12/10/2014
Acids and Bases
Kb
December 2014
Kb
• When using weak bases, the same rules apply as with weak acids, except you are solving for pOH and using [OH‐] instead of pH and [H3O+].
• Smaller Kb = less dissociated base
• Larger Kb = more dissociated and stronger
Kb
• Kb is similar to Ka except b stands for base
• Kb= Base dissociation constant
• The general reaction involving a base can be written as B(aq) + H2O ↔ BH+(aq) + OH–(aq)
• Thus Kb = [BH+] [OH–] / [B]
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Steps/Rules
1.
2.
3.
4
4.
Write balanced chemical equation
Set up an ICE table
Enter any values that are given
For problems that give initial [B] compare it to
For problems that give initial [B] compare it to Kb;
a) If [B]/Kb> 500, x is negligible and can be ignored
b) If [B]/Kb < 500, might need to use quad formula to calculate x.
5. Solve using Kb expression
Kb Example
• Pyridine C5H5N, is used to manufacture medications and vitamins. Calculate the Kb for pyridine if a 0.125mol/L aqueous solution has a pH of 9 10
a pH of 9.10.
Answer
1. C5H5N (aq) + H2O (l) ÅÆC5H5NH+ (aq) + HO‐(aq)
2. pOH=14‐9.10=4.90 ; [OH‐]=10‐4.90=1.3x10‐5mol/L
3.
Set up table:
I
C
E
ÅÆ
C5H5NH+
OH-
0.125
0
~0
-x
+x
+x
0.125-x
x
x
C5 H5 N
H2O
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4. Kb = [BH+][ OH‐] = (x)(x)/(0.125‐x)
[B]
Kb = (1.3x10‐5)2
(0.125)
Kb = 1.4 x 10 ‐9
Example
• The characteristic taste of tonic water is due to the addition of quinine. Quinine is a naturally occurring compound that is also used to treat malaria The base dissociation
used to treat malaria. The base dissociation constant, Kb, for quinine is 3.3 x 10‐6. Calculate the [OH‐] and the pH of a 1.7 x 10‐3 mol/L solution of quinine.
Answer
1. Q (aq) + H2O (l) ÅÆHQ+ (aq) + OH‐(aq)
2. [Q]/Kb= (1.7x10‐3)/(3.3x10‐6)=515 (x negligible)
I
C
E
H2O ÅÆ
HQ+
OH-
1.7x10-3
0
~0
-x
+x
+x
1.7x10-3
x
x
Q
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4. Kb = [HQ+][ OH‐] = (x)(x)/(1.7x10‐3)
[Q]
3.3 x 10‐6 =
(x)2 (1.7x10‐3)
x = 7.5 x 10‐5 M = [OH‐]
5. pOH= ‐ log (7.5 x 10‐5) = 4.13
pH= 9.87
Try it
• Worksheet!!
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