Answer:
The net ionic equation for the overall reaction is
![H_2C_2O_4{(aq)} + 2OH^-_((aq)) ----> C_2O_4^(2-)_((aq)) + 2H_2O_((l))](https://img.qammunity.org/2020/formulas/chemistry/college/xsobcdamjg05eg4xotlvmpmt8db4mtwntu.png)
Step-by-step explanation:
The molecular formula for sodium hydroxide is
while the molecular formula for oxalic acid is given as (
)
Now the balanced chemical equation for this reaction is
![H_2C_2 O_4_((aq)) + 2NaOH_((aq)) ----> Na_2C_2O_4_((aq)) + 2H_2O_((l))](https://img.qammunity.org/2020/formulas/chemistry/college/u8rs1i9xd8pxxsyndvnej3te6b1r6o37pb.png)
The procedure for the ionic equation is
![H_2 C_2O_4 _((aq)) + 2 Na^+ _((aq)) + 2OH^-_((aq)) ---->2Na^+ _((aq)) + C_2O_4^(2- )_((aq)) + 2H_2O_((l))](https://img.qammunity.org/2020/formulas/chemistry/college/qw8q2gyleol3vddy97jcscsovbj5ggxv4c.png)
Here
is the sodium ion
is the hydroxide ion
is the oxalate ion
Note: The dissociation of oxalic equation is not complete because it is a weak acid
Looking at this equation we see that
is common on both sides of the equation so we cancel it out so the equation becomes
![H_2C_2O_4{(aq)} + 2OH^-_((aq)) ----> C_2O_4^(2-)_((aq)) + 2H_2O_((l))](https://img.qammunity.org/2020/formulas/chemistry/college/xsobcdamjg05eg4xotlvmpmt8db4mtwntu.png)