Item 1

Maleic acid is a carbon-hydrogen-oxygen compound used in dyeing and finishing fabrics and as a preservative of oils and fats. •In a combustion analysis, a 1.054-g sample of maleic acid yields 1.599 g of CO2 and 0.327 g of H2O.
•When 0.609 g of maleic acid is dissolved in 25.00 g of glacial acetic acid, the freezing point drops by 0.82 ∘C. Maleic acid does not ionize in glacial acetic acid, which has a freezing-point depression constant of Kf=3.90 ∘C/m.
•In a titration experiment, a 0.4250-g sample of maleic acid is dissolved in water and requires 34.03 mL of 0.2152 M KOH for its complete neutralization.
•The pH of 0.215 g of maleic acid in 50.00 mL of aqueous solution is found to be 1.80.

Determine the first ionization constant, Ka1, of maleic acid

This is a workaholic problem.

1. Determine empirical formula.
2. Determine molar mass
3. Determine molecular formula.
4. Determine the molarity of the maleic acid in the titration experiment.
5. Determine Ka1.

And you need help with which one. Show your work for what you have and I can help you through.

I have the molecular formula C4H4O4. and I know that there is 2 ionisable H ions but I don't know where to go from there.

So you have 1,2,3.

4. C4H4O4 which since it is diprotic we can write H2M.
H2M + 2KOH ==> 2H2O + K2M
mols KOH = M x L = ?
mols H2M = 1/2 that from the coefficients
M H2M = mols H2M/L H2M. That give you the M of the solution but I don't understand why this information is needed.
5. M H2M = grams/molar mass = ? which I will call Y, whatever that is. Also convert pH to H^+

.....H2M ==> H^+ + HM^-
I.....Y......0.....0
C.....-x.....x.....x
E......Y-x...x.....x

k1 = (H^+)(HM^-)/(HM)
You know H^+ (from pH) and HM and HM, solve for k1.
Post your work if you get stuck.

I still cant figure this out.

For 4, what is your problem? What have you done and what don't you understand.

For 5, what is your problem? What have you done? Where are you stuck? What do you not understand.?
Still can't figure it out doesn't help me know what's wrong.

To determine the first ionization constant, Ka1, of maleic acid, we need to first understand the concept of ionization and the equation involved.

Maleic acid (H2C4H2O4) can be represented by the following equilibrium equation:

H2C4H2O4 ⇌ H+ + HC4H2O4-

The first ionization constant, Ka1, represents the equilibrium constant for the dissociation of maleic acid into its ions with the release of one hydrogen ion (H+).

To calculate Ka1, we need to use the given data from the titration experiment:

- A 0.4250 g sample of maleic acid is dissolved in water.
- It requires 34.03 mL of 0.2152 M KOH for its complete neutralization.

First, we need to convert the volume of KOH used to moles:

The number of moles of KOH used = volume (L) × concentration (M)
= 0.03403 L × 0.2152 M
≈ 0.00733 moles

Since maleic acid requires one mole of KOH for its neutralization (as per the balanced chemical equation), we can conclude that 0.00733 moles of maleic acid were present in the 0.4250 g sample.

Now, we can calculate the concentration of maleic acid:

Concentration of maleic acid = moles / volume (L)
= 0.00733 moles / 0.05000 L
= 0.1466 M

Now, let's set up the equilibrium expression for the ionization of maleic acid:

Ka1 = [H+] [HC4H2O4-] / [H2C4H2O4]

Since we know that maleic acid does not significantly ionize in water, we can assume that the concentration of maleic acid remaining in solution after ionization is equal to the initial concentration.

Therefore, [H2C4H2O4] = 0.1466 M

We also know that for every mole of maleic acid that ionizes, one mole of hydrogen ions (H+) is produced, so [H+] = 0.00733 M.

Since the concentration of the conjugate base, HC4H2O4-, is negligible compared to the concentration of maleic acid, we can ignore its contribution to the equilibrium expression.

Now, we can substitute the values into the equilibrium expression and solve for Ka1:

Ka1 = (0.00733 M) (0.00733 M) / 0.1466 M
= 3.67 × 10^-4

Therefore, the first ionization constant, Ka1, of maleic acid is approximately 3.67 × 10^-4.