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Salts, Acids, and Bases

Chapter 11

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Solutions• Solutions are homogeneous mixtures• Solute is the dissolved substance

– Seems to “disappear” or “Takes on the state” of the solvent

• Solvent is the substance the solute dissolves in– Does not appear to change state– When both solute and solvent have the same state, the solvent

is the component present in the highest percentage

• Solutions in which the solvent is water are called aqueous solutions– Water is often called the universal solvent

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Solubility• When one substance (solute) dissolves in another

(solvent) it is said to be soluble– Salt is soluble in Water, – Bromine is soluble in methylene chloride

• When one substance does not dissolve in another they are said to be insoluble– Oil is insoluble in Water

• There is usually a limit to the solubility of one substance in another– Gases are always soluble in each other– Some liquids are always mutually soluble

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Solutions & Solubility

• Molecules that are similar in structure tend to form solutions– Like dissolves like

• The solubility of the solute in the solvent depends on the temperature– Higher Temp = Larger solubility of solid in liquid– Lower Temp =Larger solubility of gas in liquid

• The solubility of gases depends on the pressure– Higher pressure = Larger solubility

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The precipitation reaction that occurs when yellow potassium chormate, K2CrO4 (aq), is mixed with a colorless barium nitrate solution, Ba(NO3)2 (aq)

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The Solution Process - Ionic Compounds

• When ionic compounds dissolve in water they dissociate into ions– ions become surrounded by water molecules -

hydrated

• When solute particles are surrounded by solvent molecules we say they are solvated

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When solid sodium chloride dissolves, the ions are dispersed randomly throughout the solution

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Polar water molecules interact with the positive and negative ions of a salt

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Describing Solutions - Qualitatively

• A concentrated solution has a high proportion of solute to solution

• A dilute solution has a low proportion of solute to solution

• A saturated solution has the maximum amount of solute that will dissolve in the solvent– Depends on temp

• An unsaturated solution has less than the saturation limit• A supersaturated solution has more than the saturation

limit– Unstable

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Solution Concentration

• Parts Per Million

• PPM = grams of solute per 1,000,000 g of solution

• PPM = mg of solute per liter of solution– 1000 mg = 1 g

• Mass of Solution = Mass of Solute + Mass of Solvent

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Chemical Packages - Moles• We use a package for atoms and molecules

called a mole• A mole is the number of particles equal to the

number of Carbon atoms in 12 g of C-12 • One mole = 6.022 x 1023 units• The number of particles in 1 mole is called

Avogadro’s Number• 1 mole of C atoms weighs 12.01 g and has 6.02 x 1023 atoms

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One-mole samples of iron (nails), iodine crystals, liquid mercury, and powdered sulfur

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Example #1

Use the Periodic Table to determine the mass of 1 mole of Al

1 mole Al = 26.98 g Use this as a conversion factor for

grams-to-moles

Al mol 0.371 g 26.98

Al mol 1x Al g 10.0

Compute the number of moles and number of atoms in 10.0 g of Al

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Use Avogadro’s Number to determine the number of atoms in 1 mole

1 mole Al = 6.02 x 1023 atoms Use this as a conversion factor for

moles-to-atoms

atoms Al 10x 2.23 Al mol 1atoms 10x 6.02

x Al mol 0.371 2323

Example #1

Compute the number of moles and number of atoms in 10.0 g of Al

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Use Avogadro’s Number to determine the number of atoms in 1 mole

1 mole Al = 6.02 x 1023 atoms Use this as a conversion factor for

atoms-to-moles

Al mol 0.370 atoms 10x 6.02

Al mol 1x atoms Al 10x 2.23 23

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Compute the number of moles and mass of 2.23 x 1023 atoms of Al

Example #2

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Use the Periodic Table to determine the mass of 1 mole of Al

1 mole Al = 26.98 g Use this as a conversion factor for

moles-to-grams

Al g 9.99 Al mol 1g 26.98

x Al mol 0.370

Compute the number of moles and mass of 2.23 x 1023 atoms of Al

Example #2

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Molar Mass• The molar mass is the mass in grams of one mole of a

compound• The relative weights of molecules can be calculated from

atomic masses

water = H2O = 2(1.008 amu) + 16.00 amu = 18.02 amu

• 1 mole of H2O will weigh 18.02 g, therefore the molar mass of H2O is 18.02 g

• 1 mole of H2O will contain 16.00 g of oxygen and 2.02 g of hydrogen

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Solution ConcentrationMolarity

• moles of solute per 1 liter of solution

• used because it describes how many molecules of solute in each liter of solution

• If a sugar solution concentration is 2.0 M , 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar, etc.

molarity = moles of soluteliters of solution

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Properties of Acids• Sour taste• Change color of vegetable dyes• React with “active” metals

– Like Al, Zn, Fe, but not Cu, Ag or Au

Zn + 2 HCl ZnCl2 + H2

– Corrosive

• React with carbonates, producing CO2

– Marble, baking soda, chalk

CaCO3 + 2 HCl CaCl2 + CO2 + H2O

• React with bases to form ionic salts – And often water

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Properties of Bases

• Also Known As Alkalis• Taste bitter• Feel slippery• Change color of vegetable dyes

– Different color than acid

– Litmus = blue

• React with acids to form ionic salts– And often water

– Neutralization

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Arrhenius Theory

• Acids ionize in water to H+1 ions and anions

• Bases ionize in water to OH-1 ions and cations

• Neutralization reaction involves H+1 combining with OH-1 to make water

• H+ ions are protons

• OH- ions are called hydroxide ions

• Definition only good in water solution

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Strength of Acids & Bases• The stronger the acid, the more willing it is to donate H• Strong acids donate practically all their H’s

HCl + H2O H3O+1 + Cl-1

• Strong bases will react completely with water to form hydroxidesCO3

-2 + H2O HCO3-1 + OH-1

• Weak acids donate a small fraction of their H’s– The process is reversible, the conjugate acid and conjugate base can react

to form the original acid and base

HC2H3O2 + H2O H3O+1 + C2H3O2-1

• Only small fraction of weak base molecules pull H off waterHCO3

-1 + H2O H2CO3 + OH-1

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Graphical representation of the behavior of acids in aqueous solution

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Multiprotic Acids

• Monoprotic acids have 1 acid H, diprotic 2, etc.– In oxyacids only the H on the O is

acidic• In strong multiprotic acids, like

H2SO4, only the first H is strong; transferring the second H is usually weak

H2SO4 + H2O H3O+1 + HSO4-1

HSO4-1 + H2O H3O+1 + SO4

-2

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Water as an Acid and a Base

• Amphoteric substances can act as either an acid or a base

– Water as an acid, NH3 + H2O NH4+1 + OH-1

– Water as a base, HCl + H2O H3O+1 + Cl-1

• Water can even react with itself

H2O + H2O H3O +1 + OH-1

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Autoionization of Water• Water is an extremely weak electrolyte

– therefore there must be a few ions present

H2O + H2O H3O+1 + OH-1

• all water solutions contain both H3O+1 and OH-1

– the concentration of H3O+1 and OH-1 are equal

– [H3O+1] = [OH-1] = 10-7M @ 25°C

• Kw = [H3O+1] x [OH-1] = 1 x 10-14 @ 25°C

– Kw is called the ion product constant for water

– as [H3O+1] increases, [OH-] decreases

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Acidic and Basic Solutions

• acidic solutions have a larger [H+1] than [OH-1]

• basic solutions have a larger [OH-1] than [H+1]

• neutral solutions have [H+1]=[OH-1]= 1 x 10-7 M

[H+1] = 1 x 10-14

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

[H+1]

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Example #2

Determine the given information and the information you need to find

Given [H+1] = 10.0 M Find [OH-1]

Solve the Equation for the Unknown Amount

][H

K ]OH[

][OHx ][H K

1w1-

1-1w

Determine the [H+1] and [OH-1] in a 10.0 M H+1 solution

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Convert all the information to Scientific Notation and Plug the given information into the equation.

Given [H+1] = 10.0 M = 1.00 x 101 M

Kw = 1.0 x 10-14

M 10x 1.0 10x 1.00

10x 1.0 ]OH[

][H

K ]OH[

15-1

14-1-

1w1-

Example #2Determine the [H+1] and [OH-1] in a

10.0 M H+1 solution

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pH & pOH• The acidity/basicity of a solution is often expressed as pH or

pOH• pH = -log[H3O+1] pOH = -log[OH-1]

– pHwater = -log[10-7] = 7 = pOHwater

• [H+1] = 10-pH [OH-1] = 10-pOH

• pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral• The lower the pH, the more acidic the solution; The higher the pH,

the more basic the solution• 1 pH unit corresponds to a factor of 10 difference in acidity • pOH = 14 - pH

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The pH scale and pH values of some common substances

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A pH meter

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Indicator paper being used to measure the pH of a solution

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Example #3

Find the concentration of [H+1]

M 10x 1.0 10x 1.0

10x 1.0 ]H[

][OH

K ]H[

8-6-

14-1

1w1

Calculate the pH of a solution with a [OH-1] = 1.0 x 10-6 M

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Enter the [H+1] concentration into your calculator and press the log key

log(1.0 x 10-8) = -8.0

Change the sign to get the pHpH = -(-8.0) = 8.0

Example #3

Calculate the pH of a solution with a [OH-1] = 1.0 x 10-6 M

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Enter the [H+1] or [OH-1]concentration into your calculator and press the log key

log(1.0 x 10-3) = -3.0Change the sign to get the pH or pOH

pOH = -(-3) = 3.0Subtract the calculated pH or pOH from

14.00 to get the other valuepH = 14.00 – 3.0 = 11.0

Calculate the pH and pOH of a solution with a [OH-1] = 1.0 x 10-3 M

Example #4

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If you want to calculate [OH-1] use pOH, if you want [H+1] use pH. It may be necessary to convert one to the other using 14 = [H+1] + [OH-1]

pOH = 14.00 – 7.41 = 6.59Enter the pH or pOH concentration into your

calculatorChange the sign of the pH or pOH

-pOH = -(6.59)Press the button(s) on you calculator to take the

inverse log or 10x

[OH-1] = 10-6.59 = 2.6 x 10-7

Example #5Calculate the [OH-1] of a solution with a pH of 7.41

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Calculating the pH of a Strong, Monoprotic Acid

• A strong acid will dissociate 100%

HA H+1 + A-1

• Therefore the molarity of H+1 ions will be the same as the molarity of the acid

• Once the H+1 molarity is determined, the pH can be determined

pH = -log[H+1]

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Example #6

Determine the [H+1] from the acid concentrationHNO3 H+1 + NO3

-1

0.10 M HNO3 = 0.10 M H+1

Enter the [H+1] concentration into your calculator and press the log key

log(0.10) = -1.00Change the sign to get the pH

pH = -(-1.00) = 1.00

Calculate the pH of a 0.10 M HNO3 solution

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Neutralization Reactions• Acid-Base reactions are also called

Neutralization Reactions• Often we use neutralization reactions to

determine the concentration of an unknown acid or base

• The procedure is called a titration. With this procedure we can add just enough acid solution to neutralize a known volume of a base solution– Or visa versa

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Buffered Solutions• Buffered Solutions resist change in pH when an acid

or base is added to it.• Used when need to maintain a certain pH in the

system– Blood

• A buffer solution contains a weak acid and its conjugate base

• Buffers work by reacting with added H+1 or OH-1 ions so they do not accumulate and change the pH

• Buffers will only work as long as there is sufficient weak acid and conjugate base molecules present