acids & bases lesson 9 ph, poh, kb for weak bases
TRANSCRIPT
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Acids & BasesLesson 9
pH, pOH, Kb for Weak Bases
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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
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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)
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Weak Bases calculations
Two types of questions
-calculate pOH, or [OH-]
-calculate Kb
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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
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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
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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
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Kb = [HCO3- ][OH-]
[CO32-]
= x2 = 1.786 x 10-4
0.20
x = 0.005976 M
pOH = 2.22
pH = 11.78
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TRY: calculate [H+], [OH-], pH, pOH for a 0.20 M solution of NH3.
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3. Calculate the pH of 0.20 M NaCl.
NaCl is a neutral salt
pH = 7.00
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What if asked to calculate Kb?
Kb indicates weak bases, then use ICE tables.
need to know [ ]’s to plug in the equation.
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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 = =
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TOGETHER: The pOH of a 0.50 M solution of the weak acid HA is 10.64. what is Kb for A-?
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TRY: A 0.600 M solution of the weak base hydroxylamine, NH2OH, has a pH of 9.904. What is Ka for NH3OH+ ?
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Next question
• Strong acid and a strong base
• Calculate excess
• No ICE tables
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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.
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Next question
• It says saturated solution = equilibrium• This is a solubility equilibrium- no ICE• Remember, unit 3!
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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
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2s = [OH-] = 2.237 x 10-4 M
pOH = 3.65
pH = 10.35
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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!
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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
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Homework
p.153
84, 86, 87, 88,
p.154
90, 91, 93.