newburyparkhighschool.netnewburyparkhighschool.net/dogancay/ibchem/08_acids_and... · Web viewsubstances which can act as Bronsted-Lowry acids and bases, meaning they can accept or

Unit 8: Acids & Bases IB Topics 8 & 18

NOTES - Unit 8: Acids & Bases

PART 1: Acid/Base Theory & PropertiesBr Ø nsted-Lowry: a theory of proton transfer

o A Bronsted-Lowry ACID is a _______________________________________________.o A Bronsted-Lowry BASE is a _______________________________________________.

Conjugate pairs: Acids react to form bases and vice versa. The acid-base pairs related to each other in this way are called conjugate acid-base pairs. They differ by just one proton.

HA + B A- + BH+

Ex) List the conjugate acid-base pairs in the following reaction: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Ex) Write the conjugate base for each of the following.a) H3O+ b) NH3 c) H2CO3

Ex) Write the conjugate acid for each of the following.a) NO2

- b) OH- c) CO32-

Amphoteric / __________________________ substances: substances which can act as Bronsted-Lowry acids and bases, meaning they can accept or donate a proton (capable of both). These features enable them to have a “double-identity:”

1) To act as a Bronsted-Lowry acid, they must be able to dissociate and ________________________________.2) To act as a B-L base, they must be able to accept H+, which means they must have a lone pair of electrons.

Water is a prime example – it can donate H+ and it has two lone pairs of electrons. Auto-ionization of water: H2O + H2O H3O+ + OH-

Water reacting as a base with CH3COOH: CH3COOH(aq) + H2O(l) CH3COO- (aq) + H3O+ (aq)

Water reacting as an acid with NH3: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Ex) Write equations to show HCO3- reacting with water with (a) HCO3

- as an acid and (b) HCO3- as a base.

Lewis: a theory of electron pairso A Lewis ACID is an _______________________________________________.o A Lewis BASE is an _______________________________________________.

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Lewis acid-base reactions result in the formation of a covalent bond, which will always be a _______________________ bond (a.k.a. ___________________________________________ bond) because both the electrons come from the base.Example: (note – the “curly arrow” is a convention used to show donation of electons.)

Example: (note – boron has an incomplete octet, so it is able to accept an electron pair)

Example: (note – metals in the middle of the periodic table often form ions with vacant orbitals in their d subshell, so they are able to act as Lewis acids and accept lone pairs of electrons when they bond with ligands to form complex ions. Ligands, as donors of lone pairs, are therefore acting as Lewis bases)

Typical ligands found in complex ions include H2O, CN- and NH3. Note that they all have lone pairs of electrons, the defining feature of their Lewis base properties.

Table 8.1: Acid-base theory comparison Theory Definition of acid Definition of baseBronsted-Lowry Proton donor Proton acceptor

Lewis Electron pair acceptor Electron pair donor

Ex: For each of the following reactions, identify the Lewis acid and the Lewis base. a) 4NH3(aq) + Zn2+(aq) [Zn(NH3)4]2+(aq)

b) 2Cl-(aq) + BeCl2 (aq) + [BeCl4]2- (aq)

c) Mg2+(aq) + 6H2O(l) [Mg(H2O)6]2+(aq)

Ex: Which of the following could not act as a ligand in a complex ion of a transition metal?

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a) Cl- b) NCl3 PCl3 d) CH4

Properties of acids and basesFor acids and bases here, we will use the following definitions:

Acid: a substance that donates H+ in solution Base: a substance that can neutralize an acid to produce water --- includes metal oxides, hydroxides, ammonia,

soluble carbonates (Na2CO3 and K2CO3) and hydrogencarbonates (NaHCO3 and KHCO3) Alkali: a soluble base. When dissolved in water, alkalis all release the hydroxide ion, OH -. For example:

o K2O(s) + H2O(l) 2K+(aq) + 2OH-(aq)o NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)o CO3

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

o HCO3-(aq) CO2(g) + OH-(aq)

Neutralization: net ionic equation =

Acid-Base IndicatorsAcid-Base indicators change color reversibly according to the concentration of H+ ions in solution.

Many indicators are derived from natural substances such as extracts from flower petals and berries. ______________, a dye derived from lichens, can distinguish between acids and alkalis, but cannot indicate a particular pH. For this purpose, _____________________________________ was created by mixing together several indicators; thus universal indicator changes color many times across a range of pH levels.

Table 8.2: Some common acid-base indicatorsIndicator Color in acid Color in alkaliLitmus Pink bluemethyl orange Red yellowphenolphthalein Colorless pink

Acids react with metals, bases and carbonates to form salts1. Neutralization reactions with bases: acid + base salt + water

a. With hydroxide bases

b. With metal oxide bases

c. With ammonia (via ammonium hydroxide)

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Figure 8.1: Universal Indicator (pH 014)


2. With reactive metals (those above copper in the reactivity series): acid + metal salt + hydrogen

3. With carbonates (soluble or insoluble) / hydrogencarbonates: acid + carbonate salt + water + carbon dioxideStrong, Concentrated and CorrosiveIn everyday English, strong and concentrated are often used interchangeably. In chemistry, they have distinct meanings:

strong: completely _______________________________ into ions concentrated: high number of __________________ of solute per liter (dm3) of solution corrosive: chemically ___________________________

Similarly, weak and dilute also have very different chemical meanings: weak: only slightly dissociated into ions dilute: a low number of moles of solute per liter (dm3) of solution

Strong and weak acids and basesConsider the acid dissociation reaction: HA(aq) H+(aq) + A-(aq)Strong acid: equilibrium lies to the right (acid dissociates fully) reversible rxn is negligible exists entirely as ions

Ex:

Weak acid: equilibrium lies to the left (partial dissociation) exists almost entirely in the undissociated form Ex:

Similarly, the strength of a base refers to its degree of dissociation in water.Strong base ex:

Weak base ex:

NOTE: Weak acids and bases are much more common than strong acids and bases.Table 8.3: Strong and Weak Acids and Bases you should know

Strong Acids(only six; know 1st three for IB)

Strong Bases (Grp 1 hydroxides & barium hydroxide)

Weak Acidscarboxylic and carbonic acids

Weak Basesammonia and amines

H2SO4, sulfuric acid* LiOH, lithium hydroxide CH3COOH, ethanoic acid and other organic acids

C2H5NH2, ethylamineand other amines

HNO3, nitric acid NaOH, sodium hydroxide H2CO3, carbonic acidNote CO2(aq) = H2CO3(aq)

NH3, ammoniaNote NH3(aq) = NH4OH(aq)

HCl, hydrochloric acid KOH, potassium hydroxide H3PO4, phosphoric acidHI, hydroiodic acid Ba(OH)2, barium hydroxideHBr, hydrobromic acidHClO4, perchloric acid*NOTE: Sulfuric acid, H2SO4, is a diprotic acid which is strong in the dissociation of the first H+ and weak in the dissociation of the second H+.

For purposes of IB, only monoprotic dissociations are considered.

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Experimental methods for distinguishing between strong and weak acids and bases1. Electrical conductivity: strong acids and bases will have a higher conductivity (higher concentration of mobile ions)2. Rate of reaction: faster rate of rxn with strong acids (higher concentration of ions)3. pH: measure of H+ concentration in sol’n. A 1.0 M sol’n of strong acid will have lower pH than 1.0 M sol’n of weak

acid; 1.0 M sol’n of strong base will have higher pH than 1.0 M sol’n of weak base

PART 2: pH, pOH & pKw

The pH Scale pH is a value chemists use to give a measure of the acidity or alkalinity of a solution. Used because [H+] is usually very small pH stands for pouvoire of hydrogen.

o Pouvoire is French for “power.”o The normal range of the pH scale is 0-14.o However, it is possible (if the hydronium or hydroxide concentrations get above 1 Molar) for the pH to go

beyond those values. pH= -log[H+] [H+] = 10-pH

As pH decreases, [H+] increases exponentially (a change of one pH unit represents a 10-fold change in [H+]Example: If the pH of a solution is changed from 3 to 5, deduce how the hydrogen ion concentration changes.

Calculations involving acids and bases Sig figs for Logarithms (see page 631): The rule is that the number of decimal places in the log is equal to the number

of significant figures in the original number. Another way of saying this is only numbers after decimal in pH are significant.Example: [H+] = 1.0 x 10-9 M (2 significant figures) pH = -log(1.0 x 10-9) = 9.00 (2 decimal places)

Ion product constant of water, Kw

o Recall that water autoionizes: H2O(l) H+(aq) + OH-(aq) (endothermic)o Therefore Kc =

o The concentration of water can be considered to be constant because so little of it ionizes, and it can therefore be combined with Kc to produce a modified equilibrium constant known as kw. In fact, liquids and solids never appear in equilibrium expressions for this reason.

o Therefore, Kw =

o At 25C, Kw = 1.00 x 10-14

o In pure water, because [H+]=[OH-], it follows that [H+]=o So at 25C, [H+] = 1.0 x 10-7, which gives pH = 7.00

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Kw is temperature dependento Since the dissociation of water reaction in endothermic (bonds breaking), an increase in temperature will

shift the equilibrium to the ________________, thus ___________________________ the value of Kw.o As Kw increases, so do the concentrations of H+(aq) and OH-(aq) pH decreaseso However, since hydronium and hydroxide concentrations remain equal, water does not become acidic or

basic as temperature changes, but the measure of its pH does change. o Example: Fill in the rest of the table below

Table 8.4: Kw is temperature dependent

Temp (C) Kw [H+] in pure water pH of pure water

0 1.5 x 10-15 0.39 x 10-7 7.4710 3.0 x 10-15

20 6.8 x 10-15 0.82 x 10-7 7.0825 1.0 x 10-14

30 1.5 x 10-14 1.22 x 10-7 6.9240 3.0 x 10-14 1.73 x 10-7 6.7750 5.5 x 10-14

H+ and OH- are inversely relatedo Because the product [H+] x [OH-] is constant at a given temperature, it follows that as one goes up, the other

must go down (since Kw = [H+][OH-])Table 8.5 Solutions are defined as acidic, basic, or neutral based on the relative concentrations of H+ and OH-

Type of sol’n Relative concentrations pH at 25CAcid

Neutral

Alkaline

o Example: A sample of blood at 25C has [H+]=4.60 x 10-8 mol dm-3. Calculate the concentration of OH- and state whether the blood is acidic, neutral or basic.

How would you expect its pH to be altered at body temperature (37C)?

pH and pOH scales are inter-relatedo pOH=

From the relationship: KW = [H+][OH-] -log KW = -log([H+][OH-])

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-log KW = (-log[H+]) + (-log[OH-])pKW = pH + pOH

at 25C, KW = 1.0 x10-14, thus 14.00 = pH + pOH at 25C Given any one of the following we can find the other three: [H+],[OH-],pH and pOH Example: Lemon juice has a pH of 2.90 at 25C. Calculate its [H+],[OH-], and pOH.

PART 3: Weak Acids & BasesStrong acids and bases: pH and pOH can be deduced from their concentrations: since we assume strong acids and bases dissociate completely, pH and pOH can be calculated directly from the initial concentration of solution.Example: Calculate the pH of a 0.10 M solution of NaOH at 298 K.

Example: Calculate the pH of a 0.15 M solution of HNO3 at 298 K.

Dissociation constants express the strength of weak acids and basesSince equilibrium for weak acids and bases lies far to the left (they do not dissociate fully), concentrations of ions in solution cannot be determined by the initial concentrations without knowing the extent of dissociation. Consider the equilibrium expression for the dissociation of any weak acid in water:

o Ka is known as the acid dissociation constant. o It has a fixed value for a particular acid at a specified temperature. o Since the value of Ka depends on the position of equilibrium of acid dissociation, it gives us a direct measure of the

strength of an acid. o The higher the value of Ka at a particular temperature, the greater the dissociation and so the stronger the acid.o Note: because Ka is an equilibrium constant, its value does not change with the concentration of the acid or in the

presence of other ions.Consider the equilibrium expression for the dissociation of any weak base in water:

Kb is known as the base dissociation constant. It has the same characteristics as those described above for Ka.

Calculations involving Ka and Kb

The values of Ka and Kb enable us to compare the strengths of weak acids and bases and to calculate ion concentrations present at equilibrium (and therefore the pH and pOH values). Keep the following in mind:

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o The given concentration of an acid or base is its initial concentration (before dissociation occurs).o The pH (or pOH) of a solution refers to the concentration of H+ ions (or OH-ions) at equilibrium.o The concentration values substituted into the expressions for Ka and Kb must be the equilibrium values for all

reactants and products.o When the extent of dissociation is very small (very low value for Ka or Kb) it is appropriate you use the

approximations [acid]initial [acid]equilibrium and [base]initial [base]equilibrium.

Calculation of Ka and Kb from pH and initial concentrationExample: Calculate Ka at 25C for a 0.0100 mol dm-3 solution of ethanoic acid, CH3COOH. It has a pH value of 3.40 at this temperature.

Example: Calculate Kb at 25C for a 0.100 mol dm-3 solution of methylamine, CH3NH2. Its pH value is 11.80 at this temperature.

Calculation of [H + ] and pH, [OH - ] and pOH from K a and Kb

A real, but ugly example: Calculate the pH of a 0.10M solution of HNO2 (Ka = 4.0 x 10-4)Step 1: Write out the dissolving equation & the equilibrium law expression

Step 2: Set up ICE and let x = [H+] Step 3: Substitute equilibrium values into the equilibrium law expression.

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Step 4: Solve the problem using the assumed values. Step 4¼: Check assumption for validity (5% rule)

Step 4½: Solve the problem again using quadratic since the assumed values made an __ __ __ out of U and ME.

Step 5: Solve for pH

NOTE: Due to the nature of the timed exams you will be taking (IB and/or AP), assumptions of weak acids and bases dissociating less than 5% will always be considered valid. ASSUME away and do not use the quadratic formula. You must state your assumption, but you need not check on the validity of said assumption.

Pretty little AP/IB example: Determine the pH of a 0.75 mol dm-3 solution of ethanoic acid (Ka = 1.8 x 10-5).

Another pretty little AP/IB example: Determine the pH of a 0.20 mol dm-3 solution of ammonia (Kb = 1.8 x 10-5).

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pKa and pKb:

1) pKa and pKb numbers are usually positive and have no units2) The relationship between Ka and pKa and between Kb and pKb is inverse. Stronger acids or bases with higher

values for Ka or Kb have lower values for pKa or pKb.

3) A change of one unit in pKa or pKb represents a 10-fold change in the value of Ka or Kb.4) pKa and pKb must be quoted at a specified temperature (they are derived from temp-dependent Ka and Kb).

Relationship between Ka and Kb, pKa and pKb for a conjugate pair: Consider the Ka and Kb expressions for a conjugate acid-base pair HA and A-.

Therefore, Ka Kb = [H+] [OH-] = Kw

By taking the negative logarithms of both sides: pKa + pKb = Kw

At 25C Kw = 1.0 x 10-14, so pKw = 14.00 therefore pKa + pKb = 14.00 (at 25C)

Thus, the higher the Ka for the acid, the lower the value of Kb for its conjugate base. In other words, stronger acids have weaker conjugate bases and vice versa.

Practice Problems (Parts 1-3):1) Write the dissociation rxn and corresponding Ka equilibrium expression for the following acids in water: a) HCN and b) HF2) Identify the acid, base, conjugate base and conjugate acid:

a) C5H5NH+ + H2O C5H5N + H3O+ b) Al(H2O)63+ + H2O H3O+ + Al(H2O)5(OH)2+

3) List the 6 strong acids.

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4) List the four strong bases you should know.5) List the weak acids you should know.6) List the weak bases you should know.7) Fill in the chart below, assuming solutions are at 25 °C:

pH pOH [H+] [OH-] Acidic, alkaline, or neutral?5.02

3.9E-9 M0.27 M

8.5

8) Calculate the Kb of ethylamine, C2H5NH2 given that a 0.10 mol dm-3 solution has a pH of 11.86.9) What are the [H+] and [OH-] in a 0.10 mol dm-3 solution of an acid that has a Ka = 1.0 x 10-7?10) The pKa of ethanoic acid, CH3COOH, at 25 C is 4.76. What is the pKb of its conjugate base, CH3COO-?

PART 4: Salt Hydrolysis & Buffer SolutionsBUFFER SOLUTIONS: solutions that resists a change in pH on the addition of small amounts of acid or alkali.

Two types:o acidic buffers – maintain a pH value _______________________o basic buffers – maintain a pH value _______________________

Mixture of two solutions, combined in such a way that they each contain the two species of a conjugate acid-base pair.

1) Acidic BuffersComposition of the buffer solution:

Made by mixing an aqueous solution of a weak acid with a solution of its salt of a strong alkali. For example:

The following equilibria exist in a solution of this mixture:(weak acid – eq’m lies to the left)

(soluble salt, fully dissociated in solution)

So the mixture contains relatively high concentrations of both CH3COOH and CH3COO-. Note that this is an acid and its conjugate base. These can be considered “reservoirs,” ready to react with added OH - and H+ respectively in neutralization reactions.

Response to added acid and base:Addition of acid (H+): H+ combines with the base (CH3COO-) to form CH3COOH, removing most of the added H+.

Addition of base (OH-): OH- combines with the acid (CH3COOH) to form CH3COO- and H2O, thereby removing most of the added OH-.

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Consequently, as the added H+ and OH- are used in these rxns, they do not persist in solution and so pH does not change significantly.

2) Basic BuffersComposition of the buffer solution:

Made by mixing an aqueous solution of a weak base with a solution of its salt of a strong acid. For example:

The following equilibria exist in solution:(weak base – eq’m lies to the left)

(soluble salt, fully dissociated in solution)

So the mixture contains relatively high concentrations of both NH3 and NH4+. Note that this is a base and its

conjugate acid. These act as “reservoirs,” ready to react with added H+ and OH- respectively in neutralization reactions.

Response to added acid and base:Addition of acid (H+): H+ combines with the base (NH3) to form NH3

+, removing most of the added H+.

Addition of base (OH-): OH- combines with acid (NH4+) to form NH3 and H2O, removing most of the added OH-.

So, as with the acidic buffer, the removal of the added H+ and OH- by reactions with components of the buffer solution keeps the pH of the buffered solution relatively constant. (For further proof that buffers resist pH change: see Sample Exercise 15.3, p. 685)

Determining the pH of a buffer solutionConsider an acidic buffer made of the generic weak acid HA and its salt MA.

We can make two approximations based on some assumptions about these reactions:1. The dissociation of the weak acid is so small that it can be considered negligible thus [HA]initial [HA]eq’m

2. The salt is considered to be fully dissociated into its ions thus [MA]initial [A-]eq’m

Henderson-Hasselbalch Equation(s):pH = pOH =

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Example: Calculate the pH of a buffer solution at 298 K, prepared by mixing 25 cm3 of 0.10 mol dm-3 ethanoic acid, CH3COOH, with 25 cm3 of 0.10 mol dm-3 sodium ethanoate, NaCH3COO. Ka of CH3COOH = 1.8 x 10-5 at 298 K.

Note: when a buffer solution contains equal amounts in moles of acid and salt (or base and salt), the last term in the Henderson-Hasselbalch expression becomes 0, so pH = pKa (or pOH = pKb).

Making buffer solutionspH of a buffer depends on:

1. The pKa (or pKb) of its acid or base2. The ratio of the initial concentration of acid and salt (or base and salt) used in its preparation

Acidic buffer: React a weak acid (whose pKa value is as close as possible to the desired pH of the buffer) with enough strong alkali such that one half of it is converted into salt.For example:

CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O(l)Initial amt. (mol)Change (mol) Eq’m (mol)

Basic buffer: React a weak base (whose pKb value is as close as possible to the desired pOH of the buffer) with enough strong acid such that one half of it is converted into salt.

Example: How would you prepare a buffer solution of pH 3.75? (See IB Data Booklet)

Factors that can influence buffers _______________: does not change pH (Ka & Kb are constant & ratio of acid or base to salt does not change), but

does decrease buffering capacity (amt. of acid or base the sol’n can absorb without a significant change in pH) _______________: As temp. affects the values of Ka and Kb, the pH of the buffer is affected accordingly.

SALT HYDROLYSISNeutralization rxn: acid + base salt + water (pH of resulting solution depends on relative strengths of acid and base)

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Anion hydrolysis: The anion (A-) is a conj. base of the parent acid. When the acid is weak, the conj. base is strong enough to hydrolyze water (and the release of OH- causes the pH of the sol’n to ↑):

Cation hydrolysis: The cation (M+) is a conj. acid of the parent base. When the base is weak and this conj. is a non-metal (i.e. NH4

+), the conj. acid is strong enough to hydrolyze water (and the release of H+ causes the pH of the sol’n to ↓).

When the cation is a metal, the outcome depends on charge density. Metal ions that are either small with a +2 or +3 charge, or are transition metals (i.e. Fe3+) hydrolyze water (attracting OH-; releasing H+ into sol’n). Group 1 cations and Group 2 cations below Be on the P.T. do not have sufficient charge density to hydrolyze water.

Neutralization Rxn Ex: parent acid & base Salt formed Hydrolysis of ions Type of salt sol’n pH of salt sol’nstrong acid and

strong baseHCl + NaOH

neither ion hydrolyzes

weak acid andstrong base

CH3COOH + NaOH anion hydrolysis

strong acid andweak base

HCl + NH3

HCl + Al(OH)3cation hydrolysis

weak acid and weak base

CH3COOH + NH3anion and cation

hydrolyzeDepends on relative

strengths of conjugatescannot

generalize

PART 5: Titration Curves & IndicatorsACID-BASE TITRATIONS

Commonly used to determine the amount of acid or base in a solution. Complete by adding a solution of known concentration until the substance being tested is consumed.

This is called the _____________________ point (a.k.a.__________________________ point) Titration curve - graph of pH vs. volume of added acid/base

Units millimole (mmol) = 1/1000 mol = 10-3 mol Molarity = mol/L = mmol/mL This makes calculations easier because we will rarely add liters of solution.

Strong acid & Strong Base: Do the stoichiometry. There is no equilibrium. They both dissociate completely.Weak acid & Strong base: There is an equilibrium. Do stoichiometry. Then do equilibrium (ICE or H-H).

Titration Curves1) Strong acid and strong base

2) Weak acid and strong baseTitration Curves: Weak Acid and Strong Base (Comparing Weak Acids)

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3) Weak base and strong acid

4) Weak acid and weak base

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1. The solution of a weak acid has a higher initial pH than a solution of a strong acid of the same concentration

2. The weaker the acid the more rapidly the pH rises in the early part of the titration, but more slowly near the equivalence point. This is because of the buffering action of the formed salt.

3. The pH at the equivalence point is not 7.00. The weaker the acid, the higher the pH at the EQ point.


Calculations involving Titrations of Weak Acids/BasesDivide the titration into the following points:

1.

2. Use a two step procedure when calculating pH prior to the equivalence point whenever a strong acid/base is added

to a weak acid/base.o 1. Stoichiometric Calculation: Allow the strong acid/base to react to completion with the weak acid/base,

producing a solution that contains the weak acid and its common ion. (its conjugate base/acid)o 2. Equilibrium Calculation: Use the value of Ka/Kb and the equilibrium expression to calculate the

equilibrium concentrations of the weak acid/base, and the common ion, and H+ (ICE notation or HH Eq’n)

3.

4.

Example: Titration of a weak acid with a strong baseCalculate the pH when the following quantities of 0.050M KOH solution have been added to 50.0 ml of a 0.025M solution of benzoic acid (HC7H5O2 Ka = 6.5x10-5). A) 20.0ml B) 25.0ml C) 30.0ml

Step 0: Find the EQ point

Step 1: Starting pointThis is just a “weak acid in water” problem…..

Step 2: Before the EQ point (20.0 ml KOH added)First perform stoichiometric calculations…Calculate moles of OH- present… Calculate moles HC7H5O2 present…

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Next, using these values, perform equilibrium calculations…

Step 3: At the Eq point (25.0 ml KOH added)

Step 4: Beyond the Eq point (30.0 ml KOH added)Beyond the equivalence point, there is no longer any benzoic acid to neutralize the additional KOH added. The pH of the solution will be determined by the amount of excess KOH (the effect of the salt is negligible compared to the KOH)

We have used three different methods of calculating the pH during a titration….1. Before the equivalence point has been reached. (Weak acid/base + common ion, buffered region)2. At the equivalence point. (hydrolysis of a salt)3. Beyond the equivalence point. (Strong Acid/Base in water)

INDICATORSSignal change in pH; change color when the pH = their pKa; can be used to signal the equivalence point in titrationsAn indicator is a weak acid (or base) in which the dissociated form is a different color than the undissociated form.

Color changes when [I-] = [HIn]. Thus, indicator changes color when Kin = [H+] …or when pKin = pHIndicator pKin pH range Usemethyl orange 3.7 3.1-4.4 Titrations with strong acidsphenolphthalein 9.6 8.3-10.0 Titrations with strong bases

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Part 6: Acid Deposition

Causes of acid depositionRain is naturally _________________ because of dissolved __________, but acid rain has a pH of < ______.

Acid deposition refers to the process by which acidic particles, gases and precipitation leave the atmosphere. Both wet deposition (________________________________) and dry deposition (_________________________________) occur.

True “acid deposition” is caused by oxides of nitrogen and oxides of sulfur.

Primary air pollutants – harmful substances released into the air that are not normally present Ex: SOx and NOx

Secondary air pollutants – harmful compounds formed when primary pollutants react in air Ex: acid rain

Sulfur oxides (SOx)Sulfur dioxide occurs naturally from volcanoes and is produced industrially from the combustion of sulfur-containing fossil fuels and the smelting of sulfide ores.

In the presence of sunlight, sulfur dioxide is oxidized to sulfur trioxide.

The oxides can react with water in the air to form sulfurous acid and sulfuric acid:

Reduction of SO2 emissionsi) Pre-combustion methods

Hydrodesulfurization: Some sulfur is present in _________________ as metal sulfides (i.e. FeS) and can be physically removed by crushing coal and mixing with water. The more dense sulfides sink to the bottom and the cleaned coal can be skimmed off. Sulfur is also removed from oil before it is refined by converting it into _____________________.

Flue-gas desulfurization: ______________________________ can be removed from the exhaust of coal burning plants by “scrubbing” with an alkaline slurry of limestone (_________) and lime (________). The resulting sludge is used for landfill or as gypsum (___________________) to make plasterboard (drywall). Rxns:

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Unit 8: Acids & Bases IB Topics 8 & 18 AP Chapters (Zumdahl): 4.8; 14(all); 15(all) AP Chapters 1-5

ii) Post-combustion methodsA more modern method known as fluidized bed combustion involves burning the coal on a bed of limestone which removes the sulfur as __________________ or _______________ as the coal burns.

Nitrogen oxides (NOx)Nitrogen oxides occur naturally from electrical storms and bacterial action. Nitrogen monoxide is produced in the internal combustion engine and in jet engines.

Oxidation to nitrogen dioxide occurs in the air.

The nitrogen dioxide then reacts with water to form nitric acid and nitrous acid:

…or is oxidized directly by to nitric acid by oxygen in the presence of water:

Reduction of NOx emissionsi) Catalytic converters in vehicles

The hot exhaust gases are passed over a catalyst of platinum, rhodium or palladium. These fully oxidize _________ and unburned ____________, and also catalyze the rxn between CO and NO.

ii) Lower temperature combustion:

iii) Other options:

Mechanism of acid deposition caused by NOx and SOx In the atmosphere, NOx and SOx are converted into acids by a free radical mechanism involving hydroxyl free radicals, OH. These hydroxyl free radicals are formed either by the reaction of water vapor with ozone

…or by the reaction of water vapor with oxygen free radicals that are formed when ozone decomposes.

The hydroxyl radicals then react directly with NOx and SOx in the presence of water to give the dissolved acids.

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Unit 8: Acids & Bases IB Topics 8 & 18 AP Chapters (Zumdahl): 4.8; 14(all); 15(all) AP Chapters 1-5

Effects of Acid Deposition

Environmental effects of acid depositionVegetation – increased acidity in the soil leaches important nutrients, such as Ca2+, Mg2+ and K+. Reduction of Mg2+ can cause reduction in chlorophyll and consequently lowers the ability of plants to photosynthesize. Many trees have been seriously affected by acid rain. Symptoms include stunted growth, thinning of tree tops, and yellowing and loss of leaves. The main cause is the aluminum leached from rocks into the groundwater. The Al3+ ion damages the roots and prevents the tree from taking up enough water and nutrients to survive.

Lakes and rivers – increased levels of aluminum ions in water can kill fish. Aquatic life is also highly sensitive to pH. Below pH 6 the number of sensitive fish, such as salmon and minnow, decline as do insect larvae and algae. Snails cannot survive a pH less than 5.2 and below pH 5.0 many microscopic animal species disappear. Below pH 4.0 lakes are effectively dead. The nitrates present in acid rain can also lead to eutrophication.

Buildings – stone, such as marble, that contains calcium carbonate is eroded by acid rain. With the sulfuric acid the calcium carbonate reacts to form calcium sulfate, which can be washed away by rainwater thus exposing more stone to corrosion. Salts can also from within the stone that can cause the stone to crack and disintegrate.

Human health – the acids formed when NOx and SOx dissolve in water irritate the mucus membranes and increase the risk of respiratory illness, such as asthma, bronchitis and emphysema. In acidic water there is more probability of poisonous ions, such as Cu2+ and Pb2+, leaching from pipes and high levels of aluminum in water may be linked to Alzheimer’s disease (the jury is still out on this one).

Methods to lower or counteract the effects of acid deposition1. Lower the amounts of NOx and SOx formed (i.e. by improved engine design, use of catalytic converters, and

removing sulfur before, during and after combustion of sulfur-containing fuels.)2. Switch to alternative methods of energy (i.e. wind and solar power) and reducing the amount of fuel burned (i.e.

by reducing private transport and increasing public transport and designing more efficient power stations)3. Liming of lakes – adding calcium oxide or calcium hydroxide (lime) neutralizes acidity, increases the amount of

calcium ions and precipitates aluminum from solution. This has been shown to be effective in many, but not all, lakes where it has been tried.

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