Question

Ascorbic acid( C6H8O6) is a diprotic acid. (ka1=6.8*10^-5, ka2=2.7*10^-12)

What is the pH of a solution that contains 4.4 mg of acid per milliliter of solution?

Answer 1

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

                                        H2ASc                       H+     HASC-

Initial                                  0.025                       0            0

change                               -x                            x           x

equilibrium                        0.025-x                    x            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

                                             HASC-                  H+       ASC-2

Initial                                  0.00127                       0            0

change                               -x                          x           x

equilibrium                        0.00127-x                    x            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