Acids & Bases

Theories of Acids and Bases:
Arrhenius Theory
 Examples:                                                                
Acid - will produce a H + ion in water            
Base - will produce a OH- ion in water		 HBr


KOH                                                          
Bronsted-Lowry Theory
Acid - a proton donor
Base - a proton acceptor		 HCl
NH3
Lewis Theory
Acid - e- pair acceptor
Base - e- pair donor
 BF3
Xe

pH Scale
pH

Formulas to Solve pH Problems
1 x 10-14 = [H+ ][OH-]  
pH = -log[H +]
[H+ ] = antilog-pH
pH + pOH = 14    

Find the [OH -]:

  1)  [H+] = 1 x 10 -3  
   2)  [H+] = 6.22 x 10 -9  
   3)  [H+] = 7.5 x 10 -5  
Find the [H +]:

  4)  [OH-] = 1 x 10-6
   5)  [OH-] = 2.8 x 10 -11  
   6)  [OH-] = 4.95 x 10 -2  
Find the pH:

  7)  [H+] = 1 x 10 -13  
   8)  [H+] = 9.25 x 10 -2  
   9)  [H+] = 5.045 x 10 -4  
Find the [H+]:

10)  pH = 6
 11)  pH = 2.35
 12)  pH = 12.62
Find the [OH-]:

13)  pH = 11
 14)   pH = 3.47
 15)   pOH = 5.4
Find the pOH:

16)  pH = 2
 17)  [H+] = 2.60 x 10 -9
 18)  [OH-] = 8.12 x 10 -12



Answers:

  1)  [OH -] = 1 x 10-14/1 x 10-3 =  1 x 10 -11
   2)  [OH -] = 1 x 10 -14 /6.22 x 10-9 = 1.61 x 10-6
   3)  [OH -] = 1 x 10 -14 /7.5 x 10-5 = 1.33 x 10-10
  4)  [H +] = 1 x 10 -14 /1 x 10-6 = 1 x 10-8
   5)  [H +] = 1 x 10 -14 /2.8 x 10-11 = 3.6 x 10-4
   6)  [H +] = 1 x 10 -14 /4.95 x 10-2 = 2.02 x 10-13
  7)  pH = -log 1 x 10 -13 = 13
   8)  pH = -log 9.25 x 10 -2 = 1.03
   9)  pH = -log 5.045 x 10 -4 = 3.297
10)  [H +] = antilog -6 = 1 x 10-6
 11)  [H +] = antilog -2.35 = 4.47 x 10 -3 12)  [H +] = antilog -12.62 = 2.399 x 10 -13
13)  pOH = 14 - 11 = 3
        [OH -] = antilog -3 = 1 x 10-3
 14)  pOH = 14 - 3.47 = 10.53
        [OH -] = antilog -10.53 = 2.95 x 10-11
 15)  [OH - ] = antilog -5.38 = 3.9 x 10 -6
16)  pOH = 14 - 2 = 12
 17)  pH = -log 2.60 x 10 -9 = 8.59
       pOH = 14 - 8.59 = 5.41
 18)  pOH = -log 8.12 x 10 -12 = 11.1

Titrations
M1V1 = M 2V2
Solve:
35. 8 ml of 2.70 M HCl is titrated to its endpoint with 22.1 ml of NaOH. Determine the [NaOH].
M1 V 1 = M2 V 2
2.70 M(35 . 8 ml) = M2(22 . 1 ml)
M2 = 4 . 37 M