Question

In: Chemistry

1) 20.00 mL of a 0.3000 M lactic acid solution is titrated with 0.1500 M NaOH....

1) 20.00 mL of a 0.3000 M lactic acid solution is titrated with 0.1500 M NaOH.

a. What is the pH of the initial solution (before any base is added)?

b. What is the pH of the solution after 20.00 mL of the base solution has been added?

c. What is the pH of the solution after 40.00 mL of the base solution has been added?

Solutions

Expert Solution

Given                                     

Lactic acid(C3H6O3)                                    

Volume           0.02     L                     

[C3H6O3]       0.3       M                    

Ka =1.40E-04                        

                                               

ICE                                        

            C3H6O3+ H2O -- > C3H5O3- +     H3O+

I           0.3                               0                      0

C         -x                                 +x                    +x

E          (0.3-x)                         x                      x

                                               

Ka expression                                     

            ka=[C3H5O3-][H3O+]/[C3H6O3]                            

            Lets plug in the ka and equilibrium concentration values.                            

            1.40E-4=x^2 / (0.3-x)                        

            x^2 + 1.40 E-4 x - 4.2E-5=0                          

            Lets solve this quadratic equation                             

            x=0.006411                            

                                               

            [H3O+]=x=0.006411                         

                                               

            pH=-log[H3O+]                                 

            pH=-log (0.006411)= 2.193074223              

pH of the solution = 2.20

b)

Addition of base

Calculation of moles of base

Mol of base = Volume of base in L x molarity of base                                 

volume            0.02     L         

Molarity          0.15     M        

mol                0.003     mol     

mol acid          0.006   mol     

Reaction between C3H6O3 and NaOH

            C3H6O3 (aq) + NaOH (aq) ---- > C3H5O3Na (aq) + H2O

I             0.006             0.003                         0                        

C            -0.003            -0.003                      +0.003                 

E            0.003                  0                             0.003

Moles of both acid and salt are same therefore,

[C3H6O3] = [C3H5O3-]   ….(since concentration of C3H5O3Na and C3H5O3- is same)

We know use Henderson equation

pH = - log ka + log ( [ conj. Base ] /   [ weak acid])

here conj base is C3H5O3- and weak acid is C3H6O3

Lets plug in the values to get pH

pH = - log ( 1.40E-4 ) + log ([C3H5O-]/[ C3H6O3])

since

[C3H6O3] = [C3H5O3-] , log ([C3H5O-]/[ C3H6O3]) = log 1 = 0

Therefore pH = - log ( 1.40E-4 )

pH = 3.854

c)

mol base          0.006                                      

mol acid          0.006                                      

This is equivalnce point reaction and there is formation 0.006 mol C3H5O3-                                                         

[C3H5O-] = mol / total volume                                            

                    = 0.006 mol / 0.06 L                  

                    = 0.1 M                                      

Reaction of C3H5O3- with water

            C3H5O3- + H2O -- > C3H6O3 + OH-

I           0.1                                           0          0

C         -x                                             +x        +x

E          (0.1-x)                                     x          x

Kb = x2 / (0.1-x)

Kb = 1.0 E-14 / ka = 1.0 E -14 / 1.40E-4 = 7.14E-11

Since the value of kb is very small we can neglect x in the denominator.

7.14E-11 = x2/ 0.1

x = 2.67 E-6

x = [OH- ] = 2.67 E-6

pOH = -log ( 2.67 E-6) = 5.57

pH = 14 – pOH = 14 -5.57 = 8.4

pH of the solution after addition of 40.00 mL of base = 8.4


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