let Ascorbic acid be H2ASc
H2ASc-<--->H+ + HASc-
Ka1= [H+] [HASC-]/ [H2ASC]
molar mass of acid (C6H8O6)= 6*12+8+6*16= 176
given [H2ASc] = 4.4*10-3/176*10-3 moles/L= 0.025M
                                      Â
H2ASc                     Â
H+Â Â Â Â HASC-
Initial                                Â
0.025Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â
0
change                             Â
-x                          Â
x          x
equilibrium                      Â
0.025-x                  Â
x          Â
x
Ka1= x2/(0.025-x)= 6.8*10-5
when solved uisng solver ,x = 0.00127 = [H+]
HASC- =0.00127
The ionization of HASC- is HASC- ----------> H+ + ASC-2
                                           Â
HASC-Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
H+Â Â Â Â Â Â ASC-2
Initial                                Â
0.00127
                    Â
0Â Â Â Â Â Â Â Â Â Â Â
0
change                             Â
-x                        Â
x          x
equilibrium                      Â
0.00127-x                  Â
x          Â
x
x2/ (0.0127-x)= 2.7*10-12
when solved using excel, x= 5.9*10-8 =[H+], total [H+] =
5.9*10-8 +0.00217 =0.001270059
pH= -log[H+]= 2.896
