If maleic acid could be represented as H2Ma, write the 2 equations for its reaction with NaOH in the space below (showing the sequential neutralization of each acidic proton):
1) H2Ma + NaOH --> H3O + MaNa (molecular)
2) H2Ma + OH- --> H2O + MaH- (net ionic)
Note: the "-" is the superscript.
1) H2Ma + NaOH --> H2O + NaHMa
2) NaHMa + NaOH --> H2O + Na2Ma
I think is what they are after so you end up with the disodium salt.
1) The molecular equation for the reaction between maleic acid (H2Ma) and sodium hydroxide (NaOH) is:
H2Ma + NaOH → H3O + MaNa
2) The net ionic equation for the reaction is:
H2Ma + OH⁻ → H2O + MaH⁻
To write the equations for the reaction between maleic acid (H2Ma) and sodium hydroxide (NaOH), we need to understand the neutralization process and the properties of the compounds involved.
Neutralization is a chemical reaction between an acid and a base, resulting in the formation of a salt and water. Maleic acid (H2Ma) is a dicarboxylic acid with two acidic protons (H+), while sodium hydroxide (NaOH) is a strong base that dissociates to produce hydroxide ions (OH-).
Here's how we can write the equations for the reaction:
1) Molecular Equation:
H2Ma + 2NaOH → Na2Ma + 2H2O
In this equation, two moles of sodium hydroxide neutralize both acidic protons of maleic acid to form sodium maleate (Na2Ma), a salt, and two moles of water.
2) Net Ionic Equation:
H2Ma + 2OH- → 2H2O + Ma2-
This equation represents the reaction in terms of the ions involved in the reaction. Both the hydrogen ions (H+) and the sodium ions (Na+) are spectator ions, meaning they do not participate in the actual chemical reaction. Therefore, they are omitted from the equation. Maleic acid (H2Ma) reacts with hydroxide ions (OH-) to form water and the maleate ion (Ma2-).
Remember that the 2+ and - signs in the equations represent the charges of the ions. The superscripts indicate the number of atoms or ions present in the formula.