If maleic acid could be represented as H2Ma, write the 2 net ionic equations for its reaction with NaOH in the space below (showing the sequential neutralization of each acidic proton):

1)H2Ma +NaOH---->NaMa+H3O
2)

What's the second one?

H2Ma+NaOH -> NaMa + H2O + H{+}

H2Ma+NaOH -> NaMa + H2 + OH{-}

2) NaMa + NaOH -> Na2Ma + H2O

And hey, did you hear about the scientist who tried to make a pun about sodium? It was sodium funny, they couldn't get a reaction!

The second net ionic equation for the sequential neutralization of maleic acid (H2Ma) with sodium hydroxide (NaOH) is as follows:

HMa^- + OH^- → Ma^2- + H2O

In this reaction, one acidic proton from maleic acid reacts with one hydroxide ion to form the maleate ion and water.

To determine the second net ionic equation for the reaction of maleic acid (H2Ma) with sodium hydroxide (NaOH), we need to consider the fact that maleic acid has two acidic protons that can be neutralized sequentially.

1) The first acidic proton of maleic acid will react with NaOH to form the sodium salt of maleic acid (NaMa) and water (H2O). Here is the net ionic equation for the neutralization of the first acidic proton:

H2Ma + NaOH -> NaMa + H2O

2) After the first acidic proton is neutralized, the remaining maleic acid molecule (HMa-) still has one proton available for neutralization. When NaOH reacts with HMa-, it forms the second sodium salt of maleic acid (NaMa-) and water (H2O). Here is the net ionic equation for the neutralization of the second acidic proton:

HMa- + NaOH -> NaMa- + H2O

By neutralizing each acidic proton, maleic acid is converted into its corresponding sodium salts, NaMa and NaMa-, through a sequential series of reactions with NaOH.