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AnnouncementsAnnouncements & Agenda& Agenda (02/23/07)(02/23/07)

You should be reading Ch 10 this weekend!You should be reading Ch 10 this weekend!

Quiz Today!Quiz Today!

Open Review Sessions @ 3pm on Wed.Open Review Sessions @ 3pm on Wed. Low attendance this week Low attendance this week

TodayToday Acid & base strength: Quantitative (8.4-8.5)Acid & base strength: Quantitative (8.4-8.5)

the pH scalethe pH scale Acid & base reactionsAcid & base reactions

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Last Time:Last Time:BrBroonsted-Lowry Acids & Basesnsted-Lowry Acids & Bases

• acids donate a proton (Hacids donate a proton (H++) ) • bases accept a proton (Hbases accept a proton (H++))

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• A A strong acid/basestrong acid/base completely ionizes (100%) completely ionizes (100%) in aqueous solutions.in aqueous solutions.

HCl(HCl(gg) + H) + H22O(O(ll) H) H33OO++ ( (aqaq) + Cl) + Cl−− ( (aqaq))

COMPARE TO STRONG ELECTROLYTESCOMPARE TO STRONG ELECTROLYTES

• A A weak acid/baseweak acid/base dissociates only slightly in dissociates only slightly in water to form a few ions in aqueous solutions. water to form a few ions in aqueous solutions.

HH22COCO33((aqaq) + H) + H22O(O(ll) H) H33OO++((aqaq) + HCO) + HCO33−− ( (aqaq))

COMPARE TO WEAK ELECTROLYTESCOMPARE TO WEAK ELECTROLYTES

Last Time:Last Time:Strengths of Acids/Bases - IonizationStrengths of Acids/Bases - Ionization

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• make up six (just a few) of all the acids. make up six (just a few) of all the acids.

• have weak conjugate bases (the product have weak conjugate bases (the product formed after the proton is transferred).formed after the proton is transferred).

Strong Acids (Know These)Strong Acids (Know These)

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Strong BasesStrong Bases

• are formed from metals of are formed from metals of Groups 1A (1) and 2A (2). Groups 1A (1) and 2A (2).

• include LiOH, NaOH, KOH, include LiOH, NaOH, KOH, and Ca(OH)and Ca(OH)22..

• dissociate completely in water.dissociate completely in water.

KOH(KOH(ss) K) K++((aqaq) + OH) + OH−−((aqaq))

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In water In water occasionallyoccasionally,,• HH++ is transferred from 1 H is transferred from 1 H22O molecule to another. O molecule to another. • one water acts an acid, the another acts as a base.one water acts an acid, the another acts as a base.

HH22O + HO + H22O HO H33OO++ + OH+ OH−− .. .. .. .. .. .. .... ::OO:: H + H + HH::OO:: H H::OO::HH ++ + + ::OO::HH−−

.. .. .. .. .. .. .... HH HH H H water water hydronium hydroxidewater water hydronium hydroxide

ion (+)ion (+) ion (-)ion (-)

Ionization of Water: A Basis for Ionization of Water: A Basis for Understanding pH (H+ concentrations)Understanding pH (H+ concentrations)

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Pure Water is Neutral Pure Water is Neutral (NOT ACIDIC OR BASIC)(NOT ACIDIC OR BASIC)

• the ionization of water the ionization of water molecules produces small, molecules produces small, but equal quantities of Hbut equal quantities of H33OO++

and OHand OH−− ions.ions.• molar concentrations are molar concentrations are

indicated in brackets as indicated in brackets as [H[H33OO++] and [OH] and [OH−−]. ].

[H[H33OO++]] = 1.0 x 10= 1.0 x 10−−7 7 MM

[OH[OH−−]] == 1.0 x 101.0 x 10−−7 7 MM

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Acidic SolutionsAcidic Solutions

Adding an acid to Adding an acid to pure water:pure water:

• increases the increases the [H[H33OO++].].

• causes the causes the [H[H33OO++]] to to

exceed 1.0 x 10exceed 1.0 x 10-7 -7 M.M.

• decreases the decreases the [OH[OH−−].].

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Basic SolutionsBasic Solutions

Adding a base to pure Adding a base to pure water:water:

• increases the increases the [OH[OH−−].].

• causes the causes the [OH[OH−−]] to to exceed 1.0 x 10exceed 1.0 x 10−− 7 7M.M.

• decreases the decreases the [H[H33OO++].].

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

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The ion product constant, KThe ion product constant, Kww, for water , for water

• is the product of the concentrations of the hydronium is the product of the concentrations of the hydronium and hydroxide ions.and hydroxide ions.

KKww = = [ H[ H33OO++]] [ OH[ OH−− ]]

• can be obtained from the concentrations in pure water.can be obtained from the concentrations in pure water.

KKww = = [ H[ H33OO++]] [ OH[ OH−− ]]

KKww = [1.0 x 10 = [1.0 x 10−− 7 7 M] x [ 1.0 x 10M] x [ 1.0 x 10−− 7 7 M] M]

= = 1.0 x 101.0 x 10−− 14 14

Ion Product of Water, KIon Product of Water, Kww

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[H[H33OO++] and [OH] and [OH−−] in Solutions] in Solutions

IMPORTANT: Kw is always 1.0 x 10−14.

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Calculating [HCalculating [H33OO++]]What is the [HWhat is the [H33OO++] of a solution if [OH] of a solution if [OH−−] is 5.0 x 10] is 5.0 x 10-8-8

M? M?

STEP 1:STEP 1: Write the K Write the Kww for water. for water.

KKww = [H = [H33OO+ + ][OH][OH−− ] = 1.0 x 10] = 1.0 x 10−−1414

STEP 2:STEP 2: Rearrange the K Rearrange the Kww expression. expression.

[H[H33OO++] = ] = 1.0 x 101.0 x 10-14-14

[OH[OH−−]]

STEP 3:STEP 3: Substitute [OH Substitute [OH−−].]. [H[H33OO++] = ] = 1.0 x 101.0 x 10-14 -14 = 2.0 x 10= 2.0 x 10-7-7 M M

5.0 x 105.0 x 10- 8 - 8

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If lemon juice has [HIf lemon juice has [H33OO++] of 2 x 10] of 2 x 10−−33 M, what M, what

is the [OHis the [OH−−] of the solution?] of the solution?

0%

0%

0% 1) 1) 2 x 102 x 10−−1111 M M

2) 2) 5 x 105 x 10−−1111 M M

3) 3) 5 x 105 x 10−−1212 M M

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3) 3) 5 x 105 x 10−−1212 M M

Rearrange the KRearrange the Kww to solve for [OH to solve for [OH- - ]]

KKww = [H= [H33OO+ + ][OH][OH−− ] = 1.0 x 10] = 1.0 x 10−−1414

[OH[OH−− ] ] = = 1.0 x 10 1.0 x 10 -14-14 = 5 x 10= 5 x 10−−1212 M M 2 x 10 2 x 10 - 3- 3

SolutionSolution

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pH ScalepH Scale

The pH of a solutionThe pH of a solution

• is used to indicate the acidity of a solution. is used to indicate the acidity of a solution.

• has values that usually range from 0 to 14.has values that usually range from 0 to 14.

• is acidic when the values are less than 7.is acidic when the values are less than 7.

• is neutral with a pH of 7.is neutral with a pH of 7.

• is basic when the values are > 7.is basic when the values are > 7.

NOTE: pH is a logarithmic scale!!!NOTE: pH is a logarithmic scale!!!

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pH of Everyday SubstancespH of Everyday Substances

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Testing the pH of SolutionsTesting the pH of Solutions

The pH of solutions can be determined using The pH of solutions can be determined using • a) pH metera) pH meter• b) pH paperb) pH paper• c) indicators that have specific colors at c) indicators that have specific colors at

different pH values.different pH values.

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pH is the negative log of the hydronium ion concentration.pH is the negative log of the hydronium ion concentration.

pH = - log [HpH = - log [H33OO++] ]

Example: Example: For a solution with [HFor a solution with [H33OO++] = 1 x 10] = 1 x 10−−4 4

pH =pH = −log [−log [1 x 101 x 10−−4 4 ]]

pH = - [-4.0]pH = - [-4.0]

pH = 4.0pH = 4.0

Note: The number of decimal places in the pH equalsNote: The number of decimal places in the pH equals

the significant figures in the coefficient of [Hthe significant figures in the coefficient of [H33OO++].].

4.4.00 1 SF in1 SF in 11 x 10x 10-4-4

Calculating pHCalculating pH

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A. The [HA. The [H33OO++] of tomato juice is 2 x 10] of tomato juice is 2 x 10−−44 M. M.

What is the pH of the solution?What is the pH of the solution?

1) 4.01) 4.0 2) 3.7 2) 3.7 3) 10.33) 10.3

B. The [OHB. The [OH−−] of a solution is 1.0 x 10] of a solution is 1.0 x 10−−3 3 M.M.

What is the pH of the solution?What is the pH of the solution?

1) 3.001) 3.00 2) 11.00 2) 11.00 3) -11.003) -11.00

Learning CheckLearning Check

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If an area received 1 inch of rain with a pH If an area received 1 inch of rain with a pH of 4, how much more neutral rain would be of 4, how much more neutral rain would be needed to have a final pH of 6?needed to have a final pH of 6?

0%

0%

0%

0% 1.1. Approximately 2 inchesApproximately 2 inches2.2. Approximately 9 inchesApproximately 9 inches3.3. Approximately 20 inchesApproximately 20 inches4.4. Approximately 100 inchesApproximately 100 inches

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[H[H33OO++], [OH], [OH--], and pH Values], and pH Values

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Calculating [HCalculating [H33OO++] from pH] from pH

The [HThe [H33OO++] can be expressed by using the pH as the ] can be expressed by using the pH as the negative power of 10.negative power of 10.

[H[H33OO++] = 1 x 10 ] = 1 x 10 --pH pH

For pH = 3.0, the [HFor pH = 3.0, the [H33OO++] = 1 x 10 ] = 1 x 10 --33

On a calculatorOn a calculator

1. Enter the pH value 1. Enter the pH value 3.0 3.0

2. Change sign 2. Change sign -3.0-3.0

3. Use the inverse log key (or 103. Use the inverse log key (or 10xx) to obtain) to obtain

the [Hthe [H3300++]. ]. = 1 x 10 = 1 x 10 --3 3 MM

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In a neutralization reaction: In a neutralization reaction: • a base such as NaOH reacts with an acid such as a base such as NaOH reacts with an acid such as

HCl.HCl.

HCl + HHCl + H22OO HH33OO++ + Cl + Cl−−

NaOHNaOH NaNa++ + OH + OH−−

• the Hthe H33OO+ + from the acid and the OHfrom the acid and the OH− − from the basefrom the base

form water.form water.

HH33OO++ + OH + OH−− 2 H 2 H22OO

Neutralization Rxns of Acids & BasesNeutralization Rxns of Acids & Bases

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Bases Used in Some AntacidsBases Used in Some Antacids

Antacids are used to neutralize stomach acid (HCl).Antacids are used to neutralize stomach acid (HCl).

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In the equation for neutralization, an acid In the equation for neutralization, an acid and a base produce a salt and water.and a base produce a salt and water.

acidacid base base saltsalt waterwater

HCl + NaOH NaCl + HHCl + NaOH NaCl + H22OO

2HCl + Ca(OH)2HCl + Ca(OH)22 CaCl CaCl22 + 2H + 2H22OO

Neutralization EquationsNeutralization Equations

Balance these like any other reaction!

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Solving Problems…Solving Problems…What is the molarity of an HCl solution if 18.5 mL of a What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl?0.225 M NaOH are required to neutralize 10.0 mL HCl?

HCl(HCl(aqaq) + NaOH() + NaOH(aqaq) NaCl() NaCl(aqaq) + H) + H22O(O(ll))

Method:Method:

Get into moles with “known”:Get into moles with “known”:

Given: Given: 18.5 mL of 0.225 M NaOH18.5 mL of 0.225 M NaOH

Do a moles-to-moles conversionDo a moles-to-moles conversion

Get out of moles with “unknown”:Get out of moles with “unknown”:

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18.5 mL NaOH x 18.5 mL NaOH x 1 L NaOH 1 L NaOH x x 0.225 mole NaOH0.225 mole NaOH

1000 mL NaOH1000 mL NaOH 1 L NaOH 1 L NaOH

x x 1 mole HCl 1 mole HCl = 0.00416 mole HCl = 0.00416 mole HCl

1 mole NaOH1 mole NaOH

MMHClHCl = = 0.00416 mole HCl0.00416 mole HCl = 0.416 M HCl = 0.416 M HCl

0.0100 L HCl0.0100 L HCl

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Calculate the mL of 2.00 M HCalculate the mL of 2.00 M H22SOSO44 required to required to

neutralize 50.0 mL of 1.00 M KOH. neutralize 50.0 mL of 1.00 M KOH. HH22SOSO44((aqaq) + 2KOH() + 2KOH(aqaq) K) K22SOSO44((aqaq) + 2H) + 2H22O(O(ll))

1) 12.5 mL1) 12.5 mL

2) 50.0 mL2) 50.0 mL

3) 200. mL3) 200. mL

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SolutionSolution

1)1) 12.5 mL12.5 mL

0.0500 L KOH x 0.0500 L KOH x 1.00 mole KOH1.00 mole KOH x x 1 mole H 1 mole H22SOSO44 x x

1 L KOH1 L KOH 2 mole KOH 2 mole KOH

1 L H1 L H22SOSO44 x x 1000 mL1000 mL = 12.5 mL = 12.5 mL

2.00 mole H2.00 mole H22SOSO44 1 L H 1 L H22SOSO44

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Two More Acid/Base Reactions…Two More Acid/Base Reactions…

1. Acids react with metals1. Acids react with metals

• such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn.such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn.

• to produce hydrogen gas and the salt of the metal.to produce hydrogen gas and the salt of the metal.

Molecular equations:Molecular equations:

2K(2K(ss) + 2HCl() + 2HCl(aqaq) ) 2KCl( 2KCl(aqaq) + H) + H22((gg))

Zn(Zn(ss) + 2HCl() + 2HCl(aqaq)) ZnCl ZnCl22((aqaq) + H) + H22((gg))

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Acids and CarbonatesAcids and Carbonates

Acids reactAcids react

• with carbonates & hydrogen carbonateswith carbonates & hydrogen carbonates

• to produce carbon dioxide gas, a salt, & to produce carbon dioxide gas, a salt, & water.water.

2HCl(2HCl(aqaq) + CaCO) + CaCO33((ss)) CO CO22((gg) + CaCl) + CaCl22((aqaq) + H) + H22O(O(ll))

HCl(HCl(aqaq) + NaHCO) + NaHCO33((ss) ) CO CO22((gg) + NaCl () + NaCl (aqaq) + H) + H22O(O(ll))