acids & bases strong & weak · 27 neutralization problems if an acid and a base combine in...

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Acids & BasesStrong & weak

Thursday, October 20, 2011

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Acid Base DissociationAcid-base reactions are equilibrium processes.

The relationship between the relative concentrations of the reactants and products is a constant for a given temperature. It is known as the Acid or Base Dissociation Constant (Ka or Kb).The stronger the acid or base, the larger the value of the dissociation constant

Write the K for this equation:


Generally...

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Acid Strength Acid strength depends on how much an acid

dissociates (breaks up into ions). The more it dissociates (turns into H3O+), the

stronger it is The larger the Ka, the stronger the acid

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Concentration vs. Strength Concentrations refers to how many moles are

in a volume Strength refers to how much of the substance

has ionized

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Acid Strength Strong Acids - Transfers all of its protons to water; - Completely ionized; - Strong electrolyte; - The conjugate base is weaker and has a negligible tendency to be protonated. - K is so large it is often not written

Identify the conjugate acid-base pairs.

Examples of Strong AcidsHydrochloric Acid Perchloric AcidNitric Acid Sulfuric Acid


Weak acids

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Transfers only a fraction of its protons to water; - Partly ionized;

- Weak electrolyte; - The conjugate base is stronger, readily accepting protons from water


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As acid strength decreases, base strength increases. The stronger the acid, the weaker its conjugate base The weaker the acid, the stronger its conjugate base


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Base Strength Strong Base - all molecules accept a proton; - completely ionizes; - strong electrolyte; - conjugate acid is very weak, negligible tendency to donate protons.

Weak Base - fraction of molecules accept proton; - partly ionized; - weak electrolyte; - the conjugate acid is stronger. It more readily donates protons.

As base strength decreases, acid strength increases. The stronger the base, the weaker its conjugate acid. The weaker the base the stronger its conjugate acid.

Sodium Hydroxide

Potassium Hydroxide

*Barium Hydroxide

*Calcium Hydroxide


Try it! You want to know which is the stronger acid:

HCl or CH3COOH. Here is your data.

Write the equation for each acid in water Write the Ka for both acids Explain which is stronger, using your data

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Try it!

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Acids & BasespH Scale


pH Scale

Acid Base

0

7

14

[H+] pH

10-14 14

10-13 13

10-12 12

10-11 11

10-10 10

10-9 9

10-8 8

10-7 7

10-6 6

10-5 5

10-4 4

10-3 3

10-2 2

10-1 1

100 0

1 M NaOH

Ammonia(householdcleaner)

BloodPure waterMilk

VinegarLemon juiceStomach acid

1 M HClAci

dic

N

eutra

l

Bas

ic

The pH scale is a logarithmic scale used to assess acidity based on the equilibrium concentration of hydrogen or hydronium ion


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pH and acidity

The pH values of several common substances are shown at the right.

Many common foods are weak acids

Some medicines and many household cleaners are bases.


pH of Common Substance 14 1 x 10-14 1 x 10-0 0 13 1 x 10-13 1 x 10-1 1 12 1 x 10-12 1 x 10-2 2 11 1 x 10-11 1 x 10-3 3 10 1 x 10-10 1 x 10-4 4 9 1 x 10-9 1 x 10-5 5 8 1 x 10-8 1 x 10-6 6

6 1 x 10-6 1 x 10-8 8 5 1 x 10-5 1 x 10-9 9 4 1 x 10-4 1 x 10-10 10 3 1 x 10-3 1 x 10-11 11 2 1 x 10-2 1 x 10-12 12 1 1 x 10-1 1 x 10-13 13 0 1 x 100 1 x 10-14 14

NaOH, 0.1 MHousehold bleachHousehold ammonia

Lime waterMilk of magnesia

Borax

Baking sodaEgg white, seawaterHuman blood, tearsMilkSalivaRain

Black coffeeBananaTomatoesWineCola, vinegarLemon juice

Gastric juice

Mor

e ba

sic

Mor

e ac

idic

pH [H3O+] [OH-] pOH

7 1 x 10-7 1 x 10-7 7


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Calculating the pHThe pH system is a logarithmic representation of the Hydrogen Ion concentration as a means of avoiding using large numbers and powers.

pH = - log [H3O+]Example 1: If [H3O+] = 1 X 10-10

pH = - log 1 X 10-10

pH = 10

Example 2: If [H3O+] = 1.8 X 10-5

pH = - log 1.8 X 10-5

pH = - (- 4.74) pH = 4.74


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pH and Water

Kw = [H3O+] [OH-] = 1.0 x10-14

In pure water [H3O+] = 1 x 10-7 mol / L (at 20oC)

∴ pH = - log(1 x 10-7) = - (0 - 7) = 7

pH < 7 (Acidic) [H3O+] > 1 x 10-7 M

pH > 7 (Basic) [H3O+] < 1 x 10-7 M


pOH and basicity

The pOH system is a logarithmic representation of the Hydroxide Ion concentration as a means of avoiding using large numbers and powers.

pOH = - log [OH-]

The relationship between pH and pOH: • pH + pOH = 14

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pH Calculations

pH

pOH

[H3O+]

[OH-]

pH + pOH = 14

pH = -log[H3O+]

[H3O+] = 10-pH

pOH = -log[OH-]

[OH-] = 10-pOH

[H3O+] [OH-] = 1.0 x10-14


pH = - log [H3O+]pH = 4.6

pH = - log [H3O+]

4.6 = - log [H3O+]

- 4.6 = log [H3O+]

- 4.6 = log [H3O+]

Given:

2nd log

10x

antilog

multiply both sides by -1

substitute pH value in equation

take antilog of both sides

determine the [hydronium ion]

choose proper equation

[H+] = 2.51x10-5 M

You can check your answer by working backwards.

pH = - log [H+]

pH = - log [2.51x10-5 M]

pH = 4.6

Recall, [H3O+] = [H+]


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Indicators


Hydrangeas

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Try it! 1. A certain solution has [H3O+] = 2 x 10-5. Calculate the concentration of [OH-] and the pH of the solution

2. A certain solution has [H3O+] = 5 x 10-9. Calculate the concentration of [OH-] and the pH of the solution

3. A certain solution has [OH-] = 4 x 10-3. Calculate the concentration of [H3O+] and the pH of the solution

4. A certain solution has [OH-] = 3 x 10-11. Calculate the concentration of [H3O+] and the pH of the solution

5. A certain solution has pH = 3.0. Calculate the pOH, the concentration of [H3O+] and the [OH-] of the solution

6. A certain solution has pH = 2.5. Calculate the pOH, the concentration of [H3O+] and the [OH-] of the solution

7. A certain solution has pOH = 2.0. Calculate the pH, the concentration of [H3O+] and the [OH-] of the solution

8. A certain solution has pOH = 9.3. Calculate the pH, the concentration of [H3O+] and the [OH-] of the solution 23


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Neutralization An acid will neutralize a base, giving a salt

and water as products Examples

HCl + NaOH NaCl + H2O H2SO4 + 2 NaOH Na2SO4 + 2 H2O H3PO4 + 3 KOH K3PO4 + 3 H2O 2 HCl + Ca(OH) 2 CaCl2 + 2 H2O


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Neutralization Calculations Recall: n = v x c

Where: n represents the number of moles

in mol

v represents volume in dm3

c represents concentration in mol/dm3 or mol dm-3


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Neutralization Problems If an acid and a base combine in a 1 to 1 ratio, the

moles of acid will equal the moles of base• nacid = nbase

Therefore the volume of the acid multiplied by the concentration of the acid is equal to the volume of the base multiplied by the concentration of the base

Vacid C acid = V base C base

If any three of the variables are known it is possible to determine the fourth


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Neutralization Problems Example 1: Hydrochloric acid reacts with potassium

hydroxide according to the following reaction: HCl + KOH KCl + H2O

If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of KOH solution, what is the concentration of the KOH solution?

Solution: Vacid Cacid = Vbase Cbase

(15.00 cm3 )(0.500 M) = (24.00 cm3 ) Cbase

Cbase = (15.00 cm3 )(0.500 M) (24.00 cm3 ) Cbase = 0.313 M


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Neutralization Problems Whenever an acid and a base do not combine in a 1 to 1 ratio, a

mole factor must be added to the neutralization equation n Vacid C acid = V base C base The mole factor (n) is the number of times the moles the acid

side of the above equation must be multiplied so as to equal the base side. (or vice versa)

ExampleH2SO4 + 2 NaOH Na2SO4 + 2 H2O

The mole factor is 2 and goes on the acid side of the equation. The number of moles of H2SO4 is one half that of NaOH. Therefore the moles of H2SO4 are multiplied by 2 to equal the moles of NaOH.


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Neutralization Problems Example 2: Sulfuric acid reacts with sodium hydroxide

according to the following reaction: H2SO4 + 2 NaOH Na2SO4 + 2 H2O

If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of NaOH solution, what is the concentration of the NaOH solution?

Solution: In this case the mole factor is 2 and it goes on the acid side,

since the mole ratio of acid to base is 1 to 2. Therefore 2 Vacid Cacid = Vbase Cbase

2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase

Cbase = (2) (20.00 cm3 )(0.400 M) (32.00 cm3 ) Cbase = 0.500 M


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Neutralization Problems Example 3: Phosphoric acid reacts with potassium hydroxide

according to the following reaction: H3PO4 + 3 KOH K3PO4 + 3 H2O

If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of H3PO4 solution, what is the concentration of the H3PO4 solution?

Solution: In this case the mole factor is 3 and it goes on the acid side,

since the mole ratio of acid to base is 1 to 3. Therefore 3 Vacid Cacid = Vbase Cbase

3 (15.00 cm3 )(Cacid) = (30.00 cm3 ) (0.300 M) Cacid = (30.00 cm3 )(0.300 M) (3) (15.00 cm3 ) Cacid = 0.200 M


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Neutralization Problems Example 4: Hydrochloric acid reacts with calcium hydroxide

according to the following reaction: 2 HCl + Ca(OH)2 CaCl2 + 2 H2O

If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of Ca(OH)2 solution, what is the concentration of the Ca(OH)2 solution?

Solution: In this case the mole factor is 2 and it goes on the base side,

since the mole ratio of acid to base is 2 to 1. Therefore Vacid Cacid = 2 Vbase Cbase

(25.00 cm3) (0.400) = (2) (20.00 cm3) (Cbase) Cbase = (25.00 cm3 ) (0.400 M) (2) (20.00 cm3 ) Cbase = 0.250 M


acids & bases strong & weak · 27 neutralization problems if an acid and a base combine in...

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