[NewtoAtoms]

Hello,

I am trying to calculate the pH for a titration experiment after neutralization, and I can't find anywhere the ionization constant (Kb) for C3H7COO- 

pH of 50 mL of KOH added   
         
   # of moles KOH      
         
   X   =          0.10    = 5.0 X 10-3 moles         
   0.050 L      1L            
                     
   # of moles C3H7COOH         
               
   X               =   0.05     = 5.0 x 10-3 moles      
   0.100 L      1L            
         
            C3H7COOH      KOH   ---------->   C3H7COO-(K)      H20
Initiation   5.00 x 10-3      5.00 X 10-3      O      O
Change   5.00 X 10-3      5.00 X 10-3      5.00 X 10-3       O
Final                  O          O              5.00 X 10-3      O
   COMPLETELY NEUTRALIZED            
                     
Molarity C3H7COO-(K)   5.00 x 10-3   = 0.1 M         
                               0.050 L               
            
          C3H7COOH    KOH    ---------->    C3H7COO-     H30+
Initiation   0.00      0.00                         0.10      O
Change   x       x                               (x)               O
Equilibrium   x      x                     0.10 -x      O
                     
Kb   [C3H7COOH] [KOH]                  
   [C3H7COO-]                  
      
WHAT IS Kb =   (x](x]   
                  0.10 - x    
      
Kb    =X2   
Kb   = X2   
Kb = X or OH   
      
pOH =   log (X)      
pOH =      
pH + X = 14      
pH =       

[Borek]

Are you given pKa?

And what is C3H7COOH? Butyric acid? Isobutyric acid?

http://www.titrations.info/acid-base-titration-equivalence-point-calculation

[NewtoAtoms]

C3H7COOH is butyric acid and has a Ka=1.54 x 10-5.

The full question is:

Construct a pH titration curve for 100 mL of 0.050 M butryic acid, C3H7COOH (for which Ka= 1.54 x 10-5) against added 0.10 M KOH solution.  Show all the calculations for the individual points and draw the curve on proper graph paper.

Now I have calculated all my points however when it comes to the neutralization point, at which 50 mL KOH is added, I have to calculate Kb according to my question set up: (found below)

pH of 50 mL of KOH added   
            
      # of moles KOH      
            
       X / 0.050 L    =   0.10 / 1L =   5.0 X 10-3 moles         
                  
                        
      # of moles C3H7COOH               
                        
      X / 0.100 L        =   0.05/ 1L =   5.0 x 10-3 moles      
      
            
               C3H7COOH      KOH   ---------->   C3H7COO-(K)      H20
   Initiation   5.00 x 10-3      5.00 X 10-3      O                      O
   Change   5.00 X 10-3      5.00 X 10-3      5.00 X 10-3               O
   Final               O               O                5.00 X 10-3             O
      COMPLETELY NEUTRALIZED               
                        
   Molarity  C3H7COO-(K)   5.00 x 10-3 / 0.050L   = 0.1 M               
         
                        
              C3H7COOH   KOH    ---------->    C3H7COO-        H30+
   Initiation   0.00      0.00                       0.10                      O
   Change        x            x                         (x)                       O
   Equilibrium   x      x                            0.10 -x              O
                        
   Kb   [C3H7COOH] [KOH] / [C3H7COO-]                  
            
   Kb =   (x][x)   / 0.10 - x       
            
(0.10) Kb    =X2   

I have typed the ion into google a million times but I can't find it's value.

ANY help would be gratefully appreciated!!

New to atoms

[Borek]

Instead of typing it millions times into google it will be enough to follow the link I posted, as it shows how to calculate pKb from known pKa.

You will find this information also here:

http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory

and speaking of titration curve

http://www.titrations.info/acid-base-titration-curve-calculation

[NewtoAtoms]

Thank you Borek,

Therefore according to the website you gave me:

ka = 1.54 x 10^-5
pka = -log(1.54 x 10-5) = 4.81

pKa + pKb = pKw
4.81 + pKb = 14
pkb = 9.19

9.19 = -log (Kb)
Kb = 10^-9.19
kb = 6.46 x 10^-10

Equating this back to my equation then:

6.46 x 10^-10 = x² / 0.10
x² = 6.46 x 10^-11
x (OH-) = 8.04 x 10^-6
pOH = - log (8.04 x 10^-6)
pOH = 5.09

pH + 5.09 = 14
pH = 8.91

THANK YOU BOREK!