Question

Determine the concentration of H3O+ in a 0.064 M solution of oxalic acid, H2C2O4.

H2C2O4(aq)+H2O(l) ⇌ H3O+(aq)+HC2O−4(aq)    Ka1=0.054

HC2O−4(aq)+H2O(l) ⇌ H3O+(aq)+C2O2−4(aq)    Ka2=5.4×10−5

Answer 1

The first dissociation of oxalic acid is written as

Given the starting concentration of 0.064 M, we can create the following ICE table to calculate the equilibrium concentrations after first dissociation.

Initial , M	0.064	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.064-x	x	x

Now, we can write the expression of K_a1 using the ICE table as follows:

Hence, the equilibrium concentrations after first dissociation are

Now, we can write the second dissociation as

Now, we can create the second dissociation ICE table as follows:

Initial , M	0.0377	0.0377	0
Change, M	-x	+x	+x
Equilibrium, M	0.0377-x	0.0377+x	x

Now, we can write the expression of K_a2 using the ICE table as follows:

Note that x << 0.0377.

Hence, the equilibrium concentration of H₃O⁺ after second dissociation is

Hence, the concentration of H₃O⁺ in the solution is 0.0377 M.