Calculate the pH and the equilibrium
concentrations of
HC2O4- and
C2O42- in a
0.0782 M oxalic acid solution,
H2C2O4
(aq).
For H2C2O4,
Ka1 = 5.9×10-2 and
Ka2 = 6.4×10-5
| pH =_______ | |
| [HC2O4-] =___________ | M |
| [C2O42-] =___________ | M |
Oxalic acid ionizes as
HCOOH HCOOH + H2O ----------> HC2O4- + H3O+
Ka1= [HC2O4-][H3O+]/[H2C2O4]=5.9*10-2
preparing the ICE table
componen initial concentration (M) change in concentration(M) Equilibrium concentration(M)
H2C2O4 0.0782 -x 0.0782-x
HC2O4- 0 x x
H3O+ 0 x x
Ka1= x2/0.0782-x)= 0.059, when sovled for x, x= [H3O+]= 0.0446, [HC2O4-]= 0.0446 and H2C2O4= 0.0782-0.0446= 0.0336
[HC2O4-] undergoes further ionization as HC2O4-+ H2O --------->C2O4-+ H3O+
Ka2= [H3O+][C2O4-]/[HC2O4-] = 6.4*10-5
let x= drop in concentration of HC2O4- to reach equilibrium, at Equilibrium,[H3O+]= [C2O4-]=x
preparing the second ICE Table , x2/(0.0446-x)= 6.4*10-5, when solved for x, x= [H3O+]= 0.00166 and [C2O4-]= 0.00166
total of [H3O+]= 0.00166+0.0446=0.04626, pH= -log [H3O+]= 1.33
Calculate the pH and the equilibrium concentrations of HC2O4- and C2O42- in a 0.0782 M oxalic...
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