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Please read ahead and don’t fall behind, one big
push at the end will help many.
Hour Exam 3 March 10 or thereabouts.
Make-up Exam 3 will be the last one.
CHEM 11 FINAL EXAM DATE
Date: March 25, 2009
Room: CTC 105
Time: 7:30 - 9:30
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Information on acid-base equilibria, including the reactions yielding salts, the nature of given solutions, and the determination of pH using Kb and initial concentrations. It covers topics such as strong and weak acids and bases, amphoteric compounds, and the calculation of pH for various solutions.
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Stuff
Please read ahead and don’t fall behind, one big
push at the end will help many.
Hour Exam 3 March 10 or thereabouts.
Make-up Exam 3 will be the last one.
CHEM 11 FINAL EXAM DATE
Date: March 25, 2009
Room: CTC 105
Time: 7:30 - 9:
There are 4 acid-base neutralization reactions
yielding salts that can react with water.
**1. Strong acid + Strong base ==> Ionic salt + H 2 O
- Strong acid + Weak base ==> Ionic salt + H 2 O
- Weak acid + Strong base ==> Ionic salt + H 2 O
- Weak acid + Weak base ==> Ionic salt + H 2 O** HCl (aq) + NaOH (aq) ==> Na + + Cl - + H 2 O HBr (aq) + NH 3 (aq) ==> NH 4
- Br
- H 2 O 2HCOOH (aq) + Ca(OH) 2 (aq) ==> Ca
- 2HCOO
- 2H 2 O CH 3 COOH (aq) + NH 3 (aq) ==> NH 4
- CH 3 COO
- H 2 O
Neutral salt
Acidic salt
Basic Salt
Depends on pKa vs pkb
NH
3 ( aq ) + H 2 O ( l ) NH 4
( aq ) + OH
- ( aq ) (^) K b = [NH 4 + ][OH - ] [NH 3 ] **We solve weak base problems just like weak acid ones, except we solve for [OH-] instead of [H
].** _For ammonia, Kb = 1.76 x 10
what is the conjugate acid and what is it’s pKa? How would you write the reaction of NH 4
with water?_
A
+ H
2
O( l ) AH
( aq ) + OH
( aq ) K
b
[AH
][OH
]
[A
]
Perhaps the hardest part for students is writing the base
reaction in water...just think proton donar and acceptor.
Generalized Base Equilibria
Determining pH from K b and initial [B]
Dimethylamine, (CH
3
2
NH, a key intermediate in
detergent manufacture, has a Kb = 5.9 x 10
. What
is the pH of a 1.5 M aqueous solution of (CH 3 ) 2 NH?
Check assumption: [3.0 x 10
- M / 1.5 M ] x 100 = 2% (error is < 5%; thus, assumption is justified)
[H
3
O
+
] = K w /[OH
-
] = 1.0 x 10
-
/3.0 x 10
-
= 3.3 x 10
-
M
pH = -log (3.3 x 10
-
Determining the pH of a solution of A
-
Sodium acetate (CH
3
COONa, abbreviated NaAc) has
applications in photographic development and textile
dyeing. What is the pH of a 0.25 M aqueous solution of
NaAc? The Ka of acetic acid (HAc) (conjugate acid of Ac)
is 1.8 x 10
- 5
PLAN: Use K a to find K b and recall pH + pOH = 14. Write balanced equation knowing that sodium salts are completely soluble in water so [Ac
- ] = 0.25 M. Be organized set up ICE table and fill in the unknowns Write the equilibrium expression--looking for pH.
K b = [HAc][OH
- ] [Ac - ] [Ac - ] = 0.25 M - x ≈ 0.25 M (since K b is small) 5.6 x 10 - 10 ≈ x 2 /0.25 M [OH - ] = x ≈ 1.2 x 10 - 5 M [H 3 O + ] = K w /[OH - ] [H 3 O + ] = K w /[OH - ] = 1.0 x 10 - 14 /1.2 x 10 - 5 = 8.3 x 10 - 10 M pH = -log (8.3 x 10 - 10 M ) = 9. Check assumption: (^) [Ac - ] in / Kb > 100 = .25/ 5.6 x 10 - 10
100