h2c4h4o4 + 2naoh --> 2h2o + na2c4h4o4

as for the previous question im sorry i forgot to plug the maleic acid in the formula properly. so this ones it ^

.2213 of c4h4o4 used (maleic acid)
and
.1 M NaOH

1. show the equation for 1st and 2nd ionization
2. find molecular mass of maleic acid

help? How do I do this?

I have a list of V (mL) and pH values...
basically titrated 50 mL

of 0.1 M NaOH with 0.2213 M maleic acid

1. First Ionization: H2C4H4O4 + 2NaOH → 2H2O + Na2C4H4O4
Second Ionization: Na2C4H4O4 + 2NaOH → 2NaOH + Na2C4H4O4

2. The molecular mass of maleic acid is 116.07 g/mol.

To answer your questions, let's break it down step-by-step:

1. Show the equation for the first and second ionization of maleic acid (C4H4O4):
Maleic acid (C4H4O4) is a weak diprotic acid, meaning it can ionize twice. Here are the equations for the first and second ionization:

First ionization:
C4H4O4 (maleic acid) + H2O ⇌ C4H3O4- (maleate ion) + H3O+

Second ionization:
C4H3O4- (maleate ion) + H2O ⇌ C4H2O42- (maleate dianion) + H3O+

These equations represent the dissociation of maleic acid and the formation of its conjugate base (maleate ion) at each ionization step.

2. Find the molecular mass of maleic acid (C4H4O4):
To find the molecular mass of maleic acid, you need to calculate the sum of the atomic masses of all the atoms in the compound. Here are the atomic masses of carbon (C), hydrogen (H), and oxygen (O):

Atomic mass of carbon (C) = 12.01 g/mol
Atomic mass of hydrogen (H) = 1.008 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol

Multiply the atomic masses by the respective subscripts in the chemical formula of maleic acid and sum them up:

Molecular mass of C4H4O4 = (4 * atomic mass of C) + (4 * atomic mass of H) + (4 * atomic mass of O)
= (4 * 12.01 g/mol) + (4 * 1.008 g/mol) + (4 * 16.00 g/mol)
= 48.04 g/mol + 4.032 g/mol + 64.00 g/mol
= 116.072 g/mol

Therefore, the molecular mass of maleic acid (C4H4O4) is approximately 116.072 g/mol.

Feel free to ask if you have any more questions!

To answer your questions, let's break it down step by step:

1. Show the equation for the 1st and 2nd ionization:
To find the equations for ionization, we need to understand the acid-base properties of maleic acid (C4H4O4) and how it reacts with NaOH.

First ionization equation:
C4H4O4 + H2O ⇌ C4H3O4- + H3O+

In this equation, maleic acid (C4H4O4) donates a proton (H+) to water (H2O) to form the maleate ion (C4H3O4-) and a hydronium ion (H3O+).

Second ionization equation:
C4H3O4- + H2O ⇌ C4H2O4^2- + H3O+

In this equation, the maleate ion (C4H3O4-) donates another proton (H+) to water (H2O) to form the hydrogen maleate ion (C4H2O4^2-) and a hydronium ion (H3O+).

2. Find the molecular mass of maleic acid (C4H4O4):
To find the molecular mass of maleic acid (C4H4O4), you need to calculate the sum of the atomic masses of all the elements in the molecule.

C4H4O4:
(4 * atomic mass of carbon) + (4 * atomic mass of hydrogen) + (4 * atomic mass of oxygen)

Using the periodic table, you can find the atomic masses:
Carbon (C): 12.01 g/mol
Hydrogen (H): 1.01 g/mol
Oxygen (O): 16.00 g/mol

Calculating the molecular mass:
(4 * 12.01) + (4 * 1.01) + (4 * 16.00) = 64.06 g/mol

Therefore, the molecular mass of maleic acid (C4H4O4) is 64.06 g/mol.

Regarding the list of V (mL) and pH values and the titration of 50 mL, it seems like you have conducted an experiment. Please provide more information and context about the experiment so that I can assist you further.