Acids & Bases Lesson 3 Levelling Effect & Ionization of Water (Kw), Ka and Kb intro.

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Acids & Bases Lesson 3 Levelling Effect & zation of Water (Kw), Ka and Kb int

Transcript of Acids & Bases Lesson 3 Levelling Effect & Ionization of Water (Kw), Ka and Kb intro.

Page 1: Acids & Bases Lesson 3 Levelling Effect & Ionization of Water (Kw), Ka and Kb intro.

Acids & Bases

Lesson 3Levelling Effect &

Ionization of Water (Kw), Ka and Kb intro.

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Review of

“Strong & Weak” Acids and Bases

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Strong acids and bases

• They 100% ionize in solution.

• The ionization is NOT at equilibrium - one-way reactions.

-In other words, reverse rxns do not occur at all

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Strong acids and bases

TOP SIX STRONG ACIDSHClO4

HI

HBr

HCl

HNO3

H2SO4

Sulphiric acid is diprotic, however, it is only strong for the first dissociation! In fact, we will always only use the first dissociation and neglect any second ones. The first one will be the most significant.

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Weak acids and bases

• They DON’T 100% ionize in solution.

• The ionization is at equilibrium -two-way reactions.

-In other words, reverse rxns do occur.

• We use Ka and Kb

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Something to think about…

Since we do not have Ka or Kb values for strong acids and bases…How do we get [OH-] or [H+] ???

There is something called “ The Levelling Effect of water”

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The Levelling Effect

Of water

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Consider these aqueous solutions of a strong and a weak acid

1 M HCl produces 1 M H3O+ and 1 M Cl-

whereas,

1 M HF produces 0.97 M HF + 0.03 M H3O+ + 0.03 M F-

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What does this indicate?

There is NONE of the strong acid left in the products, it has ALL ionized.

No un-dissociated HCl.

The concentration of H3O is EXACTLY the same as the starting concentration of HCl.

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Let us look at 2 strong acids now

1 M HCl produces 1 M H3O+ and 1 M Cl-

1 M HClO4 produces 1 M H3O+ and 1 M ClO4 -

They both have exactly the same strengths in water, they have been levelled off by water.

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The Levelling EffectAll strong acids in water are equally strong. Remember that strong means they ionize 100% in water.

In other words, in a solution of a strong acid:• No molecules of the strong acid itself remain; they all

become the hydronium ion and the anion left.

• [strong acid] = [H3O+]

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The Levelling EffectH3O+ is the strongest acid that can exist in

an undissociated form in water solution.

All strong acids ionize 100% to form H3O+.

The only way you can tell which strong acid is “stronger” is to react them in a non-aqueous (not H2O) solvent.

Ex: HClO4 + H2SO4 H3SO4+ + ClO4

-

 It is found that HClO4 donates a proton to H2SO4, not the other way around, so HClO4 is a stronger acid than H2SO4. This is not important in Chemistry 12.

This would not happen in a water solution…in H2O, they would both form H3O+.

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The Levelling EffectThe strongest base which can exist in high concentrations in water

solution is OH-

One of The two stronger bases below will react with water completely to form OH-.

Ex: O2- + H2O OH- + OH-

OR… O2- + H2O 2OH-

In other words, in a solution of a strong base:

• No molecules of the strong base itself remain; they all become the hydroxide ion and the cation left.

• [strong base] = [OH-]

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Ionization of water

Water ionizes to a small degree.

H2O(l) ⇄ H+ + OH-

 2H2O(l)⇄ H3O

+ + OH-

 Kw = [H+][OH-] = 1.0 x 10-14 @ 250 C

[H+] and [OH-] are inversely related as Kw is constant at 250 C

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Ionization of water

H2O(l) ⇄ H+ + OH-

Kw = [H+][OH-] = 1.0 x 10-14

Always assume the temperature is 25 ºC and that it is an aqueous solution to use this equation above.

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1. Calculate the [H+] and [OH-] in 1.0 M HCl(aq).

HCl H+ + Cl-

1.0 M 1.0 M 1.0 M

we know that [H+][OH-] = 1.0 x 10-14

[1.0][OH-] = 1.0 x 10-14

[OH-] = 1.0 x 10-14

Note that in acid solutions, the [H+] is large and the [OH-] is small.

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2. Calculate the [H+] and [OH-] in 1.0 M Ba(OH)2(aq)

 

Ba(OH)2 Ba2+ + 2OH-

1.0 M 1.0 M 2.0 M

[H+][OH-] = 1.0 x 10-14

[H+][2.0] = 1.0 x 10-14

[H+] = 5.0 x 10-15 M

In a basic solution the [OH-] is very large and the [H+] is very small!

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3. Calculate the [H+] and [OH-] in H2O.

H2O ⇄ H+ + OH-

x x

[H+][OH-] = 1.0 x 10-14

x2 = 1.0 x 10-14

[OH-] = [H+] = 1.0 x 10-7 M

Note that in pure water that the [H+] and the [OH-] are equal!

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Summary• A neutral solution is a solution having

[H3O+] = [OH-]

• An acidic solution is a solution having[H3O+] > [OH-]

• A basic solution is a solution having[H3O+] < [OH-]

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KaAcid ionization reaction of a WEAK acid in water.

Ka is called the acid ionization constant

Ex:

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Ka• The greater the value of Ka, the stronger the

acid.

• For a STRONG ACID, there is no Ka since the acid will ionize 100% and the [ ] of the unionized acid will be zero.

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Kb

Base Ionization reaction of a WEAK base in water.

Kb is called the base ionization constant.

Ex:

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Kb

• The greater the value of Kb, the stronger the base.

• We don’t have Kb values in the table, but we have a way to get Kb from the known Ka values.

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Relationship between Ka and KbExperimentally, it is found that the acid ionization equation

NH4+ + H2O ⇄ NH3 + H3O+

Ka= [NH3][H3O+]

[NH4+ ]

Also, the base ionization equation

NH3 + H2O ⇄ NH4+ + OH-

Kb [NH4+][OH-]

[NH3]

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Relationship between Ka and KbBoth equations involved NH3 and NH4

+ which are conjugates! You might suspect there is a relationship. Let us multiply them together and find out…

Conclusion: For a conjugate pair, Ka x Kb = Kw

This equation will always allow you to find Kb from Ka values in the table whenever you want!

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Relationship between Ka and KbFind the Ka value for: H2PO4

- and the Kb HPO4-2

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Relationship between Ka and KbFind the Ka value for: H2PO4

- and the Kb HPO4-2

Kb(HPO42-) = Kw

Ka(H2PO4-)

= 1 x 10-14

6.2 x 10-8

= 1.6 x 10-7

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You can calculate the Kb by:

using the Ka value from the table rearrange the equation below 

Ka x Kb = 1.0 x 10-14 @ 25 oC  FINAL: Kb = Kw

Ka(conjugate) 

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Calculate the Kb for CN-.

Kb(CN-) = KwKa(HCN)

= 1 x 10-14

4.9 x 10-10

= 2.0 x 10-5

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Homework:

Hebden p.125 #24, 25

p. 127 #28,30

p. 128 #31,32,33,34

p. 130 #35(a,c,e), 36, 37