acids and bases weak bases, kb calculations, and a little short cut

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Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut .

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Acids and Bases Weak Bases Since bases dissociate into different ions, the equilibrium expression is different. where K b is the base equilibrium constant.

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Page 1: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

Acidsand

Bases

Weak Bases, Kb Calculations, and a Little Short Cut

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Page 2: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

Acidsand

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Weak Bases

Arrhenius says that Bases react with water to produce hydroxide ion. What are the other definitions of Bases?

Page 3: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

Acidsand

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Weak Bases

Since bases dissociate into different ions, the equilibrium expression is different.

where Kb is the base equilibrium constant.

Page 4: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

Acidsand

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Weak BasesThere are two types of Weak bases:1)Amines: Remember these?

- Amines are molecules that have at least one nitrogen – hydrogen bond. The Nitrogen atom will donate a lone pair (Lewis Base) and accept a proton (Bronsted-Lowry Base)

2)Anions of Weak Acids: - In solution, the metal cation will be a spectator.

The anion however will be the conjugate base of a weak acid, thus the solution becomes basic.

Page 5: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Examples of Weak Bases:• NaClO + H2O What are the products? • Na+ and ClO-

• The Sodium ion will be spectator ion, but the chlorite ion is the conjugate base of HClO.• The Hydrogen isn’t there to be donated, but the presence of the ClO- ion is! Thus it

behaves like a base!• NaC2H3O2 + H2O What are the products?• Na+ = spectator• C2H3O2

- = conjugate base = BASIC Sol.

Page 6: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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pH of Basic Solutions• What is the pH of a 0.15 M solution of NH3?

• To solve this, you must realize that this is a Weak BASE, and we want the pH (H+ ions)

• Working with a weak base, we must use the KB – the base equilibrium constant.

• Where can we find the KB? Look on our tables:

Page 7: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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KA x KB = Magic!!!• We can use the charts that have

KA values to determine the KB values.

• We again can go back to the autoionization of water and prove this, but this is what you need to know:

KA x KB = 1.0 x 10-14

Page 8: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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KA + KB = Magic!!!

• Looking at the chart, We see the KA for ammonia is 5.8 x 10-10, so what is the Kb?

• KA x KB = 1.0 x 10-14

1.0 x 10-14 = 1.7 x10-5 = KB

5.8 x 10-10

Page 9: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Back to the Original Problem!What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH−]

[NH3]Kb = = 1.7 10-5

[NH3] [NH4+] [OH−]

Initial 0.15 0 0

Equilibrium 0.15 - x x x

Page 10: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Trying Something New

• For many (NOT ALL) of these weak acid/base problems, subtracting “X” from the initial concentration is SOOOO small, that it doesn’t change the calculation very much.

• So small in fact that we can ELIMINATE it from the problem.

• We will test a shortcut here, then come back and make sure it is ok to do so….

(x)2

(0.15-x)1.7 10-5 =

Page 11: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Notice the denominator! X is Gone!(x)(x)(0.15)

1.7 10-5 =

(1.7 10-5) (0.15) = x2

2.55 x10-6 = x2

2.55 x10-6 = x

1.6 x10-3 M = x = [OH-]

Page 12: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Check to see if it’s ok to eliminate X!• We need to make sure that eliminating

X from the initial concentration does not affect the answer very much.

(0.15-x) 0.15 -1.6 x10-3 = .1484

• Notice that the value didn’t change very much, but how much is acceptable?

Page 13: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Rule for Eliminating X• We need to make sure that eliminating

X from the initial concentration did not change the answer very much.

• If X / Initial Concentration is less than 5 %, then OK to eliminate.

• If X / Initial Concentration is greater than 5 %, must go back and use the quadratic equation.

Page 14: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Eliminate X Double Check!

• X = [OH-] = 1.6 x 10-3

• Initial Concentration = 0.15

(1.6 x10-3 / 0.15) x 100 = 1.06 %

1.06 % < 5 % so OK to eliminate!

Page 15: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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Oh Yeah, the Original Question…What is the pH of a 0.15 M solution of NH3?

X = [OH–] = 1.6 10-3 MpOH = –log (1.6 10-3)pOH = 2.80pH = 14.00 – 2.80pH = 11.20

Page 16: Acids and Bases Weak Bases, Kb Calculations, and a Little Short Cut

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A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, Ka, for niacin?

PRACTICE EXERCISES1. Niacin, one of the B vitamins, has the following molecular structure:

2. What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? What percentage of

the bases are ionized?

3. Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Ka for HF is 6.8 x10-4.