Question

1. Calculate the pH and [C6H5O73-] of a 0.125 M solution of the citric acid (H3C6H5O7), a triprotic acid (Ka1 = 7.4 X 10-4, Ka2 = 1.7 X 10-5, Ka3 = 4.0 X 10-7)

Answer 1

citric acid is C6H5O7H3 it undergoes 1st ionization as

H3C6H5O7+ H2O<----> H2C6H5O7- + H3O+, Ka1= 7.4*10-4

Ka1= [H2C6H5O7-] [H3O+]/[H3C6H5O7]

preparing the ICE Table, x= change in concentration of H3C6H5O7.

component                               initial (M)                              change (M)                        Equilibrium (M)

H3C6H5O7                             0.125                                        -x                                        0.125-x

H2C6H5O7+                             0                                              x                                              x

[H3O+]                                      0                                               x                                              x

hence x²/(0.125-x)= 7.4*10^-4, when solved using excel, x= 0.00925, hence [H3O+] =0.00925 (1)

[H2C6H5O7+] = 0.125-0.00925 = 0.11575, this undergoes second ionization as

H2C6H5O7- + H2O ---------> HC6H5O7-2 + H3O+, Ka2= [H3O+]/[HC6H5O7-2]/ [H2C6H5O7-]= 1.7*10^-5,

preparing the ICE table , x= change in concentration of H2C6H5O7-

component                         initial                           change                           equilibrium

H2C6H5O7-1                    0.00925                            -x                                   0.00925-x

HC6H5O7-2                         0                                      x                                          x

[H3O+]                                 0                                      x                                          x

Ka2= x^2_/(0.00925-x)= 1.7*10^-5, when solved using excel, [H3O+] =    0.00039       (2)

[HC6H5O7-2] = 0.00039

HC6H5O7-2 + H2O--------> C6H5O7-3 + H3O, Ka3= [C6H5O7-3] [H3O+]/[HC6H5O7-2]

preparing the ICE table

component                            initial                         change                Equilibrium

[HC6H5O7-2]                       0.00039                    -x                              0.00039-x

H3O+                                         0                           x                                  x

C6H5O7-3                                 0                           x                                   x

hence x²/(0.00039-x)= 4*10^-7, when solved using excel , [H3O+] = 0.0000123 (3)

[C6H5O7-3] =0.0000123

total of [H3O+]= Eq.1+Eq.2+Eq.3 = 0.00925+0.00039+0.0000123 =0.009652

pH= -log [H3O+]= -log (0.009652)= 2.01

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