Acids and Bases Chapter 15

Notes One Unit TwelveProperties of Acids

Properties of Bases

Structure of Bases

Neutralization Reactions

Properties of AcidsSour taste.

warhead

React with “active” metals produce H2. Al, Zn, and Fe

React with carbonates, producing CO2 and H2O.

Marble, baking soda, chalk, limestone.

Change color of vegetable dyes.cabbage juice / On the top

cabbage juice / baking soda (left)

cabbage juice / vinegar (right).

Acidsacid - proton donor (H+1)

strong vs. weak acids

strong - lots of (H+ )

weak - very little (H+ )

HCl is a strong acidMuriatic acid is

hydrochloric acid.Making Bleach.Making PVC pipe.Making Table Salt.Human stomach acid.Cleaning steel.Neutralize bases in

chemical plants. Chrome tanning leather.

HNO3 is a strong acid

Explosives Fertilizers etc.Nitrate salts To make H2SO4

Etching copper, brass, bronze

Dyes, perfumesPurification of Ag, Au, Pt

Properties of BasesAlso known as alkali.( Li, Na, K, Rb, Cs, Fr)

Taste bitter.caffeineoften poisonous.

Solutions feel slippery.

Change color of vegetable dyes.Different color than acid.

Red litmus turns blue.React with acids to form salt and

water(Neutralization).Acid + base salt +water

Common Base Features

Most ionic bases contain OH-1 ions.

NaOH (drain cleaner)

Some contain CO32- ions.

CaCO3(in Tums)

NaHCO3 (baking soda)

Molecular bases contain structures that react with H+.

Mostly amine groups(-NH2).

Neutralization Reactions

acid + base salt + water

Double-displacement

Some make CO2 and H2O

Acid+carbonate Salt +Water+Carbon DioxideAcid+Base Salt +WaterAcid+Metal Salt +Hydrogen

Acid Reactions

Acid and CarbonateAcid+

Na2CO3 CarbonateSalt+ Water+Carbon Dioxide

HNO3+ NaNO3 + H2O + CO22 1 2 1 1

H+1 NO3-1 Na+1 CO3

-2

NaNONaNO33

=Sodium Nitrate=Sodium Nitrate( )_( )_NaNa+1+1 NONO33

-1-11111

Salt ?Salt ?

Acid and BaseAcid+

KOH Base Salt + Water

H2SO4+ KK22SOSO44 + H2O 1 2 1 2

H+1 SO4-2 K+1 OH-1

KK22SOSO44

=Potassium Sulfate=Potassium Sulfate( )_( )_KK+1+1 SOSO44

-2-21122

Salt ?Salt ?

Acid and MetalAcid+

Mg(s) Metal Salt + Hydrogen

H2SO4 + MgMgSO4 + H2(g)1 1 1 1

H+1 SO4-2 Mg(s)

MgMgSO4

= Magnesium Sulfate= Magnesium Sulfate

( )_( )_MgMg+2+2 SO4-2

2222

Salt ?Salt ?

Titration Example One

NaOH(aq)+ HCl(aq) NaCl(aq) + H2O(l)

0.000599 m HCl x1 m NaOH

1 m HCl=

1 1 1 1

What is the molarity of a NaOH solution, if 12.01 mL is required to titrate 5.99 mL of 0.100 M HCl?1. Balance the equation.

2. Find moles used of known solution.

3. Calculate moles used of unknown (titrant).

4. Calculate the molar concentration of the titrant.

MxV=n0.00599L = 0.100M x 0.000599m HCl

n/V=M0.000599m NaOH ÷0.01201L= 0.0499M NaOH

0.000599m NaOH

Notes Two Unit Twelve -Text Pages 550-558

• Self Ionization of Water

• Brönsted-Lowrey Acid-Base Theory

• Arrhenius Theory

Self Ionization of Water

• Water is and acid and a base at the same time…amphoteric

• H2O(l) + H2O(l) H3O+ + OH-

• Mass Action Expression• Kw = [H3O+][OH-] • Or Kw = [H+][OH-] • Kw = 1.0 x 10-14

Brönsted-Lowrey Acid-Base Theory• Acid - proton donor • Base - proton acceptor• Acid-Conjugate Base / Base-Conjugate Acid

• HCl(aq) + H2O(l) Cl−1(aq) + H3O+1(aq)

• NaF(aq) + H2O(l) HF(aq) + NaOH(aq)

• OH−1(aq) + H2O(l) H2O(l)+ OH−1(aq)

• NH3(aq) + H2O(l) NH4+1(aq) + OH−1(aq)

A CB

A

A

CB

CB

CB

A

CAB

B

B

B

CA

CA

CA

H+1

Arrhenius Theory• Bases form OH- ions in water.• Acids form H+ ions in water.

Arrhenius theory

• HCl(aq) H+1(aq) + Cl−1(aq)

• HF(aq) H+1(aq) + F−1(aq)

• NaOH(aq) Na+1(aq) + OH−1(aq)

• NH3(aq) +H2O(aq) NH4+1(aq) + OH−1(aq)

Polyprotic Acids• More than one proton to donate.

• H2CO3(aq)   H+1(aq) + HCO3-1(aq)

• HCO3-1(aq)      H+1(aq)    +    CO3

-2(aq)

• H2SO4 H+1(aq) + HSO4−1(aq)

• HSO4−1 H+1(aq) + SO4

−2(aq)

Notes Three Unit Twelve

• Titration

Titration• Titration is a technique to determine

the concentration of an unknown solution.

• Titrant (unknown solution) • Phenolphthalein identifies the

Endpoint.

Titration Endpoint• Add 10mL of HCl and three drops phenolphthalein to the flask.• Add about 8mL base, swirl and add the rest of the base using

increasingly faster spins of the valve.(?????)

Titration-Acid Volume

• Acid Burette

• Initial Reading?

• 1.98mL

• Final Reading?

• 7.97mL

• Volume Used?

• 5.99mL

Titration-Volume of Base Used.

• Base Burette

• Initial Reading?

• 0.00mL

• Final Reading?

• 12.01mL

• Volume Used?

• 12.01mL

Titration Example Two• Lactic acid Concentration of Sauerkraut

• For Joe’s final in chemistry, he was asked to find the concentration of lactic acid in homemade sauerkraut.

• He did not have any home made sauerkraut.

• Therefore, Joe was left to make the home made sauerkraut.

• He looked on line and found the following recipe.

Clean and Quarter 35lb of Fresh Cabbage

Shred the Cabbage into a Crock

Add 3 Tbsp salt per 5 pounds of cabbage.

Mix salt and cabbage.

Pack the cabbage into a

crock and weight it down.

It should be fermented in one month.

Titration Example Two

NaOH + HC2H4OHCO2 NaC2H4OHCO2 + H2O

0.0899m NaOHx1m HC2H4OHCO2

1 m NaOH=

1 1 1 1

What Is the molarity of a sauerkraut juice if 10.0 mL is titrated using 89.9 mL of 1.00 M NaOH?1. Balance the equation.

2. Find moles used of known solution.

3. Calculate moles used of unknown (titrant).

4. Calculate the Molar Concentration of theTitrant.

MxV=n1.00M x 0.0899L = 0.0899m NaOH

n/V=M0.0899m HC2H4OHCO2÷0.01201L =

M=0.0899M HC2H4OHCO2

0.0899m HC2H4OHCO2

M

Titration Example Three

Ba(OH)2(q) + HCl(aq BaCl2(aq)+ H2O(l)

0.00463 m HCl x1 m Ba(OH)2

2 m HCl=

1 2 1 2

What is the volume of a 0.0622M Ba(OH)2 solution, if it is titrated using 43.8 mL of 0.1057 M HCl?1. Balance the equation.

2. Find moles used of known solution.

3. Calculate moles used of unknown (titrant).

4. Calculate volume of titrant.

MxV=n0.1057M x 0.0438L= 0.00463m HCl

n/M=V0.00234m Ba(OH)2 ÷0.0622M = 0.0376M Ba(OH)2

0.00232 m Ba(OH)2

Conclusions continued

• 2. Define these terms: standard solution; titration; endpoint.

• Standard Solution: When the concentration of a solution is known to a high degree of accuracy and precision.

• titration:  When the concentration of an acid or base is determined by neutralizing it.

• endpoint:  The point where you actually stop a titration, usually because an indicator has changed color.  This is different than the "equivalence point" because the indicator might not change colors at the exact instant that the solution is neutral.

Notes Three Unit Twelve Text Pages 559-567

Weak Acids and Bases

• A weak acid

• little H+1

• A weak base

• little OH-1

• [H+] or [OH-] from a Keq.

pH = -log[H+1]pH is a measure of the amount of hydrogen in a solution. It is based on the water.

pH Scale

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Acid BaseNeutral

1M HCl 1M NaOHAmmonia Cleaner

Blood

WaterMilk

Stomach Acid

Lemon Juice

Vinegar

pH +pOH =14 pH +pOH =14

pH of strong acid

•Find the pH of a 0.15 M Find the pH of a 0.15 M solution of Hydrochloric solution of Hydrochloric acidacid

•pH = - log 0.15pH = - log 0.15•pH = - (- 0.82)pH = - (- 0.82)•pH = 0.82pH = 0.82

pH of a strong Base

•What is the pH of the What is the pH of the 0.0010 M NaOH solution?0.0010 M NaOH solution?

•pOH = - log (0.0010)pOH = - log (0.0010)

•pOH = 3pOH = 3

•pH = 14 – 3 = 11pH = 14 – 3 = 11pH +pOH =14 pH +pOH =14

Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases

Calculating pH (pH=-log[H+1])The Ka for nitrous acid is 4.5 x 10-4. Calculate the [H+1] and pH in 0.15 M nitrous acid solution.

HNO2 NO2-1 + H+1

[HNO2 ] [NO2-1] [H+1]

Bef

ΔAt

[HNO2][H+1][NO2

-1]Ka=

-x +x +x0.15-x x x

0.15 0 0

Ka= [0.15][x][x]

= 4.5 x 10-4

X=[0.0082M]

1) Balanced Equation

2) Mass Action Expression3) What do we know?

4) Calculate the [H+1] and pH.

Very small

= [H+1] pH=-log[ ]0.0082M

pH=-log[H+1]

= 2.09

4.5 x 10-4

1.5 x 10-1

Calculating Concentrations Using Ka

The Ka for benzoic acid is 6.5 x 10-5. (a) Calculate the concentrations of C6H5COO-1 and H+ in a 0.10 M benzoic acid solution. (b) Calculate pH.

C6H5COOH C6H5COO-1 + H+1

[C6H5COOH] [C6H5COO-1] [H+1]

Bef

ΔAt

[C6H5COOH][H+1][C6H5COO-1]

Ka=

-x +x +x

0.10-x x x

0.10 0 0

Ka= [0.10][x] [x]

= 6.5 x 10-5

X=[0.0025M]

1) Balanced Equation

2) Mass Action Expression3) What do we know?

4) Calculate the [H+1] and pH.

Very small

= [H+1]pH=-log[ ]0.0025M = 2.60

6.5 x 10-5

1.0 x 10-1

End