Acids & BasesLesson 9

pH, pOH, Kb for Weak Bases

Weak Bases calculationsHelpful Hints:

-Weak bases do NOT ionize 100%

-There is an equilibrium state

-NEED….ICE tables!

-Kb can’t be used directly from table…use- Kb = kw/Ka

You can calculate the Kb using the Ka from the table and the equation below. 

Ka x Kb = 1.0 x 10-14 @ 25 oC  

Kb = KwKa(conjugate)

 or

 Ka = Kw

Kb(conjugate)

Weak Bases calculations

Two types of questions

-calculate pOH, or [OH-]

-calculate Kb

1. Calculate the pH of 0.50 M NH3.

Weak Base Need Kb(NH3) = KwKa(NH4

+)= 1 x 10-14

5.6 x 10-10

= 1.786 x 10-5

NH3 + H2O ⇄ NH4+ + OH-

I 0.50 M 0 0C -x x xE 0.50 - x x x

Small Kb0

Kb = [NH4+][OH-]

[NH3]

= x2 = 1.786 x 10-5

0.50

x = 0.002988 M = [OH-]

pOH = 2.52 pOH = -Log[OH-]pH + pOH = pKw = 14.000

pH = 11.48

2. Calculate the pH of 0.20 M Na2CO3.

Weak Base Need Kb(CO32-) = Kw

Ka(HCO3-)

= 1 x 10-14

5.6 x 10-11

= 1.786 x 10-4

CO32- + H2O ⇄ HCO3

- + OH-

I 0.20 M 0 0C -x x xE 0.20 - x x x

Small Kb0

Kb = [HCO3- ][OH-]

[CO32-]

= x2 = 1.786 x 10-4

0.20

x = 0.005976 M

pOH = 2.22

pH = 11.78

TRY: calculate [H+], [OH-], pH, pOH for a 0.20 M solution of NH3.

3. Calculate the pH of 0.20 M NaCl.

NaCl is a neutral salt

pH = 7.00

What if asked to calculate Kb?

Kb indicates weak bases, then use ICE tables.

need to know [ ]’s to plug in the equation.

EX: If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

pH = 11.427 pOH = 2.573

[OH-] = 10-2.573 [OH-] = 0.002673 M at equilibrium line!!!

[NH4+][OH-]

[NH3]

0.002673 0.0026730.002673 0.002673

0.40 0 0- 0.0026730.3973

IC E

(0.002673)2

0.3973= 1.8 x 10-5

Kb = =

TOGETHER: The pOH of a 0.50 M solution of the weak acid HA is 10.64. what is Kb for A-?

TRY: A 0.600 M solution of the weak base hydroxylamine, NH2OH, has a pH of 9.904. What is Ka for NH3OH+ ?

Next question

• Strong acid and a strong base

• Calculate excess

• No ICE tables

4. 15.0 mL of 0.20 M HNO3 reacts with 40.0 mL of

0.20 M KOH, calculate the pH of the resulting solution.

Next question

• It says saturated solution = equilibrium• This is a solubility equilibrium- no ICE• Remember, unit 3!

4. Calculate the pH of a saturated solution of Mg(OH)2.

This is a solubility equilibrium- no ICE

Mg(OH)2(s) ⇄ Mg2+ + 2OH-

s s 2s

Ksp = [Mg2+]][OH-]2 = 5.6 x 10-12

[s][2s]2= 5.6 x 10-12

4s3 = 5.6 x 10-12

s = 1.119 x 10-4 M

2s = [OH-] = 2.237 x 10-4 M

pOH = 3.65

pH = 10.35

Be carefulAre they strong or weak acids and bases?

calculations differ! Use ICE tables for only weak combinations.

Is it a Ksp question?

Always understand the question BEFORE answering it.

Remember ALL the equations!

pH EquationsYou must know the following equations, which are all based on the ionization of water at 250 C! H2O ⇄ H+ + OH-

 Kw = [H+][ OH-] = 1.00 x 10-14   pH = -Log[H+] Or pOH = -Log[OH-] [H+] = 10-pH [OH-] = 10-pOH

 pH + pOH = pKw = 14.000

Homework

p.153

84, 86, 87, 88,

p.154

90, 91, 93.