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
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 Ka1 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 Ka2 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.
Determine the concentration of H3O+ in a 0.064 M solution of oxalic acid, H2C2O4.
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