1. Calculate the pH a) Calculate the pH of a solution that contains 0.28 M oxali

Question

1. Calculate the pH a) Calculate the pH of a solution that contains 0.28 M oxalic acid. Calculate the concentration of the oxalate ion in this solution.

b) Calculate the pH of a 0.026 M solution of dimethylamine.
1. Calculate the pH a) Calculate the pH of a solution that contains 0.28 M oxalic acid. Calculate the concentration of the oxalate ion in this solution.

b) Calculate the pH of a 0.026 M solution of dimethylamine.
1. Calculate the pH a) Calculate the pH of a solution that contains 0.28 M oxalic acid. Calculate the concentration of the oxalate ion in this solution.

b) Calculate the pH of a 0.026 M solution of dimethylamine.
1. Calculate the pH a) Calculate the pH of a solution that contains 0.28 M oxalic acid. Calculate the concentration of the oxalate ion in this solution.

b) Calculate the pH of a 0.026 M solution of dimethylamine.

Explanation / Answer

Ans---- b) we know  dimethylamine (CH3)2NH is a weak base and its Kb = 5.4*10^-2 (theoritical value)

  dimethylamine (CH3)2NH dissociates as

  (CH3)2NH+ H2O <====> (CH3)2NH2+ + OH- using ICE table we get

I 0.026 0 0

C -X +X +X

E (0.026-x) x x

therefore kb for the reaction is

kb = [CH3)2NH2+ ] [OH- ]/[CH3)2NH]

5.4*10^-2= x^2/(0.026 -x)

5.4*10^-2= x^2/0.026 ( as the value of kb is very small , so x in denominator is neglected)   

=> 0.140 *10^-4 =x^2

=> x=0.003742

i,e [OH-] =0.0037

so, pOH=- log[OH-]

=- log (0.0037 )

=2.43

hence  pH = 14 – pOH

=14-2.43

=11.57

a) oxalic acid.aid is a diprotic acid and its ka1=5.9 x 10^-2 and ka2=6.4 x 10^-5

oxalic acid first dissociates as

  H2C2O4 <-----> HC2O4- + H+

Initially 0.28 0 0

at equilibrium 0.28 -x x x

so , ka1 =x^2/0.28-x

  5.9 x 10^-2=x^2 / 0.28-x

=>  5.9 x 10^-2(0.28-x) =x^2

=>0.0165 - 0.059x =x^2

solving we get x= 0.102M i,e [HC2O4-]= [ H+]=0.102M

the second dissociation is

HC2O4- <------> H+ + C2O42-  

At equilibrium , [H+]= x + 0.102
[C2O42-]= x
[HC2O4-]= 0.102-x

so, ka2 =  [H+] [C2O42-] /  [HC2O4-]

6.4 x 10^-5 = (x+0.102)( x) /( 0.102-x ) [ x in denominator is neglected]

  6.4 x 10^-5 = (x+0.102)( x) /( 0.102

=>6.5 *10^-6= x^2 +0.102x

solving we get x=0.0000638 i,e [H+]= [C2O42-]= [HC2O4-] = 0.000063 M

pH = - log[H+]

=- log (0.000063)

=4.2

  

  
  

  

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