, Mg(OH)2

Calculating pH of strong acid and strong base solutionsStrong acids and bases completely dissociate; therefore, [H+] (or [OH-]) equals the concentration of the solute.

Example: What is the pH of a 0.0142 M solution of HBr?

Since HBr is a strong acid, [H+] = 0.0142 MpH = - log (0.0142) = - (-1.848) = 1.848

Example: What the pH of 1.5 M Sr(OH)2 solution?

Since Sr(OH)2 is a strong base,

pH = - log (3.3 10-15) = 14.48

In reverse,pOH = 14.00 – pH = 14.00 – 14.48 = – 0.48

Learning objective 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations. (See SP 2.2, 6.4; Essential Knowledge 6.C.1)

WEAK ACIDS AND BASESFor weak acids and bases, the concentration of acid (base) is not the concentration of [H+] ([OH-]).

Weak AcidsDissociated ions are in equilibrium with undissociated molecule.

HC2H3O2 (aq) H+ (aq) + C2H3O2- (aq)

Equilibrium expression is

Ka – acid-dissociation constant- only difference from Kc is the label.

Each weak acid has its own Ka which does not change (at 25 C).

limited solubility

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Example: What is the pH of 2.0 M ascorbic acid (HC6H7O6)?Ka (HC6H7O6) = 8.0 10-5.

[H+] [C6H7O6-] [HC6H7O6]

Initial 0 0 2.0 MChange x x – xEquil. x x 2.0 – x

Assume that x is small.

x = 0.013 pH = 1.90

Example: What is the pH of 0.15 M iodic acid (HIO3)? Ka (HIO3) = 0.17.

[H+] [IO3-] [HIO3]

Initial 0 0 0.15 MChange x x – xEquil. x x 0.15 – x

Assume that x is small.

x = 0.16

Hmmm! Looks like x is not small. We will need the quadratic equation.

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x = [H+] = 0.097 pH = 1.01Example: Calculate the pKa of an unknown acid solution with a concentration of

0.0988 M and a pH of 4.43.

[H+] [A-] [HA]Initial 0 0 0.0988 M

Change x x – xEquil. x x 0.0988 – xEquil. 3.7×10-5 3.7×10-5 0.0988

Polyprotic Acids- Many acids have more than one proton to donate.Consider H2SO4

H2SO4 is a strong acid.H2SO4 (aq) H+ (aq) + HSO4

- (aq)

Bisulfate ion also has a proton to donate, though it is less willing to do so.

HSO4- (aq) H+ (aq) + SO4

2- (aq)

Consider H3PO4

Phosphoric acid has 3 protons to donate- before second proton leaves, almost all of the first protons have left.

2nd proton disassociated

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When we add H3PO4 to water, the disassociation isH3PO4 (aq) H+ (aq) + H2PO4

- (aq)How can we get the second proton to leave?

- Remove H+; i.e., add base.H2PO4

- (aq) H+ (aq) + HPO4

2- (aq)Weak Bases

Conc. of OH- is not concentration of weak base.

Consider ammoniaNH3 (aq) + H2O (l) NH4

+ (aq) + OH- (aq)

Equilibrium expression is

Kb – base-dissociation constant- only difference from Kc is the label.

Consider general baseB (aq) + H2O (l) BH+ (aq) + OH- (aq)

Equilibrium expression is

Example: Calculate the OH- concentration of 0.012 M solution of pyridine, C5H5N. Kb(C5H5N) is 1.7 10-9.

[C5H5N] [C5H5NH+] [OH-]Initial 0.012 0 0

Change –x x xEquil. 0.012 – x x x

Assume that x is small.

+ H2O + OH-

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x = [OH-] = 4.5 10-6 M pOH = 5.35 pH = 8.65

Example: Calculate the pH of 3.4 M solution of Na2CO3. Kb(CO32-) = 1.8 10-4

When Na2CO3 is dissolved in water, it is fully disassociated. (All sodium salts are soluble.)

Na2CO3 (aq) = 2 Na+ (aq) + CO32- (aq)

Consider that CO32- is the conjugate base of HCO3

-.CO3

2- (aq) + H2O (aq) HCO3- (aq) + OH- (aq)

[CO32-] [HCO3

-] [OH-]Initial 3.4 0 0

Change –x x xEquil. 3.4 – x x x

Assume that x is small.

x = [OH-] = 2.510-2 M pOH = 1.60 pH = 12.40

Check assumption: 3.4 – 0.025 = 3.375 3.4 Assumption is good.

Percent Ionization

While the dissociation constant gives a measure of the strength of a weak acid or base, it doesn’t give any indication of how electrolyte actually dissociates.

By definition, the percent ionization is the percentage of weak electrolyte molecules that have dissociated in solution

For a weak acid, the percent ionization is

Note the ambiguity in the definition.- For pure weak acid, either definition is valid.

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- If extra A- is added, use [H+] definition.- If extra H+ is added, use [A-] definition.- If in doubt, smaller number is correct.

For a weak base, the percent ionization is

.

- For pure weak base, either definition is valid.- If extra HB+ is added, use [OH-] definition.- If extra OH- is added, use [BH+] definition.- If in doubt, smaller number is correct.

Example: Calculate the percent ionization of hydrofluoric acid in a 1.0 M, 0.10 M and 0.010 M solutions. Ka(HF) = 6.8 10-4

To calculate percent ionization, we need [H+], therefore, we need an ICE table.

For a variable initial concentration c

[H+] [F-] [HF]Initial 0 0 c M

Change x x – xEquil. x x c – x

Assume that x is small.

For c = 1.0 M,

c = [HF]0 [H+] % ionized pH1.0 0.026 2.6 1.580.10 0.0079 7.9 2.100.010 0.0023 23 2.64

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Note that as the concentration of weak acid decreases, more of it dissociates!Relationship between Ka and Kb

Consider Kb of ammonia. Kb(NH3) = 1.8 10-5

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

Now consider Ka of conjugate acid, NH4+

NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)

Now multiply Ka and Kb.

Thus Ka Kb = Kw

Example: If Kb(NH3) = 1.8 10-5, what is Ka for NH4+?

Example: If Ka(HF) = 6.8 10-4, what is Kb for F-?

Ka Kb = Kw also implies pKa + pKb = 14.00

FR Question: Answer the following questions that relate to the chemistry of halogen oxoacids.

a) Use the information in the table below to answer part (a)(i).

Acid Ka at 298 KHOCl 2.9 10-8

HOBr 2.4 10-9

(i) Which of the two acids is stronger, HOCl or HOBr? Justify your answer in terms of Ka.

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HOCl is the stronger acid because its Ka value is greater than the Ka value of HOBr.

(ii) Draw a complete Lewis electron-dot diagram for the acid that you identified in part (a)(i).

(iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger acid or a weaker acid than the acid that you identified in part (a)(i). Justify your prediction in terms of chemical bonding.

HOI is a weaker acid than HOCl because the O–H bond in HOI is stronger than the O–H bond in HOCl. The lower electronegativity (electron-drawing ability) of I compared with that of Cl results in an electron density that is higher (hence a bond that is stronger) between the H and O atoms in HOI compared with the electron density between the H and O atoms in HOCl.

b) Write the equation for the reaction that occurs between hypochlorous acid and water.HOCl + H2O OCl− + H3O+

c) A 1.2 M NaOCl solution is prepared by dissolving solid NaOCl in distilled water at 298 K. The hydrolysis reaction OCl−(aq) + H2O(l) HOCl(aq) + OH−(aq) occurs.(i) Write the equilibrium-constant expression for the hydrolysis reaction that occurs

between OCl−(aq) and H2O(l).

(ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis reaction.

(iii) Calculate the value of [OH−] in the 1.2 M NaOCl solution at 298 K.

[OCl-] [HOCl] [OH-]Initial 1.2 0 0

Change -x x xEquil. 1.2 - x x x

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d) A buffer solution is prepared by dissolving some solid NaOCl in a solution of HOCl at 298 K. The pH of the buffer solution is determined to be 6.48.

(i) Calculate the value of [H3O+] in the buffer solution.

[H+] = 10−6.48 = 3.3 × 10−7 M

(ii) Indicate which of HOCl(aq) or OCl−(aq) is present at the higher concentration in the buffer solution. Support your answer with a calculation.

[H+] = 3.3 × 10−7 M and Ka for HOCl = 2.9 × 10−8

HYDROLYSISMany salts when dissolved in water will change its pH.

- Cations may attract the hydroxide ion.- Anions may attract the hydrogen ion.

The process where ions cause water to split apart is called hydrolysis.

The acidic or basic properties of a salt solution depend on the relative abilities of the cation and the anion to cause hydrolysis.

Cations are acidic (usually)1. Cations from strong bases are very weakly acidic.

- Li+, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, Ba2+

2. Higher cationic charge causes greater hydrolysis.- Higher charges attract OH- more effectively.- pH of TlCl soln. higher than PbCl2 soln.

Pb(H2O)62+ Pb(H2O)5(OH)+ + H+

3. Smaller cations are more acidic than larger cations.- Higher charge density attracts OH- more effectively.- pH of Ga(NO3)3 sol. higher than Al(NO3)3 sol.

4. Transition metal ions are more acidic than alkali or alkaline earth metal ions.- Incomplete d subshell attracts OH- and H2O.

- Co(ClO3)2 more acidic than Mg(ClO3)2.Co(H2O)6

2+ Co(H2O)5(OH-)+ + H+

5. Conjugate acids of nitrogen bases are acidic.

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- NH4+, C5H5NH+, (CH3)3NH+, etc…

NH4I (aq) + H2O (l) NH3 (aq) + H3O+ (aq) + I- (aq)

Anions are basic1. Anions from strong acids are not basic.

- Cl-, Br-, I-, NO3- HSO4

-, ClO3-, ClO4

-

2. Other soluble monatomic anions are basic.- K2S, Cs2O, BeH2 solutions are basic.

S2- + H2O HS- + OH-

O2- + H2O OH- + OH-

H- + H2O H2 + OH-

3. Conjugate bases of weak acids are basic.- Al(C2H3O2)3, SrF2, Li2C2O4 are basic salts.

F- + H2O HF + OH-

- Note: This is the dissociation of a weak base.Salts derived from weak acid and weak base may be acidic or basic

- Both cation and anion hydrolyze water.- pH depends on which ion has stronger acid/base properties.

- if base is stronger; i.e., Kb > Ka then pH > 7.- if acid is stronger; i.e., Ka > Kb then pH < 7.

Example: NH4NO2

Ka > Kb, thus pH < 7. NH4NO2 solution is acidic.

MC Question: A 1-molar solution of which of the following salts has the highest pH ?

(A) NaNO3

(B) Na2CO3

(C) NH4Cl(D) NaHSO4

(E) Na2SO4

PERIODIC ACID/BASE TRENDSBinary hydrides

Nonmetal hydrides tend to be acidic- Acidity as a periodic trend decreases from right to left.

Ka(HCl) > Ka(H2S) > Ka(PH3)Note NH3 and H2O are exceptions.

Metal hydrides are basic

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- Basicity as a periodic trend decreases from left to right.Kb(NaH) > Kb(MgH2)

NaH (s) + H2O (aq) H2 (g) + NaOH (aq)H- (aq) + H2O (l) H2 (g) + OH- (aq)

Binary oxidesMetal oxides are basic when dissolved in water.

- Basicity decreases from left to right- Na2O > CaO > Ga2O3

Nonmetal oxides are acidic when dissolved in water.- Acidity increases from left to right- Cl2O7 > SO3 > P2O5

Amphoteric MetalsMetals generally react with acids to form metal ion solutions.

Ni (s) + 2H+ (aq) Ni2+ (aq) + H2 (g)

However, some metals also react with bases to form metal ion solutions.Al (s) + 4 OH- (aq) Al(OH)4

- (aq)

Amphoteric metals react in both acid and bases to form ions.Amphoteric metals tend to be close to the metal/nonmetal boundary.

The oxides and hydroxides of amphoteric metals will dissolve in acid and base.

A partial list of amphoteric metals includes: Si, Al, Zn, Ga, Ge, As, Sn, Sb, Pb, etc…

MC Question: When an aqueous solution of NaOH is added to an aqueous solution of potassium dichromate, K2Cr2O7, the dichromate ion is converted to

(A) CrO42-

(B) CrO2

(C) Cr3+

(D) Cr2O3(s)(E) Cr(OH)3(s)

COORDINATION COMPLEXES

Consider Cu2+ ion.CuSO4 (s) + H2O (l) CuSO4 (aq)very pale blue bright blue

What has changed?- In the solid, Cu2+ is surrounded by SO4

2-

- In the solution, Cu2+ is surrounded by H2OO

HH

Cu2

+

O

H

H

O

HH

O

H

H

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- Cu2+ acts as Lewis acid.- H2O acts as Lewis base.

MC Question: Appropriate uses of a visible-light spectrophotometer include which of the following?

I. Determining the concentration of a solution of Cu(NO3)2

II. Measuring the conductivity of a solution of KMnO4

III. Determining which ions are present in a solution that may contain Na+, Mg2+, Al3+

(A) I only (B) II only (C) III only (D) I and II only (E) I and III onlyComplexation of Metal Ions- The addition of substances with strong Lewis base character (ligands) can react with

certain metal ions to form soluble complexes.

Consider: Silver diammine complex

Ag+ + 2NH3 [Ag(NH3)2]+

KAg NH

Ag NHf

( ).3 2

3

2717 10

- Note that Kf (formation constant) tells us that formation of the complex is overwhelmingly favored.

- The ammonia will stay bound to the silver ion until a stronger Lewis base is added to the solution.

Other ligands that form complexes with metal ionsI-, Br-, SCN-, Cl-, F-, OH-, H2O, CN-, CO

Complexes and Lewis Acid/Base Theory

Consider the following equilibrium

CoCl42- + 6 H2O Co(H2O)6

2+ + 4 Cl-

Lewis acid Lewis base

O

H

H

O

HH

Co2+ O

H

H

O

HH

Co2+ Cl-

Cl-

Cl-

Cl-

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Ag+ + 2NH3 [Ag(NH3)2]+

AgCl (s) + 2NH3 [Ag(NH3)2]+ + Cl- (aq)Ammonia is stronger Lewis base than chloride when silver ion is the Lewis acid.MC Question: Which of the following occurs when excess concentrated NH3(aq) is

mixed thoroughly 0.1 M Cu(NO3)2(aq)?

(A) A dark red precipitate forms and settles out. (B) Separate layers of immiscible liquids form with a blue layer on top. (C) The color of the solution turns from light blue to dark blue. (D) Bubbles of ammonia gas form. (E) The pH of the solution decreases.

Nomenclature of Metal Complex Ions

1. Name the ligands first, in alphabetical order, then the metal atom or ion. Note: The metal atom or ion is written before the ligands in the chemical formula.

Anionic Ligands Names Neutral Ligands NamesBr- bromo NH3 ammineCl- chloro H2O aquaF- fluoro NO nitrosylI- iodo CO carbonylO2- oxo O2 dioxygenOH- hydroxo C5H5N pyridineCN- cyano H2NCH2CH2NH2 ethylenediamine (en)CH3COO- acetato CH3COCH2COCH3 acetylacetone (acac)

2. Greek prefixes are used to designate the number of each type of ligand in the complex ion, e.g. di-, tri- and tetra-.

3. After naming the ligands, name the central metal. If the complex ion is a cation, the metal is named same as the element such as cobalt(II) or cobalt(III). If the complex

N H

H

H

Ag+NH

H

H

Lewis acid

bluepink

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ion is an anion, the name of the metal ends with the suffix –ate. For example, Co in a complex anion is called cobaltate and Pt is called platinate. For some metals, the Latin names are used in the complex anions.

Name of Metal Name in an Anionic Complex

Iron Ferrate(II) or Ferrate(III)

Copper Cuprate(I) or Cuprate(II)

Lead Plumbate(II) or Plumbate(IV)

Silver Argenate(I)

Gold Aurate(I) or Aurate(III)

Tin Stannate(II) or Stannate(IV)

  4. Following the name of the metal, the oxidation state of the metal in the complex is

given as a Roman numeral in parentheses.

Example: What is the name of the following compound: [Cr(NH3)3(H2O)3]Cl3

First name the cation, then the anion.

Cation: [Cr(NH3)3(H2O)3]3+ triamminetriaquachromium(III) ion Note: Ammine is before aqua alphabetically. (Ignore the numerical

prefixes.) Note: We know the charge of the chromium atom from the number of

chloride ions and the fact that the ligands are neutral.Anion: Cl- chloride ionCompound: triamminetriaquachromium(III) chloride

Example: What is the name of the following compound: K4[Fe(CN)6]

Cation: K+ potassium ionAnion: [Fe(CN)6]4+ hexacyanoferrate(II) ion

Note: We know the charge of the iron atom is 2+ since the overall the complex is 4- and each cyanide is 1-.

Example: Give the formula for the compound, tetraamminedichloroplatinum(IV) bromide.

The cation has four NH3 and two Cl-; therefore, its formula is [Pt(NH3)4Cl2]2+

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Note: We know the charge of the complex is 2+ since the platinum is 4+, each NH3 is neutral and each Cl- is 1-.

Therefore the compound formula is [Pt(NH3)4Cl2]Br2

Example: Give the formula for the compound, sodium tetrabromonickelate(II).

The anion has four Br-; therefore, its formula is [NiBr4]2-

Note: We know the charge of the complex is 2- since the platinum is 2+, each and each Br- is 1-.

Therefore the compound formula is Na2[NiBr4].

Example: [Cu(CO)4][CuBr4] tetracarbonylcopper(II) tetrabromocuprate(II)

Geometry of Coordination Complexes.

Most coordination complexes have a coordination number of 4 (tetrahedral or square planar) or 6 (octahedral).

Isomers of square planar coordination complexes with the general formula XA2B2 have cis and trans isomers.

Cis-platin Trans-platin

Isomers of octahedral coordination complexes have two general formulas, XA3B3 and XA4B2, where X is the central metal ion and A and B are the ligands.

Isomers of XA3B3

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Facial-triamminetrichlorocobalt(III) Meridional-triamminetrichlorocobalt(III)

Isomers of XA4B2

Cis-tetrabromodicarbonylferrate(III) Trans-tetrabromodicarbonylferrate(III)

HARD AND SOFT BASES (Supplemental Material)

The strength of a Lewis acid or a Lewis base can depend on several factors such as the electronegativities of the atoms, the charge density of the atoms, and the enthalpies of hydration (for aqueous solutions).

However, these factors alone fail to explain many patterns relating to the stability of coordination complexes (and solubility products).

Many patterns can be explained by separating both Lewis acids and Lewis bases into two categories, hard and soft.

Hard acidsHard acids have low polarizabilities.

- High charge- Small size

Examples of hard acids includeH+, Li+, Na+, K+

Be2+, Mg2+, Ca2+, Sr2+, Sn2+

Al3+, Ga3+, In3+, La3+

Cr3+, Co3+, Fe3+, Ir3+

BF3, Al(CH3)3, CO2

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Soft acidsSoft acids have high polarizabilities.

- Low charge- Large size- Often have unpaired electrons

Examples of soft acids includeCu+, Ag+, Au+, Tl+, Hg+, Cs+

Pd2+, Cd2+, Pt2+, Hg2+

I2, Br2, M0

Lewis acids on the border between hard and soft includeFe2+, Co2+, Ni2+, Cu2+, Zn2+, Pb2+

Hard basesHard bases have low polarizabilities.

Examples of hard bases includeF-, Cl-, OH-, RO-

PO43-, SO4

2-, CO32-, ClO4

-, NO3-, CH3COO-

H2O, R2O, ROH, NH3, RNH2

Soft basesSoft bases have high polarizabilities.

Examples of soft bases includeI-, SCN-, CN-, H-

S2-, RS-, S2O32-

R2S, PR3, RSH, CO, C6H6

Lewis bases on the border between hard and soft includeBr-, NO2

-, SO32-, N2

The Principle of Hard and Soft Acids and Bases (HSAB)

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Hard acids have the tendency to react faster and create stronger bonds with hard bases, likewise with soft acids and soft bases.

The role of the dispersion interaction can be important in explaining chemical reactivity.

Applications

Inorganic complexes with soft acid and soft base are generally very stable.Inorganic complexes with hard acid and hard base have moderate stability.

- Aids in the understanding of the spectrochemical series which is an approximate ordering of the binding strength of the central metal ion to the ligands in the formation of coordination complexes.

I− < Br− < S2− < SCN− < Cl− < NO3− < F− < OH− < H2O < CH3CN < NH3 < CN− < CO

Hard solvents solvate hard acids and hard bases.- nitrates, chlorates and acetates are soluble

- hard bases with higher charges have more lattice energy to overcome