1

Acids, Bases,

and SaltsAll are electrolytes

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Mr. Sharp playing with acids and bases

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The Solubility Product

Ksp

Used for sparingly or slightly soluble salts

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Insoluble Salts and Compounds

A compound may be said to be “Insoluble” in water, but there are always a few particles or ions that do dissolve.

A compound is said to be: insoluble if less that 0.1 of a gram of it will

dissolve in 100 grams of water slightly soluble if between 0.1 and 1.0 grams of it will

dissolve in 100 grams of water soluble if more that 1.0 grams of it will dissolve in

100 grams of water

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Silver chloride is a sparingly soluble salt Its equilibrium reaction is:

• AgCl(s) Ag+(aq) + Cl-(aq)

Its equilibrium expression is:• Keq = [Ag+] [Cl-] / [AgCl]

The concentration for all pure liquids and solids is constant; therefore [AgCl] becomes part of the Keg which we call Ksp, , the solubility product.• Ksp = [Ag+] [Cl-]

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Value of Ksp gives a general idea of how insoluble a substance may be in water

See figure 17-6 on page 566 in Text

In general, the larger the Ksp value the more soluble the substance will be

In general, the smaller the Ksp value the less soluble the substance will be

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General Ksp for any compound AbCd

Equilibrium equation:• AbCd bA+d + dC-b

Equilibrium expression(Ksp)• Ksp = [A+d]b [C-b]d

• Like practice problems 17-1; 2,6,10

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Practice problem 1 Text page565 Equilibrium equation:

• Ag2CrO4 2Ag+1 + CrO4-2

Ksp Expression:

• Ksp = [Ag+1]2 [CrO4-2]

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Practice Problem 2 Text page 565 Equlibrium equation:

• PbI2 Pb+2 + 2I-1

Ksp expression:

• Ksp = [Pb+2] [I-1]2

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Finding Ksp , given ion concentration Practice problem 3 text page 567

Like practice problems 17-1; 14,18 Equilibrium equation: Cd(OH)2 Cd+2 + 2OH-1

Ksp expression: Ksp = [Cd+2] [OH-1]2

Given [Cd+2] = 1.7 x 10-5; therefore, [OH-1] = 2[Cd+2] = 2(1.7 x 10-5) = 3.4 x 10-5

Ksp= (1.7 x 10-5) (3.4 x 10-5)2 = 2.0 x 10-14

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Practice problem 4 Text page 567

Equilibrium equation: Ce(OH)3 Ce+3 + 3OH-1

Ksp expression: Ksp = [Ce+3] [OH-1]3

Given: [Ce+3] = 5.2 x 10-6; therefore, [OH-] = 3[Ce+3] = 3(5.2 x 10-6) = 1.56 x 10-5

Ksp = (5.2 x 10-6)(1.56 x 10-5 )3 = 2.0 x 10-20

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Finding concentration given Ksp

Practice problem 5 Text page 570

Equilibrium equation: CaF2 Ca+2 + 2F-

Ksp expression:

Ksp = [Ca+2] [F-]2 = 3.9 x 10-11

Let [Ca+2] = x and [F-] = 2xTherefore Ksp = (x)(2x)2 = 3.9 x 10-11

4x3 = 3.9 x 10-11; x3 = (3.9 x 10-11)(4)x3 = 9.75 x 10-12; x = (9.75 x 10-12) [Ca+2] = x = 2.1 x 10 -4 ; [F-] = 2x = 4.2 x 10-4

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Practice problem 6 Text page 570 Like practice problems 17-1;

22,26,30 Equilibrium equation: BaCrO4 Ba+2 + CrO4

-2

Ksp = [Ba+2] [CrO4] = 2.0 x 10-10

Let x = [Ba+2] = [CrO4]; therefore (x) (x) = x2 = 2.0 x 10-10

x = (2.0 x 10-10)½ = 1.4 x 10-5

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Properties of Acids

Taste sour (we test not taste) Turns blue litmus red Neutralizes bases and

basic(metallic)oxides React with metals like Zn to produce

hydrogen gas Solution in water are electrolytes Examples:HCl, HNO3, H2SO4,

HCH3COO (CH3COOH) (HC2H3O2)

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Propeties of Bases

Taste bitter(we test not taste) Feel slippery Turn red litmus blue Neutralize acids and acidic(nonmetallic)

oxides Solutions in water are electrolytes Corrosive Examples: NaOH, KOH, Ca(OH)2, NH3

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Acid and Base Theories

Arrhenius - 1887

Bronsted-Lowery - 1923

Lewis - 1923

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Arrhenius Theory Acids - substances that ionize in

water to produce hydrogen ions(H+)

Bases - substances that ionize in water to produce hydroxide ions(OH-)

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Arrhenius Examples Arrhenius Acids:

HCl H+ + Cl-

HNO3 H+ + NO3-

Arrhenius Bases: NaOH Na+ + OH-

Ca(OH)2 Ca2+ + 2OH-

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Bronsted-Lowry Theory

Acids - substances that donate protons(Proton donors)

Bases - substances that accept protons(Proton acceptors)

Hydrogen ions are really the same as protons

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Hydronium Ion(hydrogen ion riding piggy-back on a water molecule)

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Bronsted-Lowry Example: Water acting as an acid

Ammonia + water yield ammonium ion plus hydroxide ion

NH3 + H2O NH4+ + OH-

base acid conjugate conjugate

acid base

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Bronsted Lowry Example:Water acting as a base

Acid base Conjugate acid conjugate base

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Amphiteric = being able to act as an acid or base

acid = proton donor

base = proton acceptor

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Lewis Theory Acids - substances that accept

electron-pairs

Bases - substances that donate electron-pairs

Must draw electron dot formula to determine

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Lewis example:

Lewis base Lewis acid

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Lewis Example

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Lewis Example:

Lewis Lewis acid base

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Lewis Theory is most inclusive, but we will use Bronsted-Lowry, mostly

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Naming Binary Acids Made up of Hydrogen and a

nonmetal (two elements) Use prefix of hydro Use the nonmetal as the root of

the name Add suffix of ic Example: HCl = hydrochloric

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Practice Naming Acids

HBr HI HF H2O H2S H2Te H2Se

hyrdobromic acid hydroiodic acid hydrofluoric acid hydroxyic acid hydrosulfuric acid hydrotelluric acid hydroselenic acid

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Naming Oxyacids Name most common acid in family

with the root of nonmetal other than oxygen then add suffix of ic

Acid of family with one more oxygen than most common add prefix of per and suffix of ic

Acid with one less oxygen than most common use suffix of ous

Acid with two less oxygens than most common add prefix of hypo and keep suffix of ous

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Naming tertiary(oxyacids) Use chlorine oxyacid family as a guide:

HClO4 (one more O) per means more than

HClO3 (most common)

HClO2 (one less O)

HClO (two less O’s) hypo means less than

perchloric acid

chloric acid

chlorous acid

hypochlorous acid

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Naming oxyacids

H2SO4(most common) H2SO3

HNO3(most common) HNO2

HBrO4

HBrO3(most common) HBrO2

HBrO

Sulfuric acid sulfurous acid nitric acid nitrous acid perbromic acid bromic acid bromous acid hypobromous acid

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Naming Arrhenius bases Name positive ion first. Then add

name hydroxide last NH4OH = ammonium hydroxide KOH = potassium hydroxide Ca(OH)2 = calcium hydroxide Mn(OH)7 = manganese(VII)

hydroxide

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Practice Naming bases LiOH

Ba(OH)2

Al(OH)3

Sn(OH)4

Lithium hydroxide

barium hydroxide

aluminum hydroxide

tin(IV) hydroxide

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Anydrides means without water

Acid anhydride = an acid without water(nonmetallic oxides SO2 + H2O H2SO3

N2O5 + H2O 2HNO3

Basic anydride = base without water(metallic oxides) BaO + H2O Ba(OH)2

Na2O + H2O 2NaOH

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Determining anhydride by subtracting water H2SO4 - H2O = SO3

2HNO3 - H2O = N2O5

2H3AsO4 - 3H2O = As2O5

Ba(OH)2 - H2O = BaO 2NaOH - H2O = Na2O 2Al(OH)3 - 3H2O = Al2O3

All H atoms must add out

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Acids and bases neutralize each other

acid + base salt + water parent parent child + water

acid base salt

1. HCl + NaOH NaCl + H2O

2. HClO4 + NaOH NaClO4 + H2O

3. HClO3 + NaOH NaClO3 + H2O

4. HClO2 + NaOH NaClO2 + H2O

5. HClO + NaOH NaClO + H2O

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Rules for naming salts Named from parent acid and base First name comes from parent base Second name comes from parent acid

Salts from binary acids end in ide Salts from oxyacids:

Salt from most common, use suffix of ate Salt from one more oxygen use per prefix and ate suffix Salt from one less oxygen use ite suffix Salt from two less oxygens use hypo prefix and ite suffix

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Practice naming salts from slide #38 NaCl

NaClO4

NaClO3

NaClO2

NaClO

sodium chloride

sodium perchlorate

sodium chlorate

sodium chlorite

sodium hypochlorite

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More Practice naming salts: Go to slide #39 to answer the following

KBrO3

Mn2(SO4)7

CuSO3

Mg(BrO4)2

BaI2 LiBrO NaNO3

KNO2

Potassium bromate manganese(VII) sulfate copper(II) sulfite magnesium perbromate barium iodide lithium hypobromite sodium nitrate potassium nitrite

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Strong acids and bases

Strong acids and bases are nearly 100% ionized HCl + H2O H3O+ + Cl-

If 0.10 mole of HCl are placed in water, we get 0.10 mole of H3O+

NaOH Na+ + OH-

If 0.02 mole of NaOH are place in water we get 0.02 mole of OH-

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Weak acids and bases Weak acids and bases are only slightly

ionized HC2H3O2 + H2O H3O+ + C2H3O2

-

Weak acid produces very few H3O+, [H3O+] must be calculated

NH3 + H2O NH4+ + OH-

Weak base produces very few OH-, [OH-] must be calculated

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45

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Self Ionization of water

H2O + H2O H3O+ + OH-

Keq = [ H3O+ ] [ OH-] / [ H2O]2

multiply each side by [H2O]2 and let Keq[H2O]2 = Kw

Kw = [ H3O+] [ OH-] = 1.00 x 10-14

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Concentration of ions in pure water

In pure water [H3O+] = [OH-] = 10-7 M

In an acid solution, [H3O+] > [OH-]

In a basic solution, [H3O+] < [OH-]

In a neutral solution [H3O+] = [OH-]

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Calculating [OH-] or [H3O+] Given that [OH-] = 4.78 x 10-12,

determine the [H3O+] and if the solution is acidic or basic.

[[H3O+] [OH-] = 1 x 10-14

[H3O+] = (1 x 10-14) [OH-] [H3O+] = (1 x 10-14) / (4.78 x 10-12) = 2.09 x 10-3

[H3O+] > [OH-] therefore solution is acidic

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Solving # 2, 19-1 Practice problems

[H3O+] [OH-] = 1 x 10-14

[OH-] = 1 x 10-8

[H3O+] = (1 x 10-14) (1 x 10-8)

[H3O+] = 1 x 10-6 M [H3O+] > [OH-], therefore solution is

acid

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pH is a method of expressing the acidity of a

water solution

pH = -log[H3O+] [H3O+] = 10-pH

In pure water, pH = 7; neutral In acid solution, pH < 7 In basic solution, pH > 7

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Logs are exponents of 10 log(1000) = 3 log(100) = 2 log(10) = 1 log(1) = 0 log(.1) = -1 log(.01) = -2 log(.001) = -3 Log(4.3 x 10-4) = -3.37 log(8.97 x 10 -12) = -11.05

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Using calculator to find logs

Put number into the calculator and push log button

Find log(2.76 x 10-8) = -7.56

Because 10-7.56 = 2.76 x 10-8

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Solving

Solving # 12 on 19-1 Practice problems

[H3O+] = 2.51 x 10-5 , given pH = -log[H3O+] pH = -log(2.51 x 10-5) pH = -(-4.60) = 4.60 pH < 7, therefore solution is acidic

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Acid-Base Indicators

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More indicators

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Most indicators are Weak organic acids, like litmus

HIn(aq) + H2O(l) H3O+(aq) + In-

(aq)

Acid Conjugate Base

Adding an acid(H3O+) will shift equilibrium to the left toward red

Adding a base (OH-) will shift equlibrium to the right toward blue

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Acid-base titration

Standard solution = solution of known concentration(M)

Titrated solution = solution of unknown concentration

Endpoint = condition where an equivalent amount of standard solution as been add to the titrated solution. [H3O+] = [OH-]

Indicator used to find endpoint

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Titration calculations

MaVa = MbVb , where Ma = molarity of the acid Mb = molarity of the base Vb = volume of base used Va = volume of acid used

When MaVa = MbVb , the endpoint has been reached and [H3O+] = [OH-], which means equivalent amounts of acid and base

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Example of titration calculation, page 638 text

Practice #5 Vb= 43.0 mL; Va = 32.0 mL Ma = 0.100; Mb = ?

MaVa = MbVb Mb=MaVa ÷ Vb

Mb= (0.100M)(32.0 mL) ÷ 43.0 mL

Mb = 0.0744 M

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Titration calculation Practice problem #6

Worked the same way as #5 except H2SO4 contains two moles of acid ions per mole of

acid the molarity of the acid must be mltiplied by 2

2MaVa = MbVb Mb= 2MaVa÷ Vb

Mb = 2(0.120M)(25mL) ÷ 40. mL Mb = 0.15 M

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Calculating Ka for a weak acid Practice problem #1, page 613 in text

HCOOH + H2O H3O+(aq) + HCOO-

(aq)

Ka= [H3O+] [HCOO-] ÷ [HCOOH] Initial conc. of acid = 0.100 M Final conc of [H3O+] = 0.0042 M Final conc. of acid is (0.100 - 0.0042) M [H3O+] = [HCOO-] = 0.0042 M Ka = [0.0042] [0.0042] ÷ [0.0958] = 1.8 x 10-4