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Acids & Bases Lesson 9 pH, pOH, Kb for Weak Bases

Weak Bases calculations

Helpful 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 o C

Kb = Kw Ka(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 NH 3 .

Weak Base Need Kb( NH 3 ) = = + H 2 O =

I NH 3 0.50 M C -x E 0.50 - x 0 Small Kb Kw Ka( NH 4 + ) 1 x 10 -14 5.6 x 10 -10 1.786 x 10 -5 NH 4 + 0 x x + OH 0 x x

Kb = = [

NH 4 +

][

OH -

]

[ NH 3 ] x 2 0.50

x = 0.002988 M pOH = 2.52

pH + pOH = pKw = 14.000

pH = 11.48

= 1.786 x 10 -5 = [OH ] pOH = -Log[OH ]

2.

I C E Calculate the pH of 0.20 M Na 2 CO 3 . Weak Base Need Kb( CO 3 2 ) = = CO 3 2 0.20 M + H 2 O -x 0.20 - x 0 Small Kb

HCO 3 0 x x = Kw Ka( HCO 3 ) 1 x 10 -14 5.6 x 10 -11 1.786 x 10 -4 + OH 0 x x

Kb = = [

HCO 3 [ CO 3 2 ] ][ OH -

]

x 2 0.20

= 1.786 x 10 -4 x = pOH = pH = 0.005976 M 2.22

11.78

TRY : calculate [H + ], [OH ], pH, pOH for a 0.20 M solution of NH 3 .

3.

Calculate the pH of 0.20 M NaCl.

NaCl is a neutral salt pH = 7.00

What if asked to calculate K

b

?

K

b

indicates weak bases, then use ICE tables.

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

EX:

If the pH of

0.40 M NH 3

@ 25 o C is

11.427

, calculate the

Kb

.

pH = 11.427

[OH ] = 10 -2.573

pOH =

[OH ]

2.573

= 0.002673 M at equilibrium line!!!

I C E NH 3 0.40 0.3973

+ H 0.002673

2 O

NH 4 + 0 0.002673

0.002673

+ OH 0 0.002673

0.002673

Kb = [NH 4 + ][OH ] [NH 3 ]

= (0.002673) 2 0.3973

= 1.8 x 10 -5

TOGETHER:

The pOH of a 0.50 M solution of the weak acid HA is 10.64. what is K b for A ?

TRY

: A 0.600 M solution of the weak base hydroxylamine, NH 2 OH, has a pH of 9.904. What is K a for NH 3 OH + ?

Next question

• • •

Strong acid and a strong base Calculate excess No ICE tables

4.

15.0 mL of 0.20 M HNO 3 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) s 2(s)

Mg s 2+ + 2OH 2s Ksp = [Mg 2+ ]][OH ] 2 = 5.6 x 10 -12 [s][2s] 2 = 5.6 x 10 -12 4s s 3 = = 5.6 x 10 -12 1.119 x 10 -4 M

2s = [OH ] = pOH = pH = 3.65

10.35

2.237 x 10 -4 M

Be careful

Are 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 Equations You must know the following equations, which are all based on the ionization of water at 25 0 C!

H 2 O

H + + OH Kw = [H + ][ OH ] = 1.00 x 10 -14 pH [H + ] = -Log[H + ] = 10 -pH Or pOH [OH ] = -Log[OH = ] 10 -pOH pH + pOH = pKw = 14.000

p.153

84, 86, 87, 88, p.154

90, 91, 93.

Homework