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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Acidsand

Bases

Weak Bases

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

Acidsand

Bases

Weak Bases

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

where Kb is the base equilibrium constant.

Acidsand

Bases

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.

Acidsand

Bases

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.

Acidsand

Bases

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:

Acidsand

Bases

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

Acidsand

Bases

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

Acidsand

Bases

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

Acidsand

Bases

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 =

Acidsand

Bases

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-]

Acidsand

Bases

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?

Acidsand

Bases

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.

Acidsand

Bases

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!

Acidsand

Bases

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

Acidsand

Bases

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.