Question

Calculate the equilibrium concentration of hc2o4− in a 0.20 m solution of oxalic acid.

Answer 1

To calculate the equilibrium concentration of Hc2o4− in a 0.20 M solution of oxalic acid, we need to know the ionization constants (Ka) of oxalic acid. The given pKa values are pKa1 = 1.25 and pKa2 = 3.81. By using the ionization reactions and equilibrium expressions, we can calculate the equilibrium concentration of HC2O4− to be 0.089 M.

Step-by-step explanation:

To calculate the equilibrium concentration of Hc2o4− in a 0.20 M solution of oxalic acid, we need to know the ionization constants (Ka) of oxalic acid. The given pKa values are pKa1 = 1.25 and pKa2 = 3.81. Oxalic acid, H2C2O4, is a diprotic acid that undergoes two successive ionizations:

1. H2C2O4 → H+ + HC2O4−
2. HC2O4− → H+ + C2O42−

We can use these ionization reactions and the equilibrium expressions to calculate the equilibrium concentration of HC2O4−. Let x be the equilibrium concentration of HC2O4−. For the first ionization:

Ka1 = [H+][HC2O4−] / [H2C2O4]. Solving for x gives:

x = sqrt(Ka1[H2C2O4]) = sqrt(10^-1.25 * 0.20) = 0.089 M

Therefore, the equilibrium concentration of HC2O4− in a 0.20 M solution of oxalic acid is 0.089 M.

Answer 2

For the first dissociation of H2C2O4 ↔ H+ + HC2O4-, pKa = 1.23
Ka = 10^-1.23
Ka = 0.0589

Ka = [H+] [ HC2O4-] / [H2C2O4]
Call [H+] = X which will also = [HC2O4-] and that [H2C2O4] = (0.2-X )
So we can write:
0.0589 = X² / (0.2-X)
X² = 0.0589( 0.2 - X)
X² = 0.01178 - 0.0589X
X² + 0.0589X - 0.01178 = 0
Solve quadratic for X: Positive root = 0.083

The molar concentration of HC2O4- at equilibrium = 0.083M

This is from yahoo. Credit to Trevor H