Question

The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (HOOCCOOH) and must be removed before the stems are used to make rhubarb pie. If pKa1 1.23 and pKa2 4.19, what is the pH of a 0.0206 Msolution of oxalic acid?

Answer 1

a be the Let, e be the ce y initial concentration af oxalic acid and dissociation for 1st dissociation. COOH CooH 1 COOH = + + ino initial concentration! equilibrium concentration: c-xe so, dissociation constant for 1st dissociation, Ka - [H+] [ cool TUOH LCOH xe, xe 1- an oxalic acid so, 1 and is an weak in i-& Nb x Ka, = Geiven, pkai = 1123 x so, – log ka, = 1.23 or Kal - 1.23 = 10 and = 0.059 e=0.0206 M in Ka, = are OY ov, May V 00262 0.059 0.0206 = - 169 1.69 so, the equilibrium concentration & [H+ ] = Гtоң 1 1 CỐC - = (1.69 x 0.0206) M = 0.0348 M

Now, for and and x dissociation d be the degree be the initial concentration of of dissociation. LOH me len COOH cool - H+ + co- Coor initial concentration : equilibrium concentration : 22 x(1- xa2 so, dissociation constant for and dissociation, kas – -] .g. xd Kan = [A +] [ bolesti LOH X (1-22) L coo- xxa As it is weak aid so, 2 and 1-2 2 1 kay = xáz Geiven, pkan = 4.19 and so – log Kap- 4.19 -4.19 or, Kaz = 10 = 6.45 X 100 ooh and x = = 0.0348 M initial in Kag = x x er, an - kan = 6.45 x 10-5 6.45 x 0.0348 10-5 - = 0.043 so, here the equilibrium concentration of [14+] a x42 = 0.0348 X0043) = (1.4964 x 10-3) M

Here, the total concentration of H+ = [147] total = (x + x42) M = [ 0.0348 + (1.4964*10-3)] M = 0.036 M Now, pH of this solution - – log [H + total - -log (0.036) - 1.44 Hence, this 0.0206 M solution of oxalic aid will the pH as be 1044