Question

This is a chemistry subject so please read before you change to chemical engineering. It is not chemical engineering. When I call and report what's going on, I will be refunded 4 questions and I will keep asking the same question. *all I ask if you don't want to do it then please just ignore it and let someone else do it* thanks

PLEASE SHOW WORK AND CALCULATIONS

I dissolved 3.010 g of maleic acid with about 5 ml of water in a 100 ml beaker. I added about 10 ml of 12M HCL, and then I placed the small beaker inside a 250 ml beaker containing about 75 ml of water to use as a hot water bath. I let the pair simmer for about 10 minutes and a precipitate of Fumaric acid formed in the smaller beaker. I cooled the beaker to room temperature and I gravity filtered the solid. I rinsed the solid with about 10-15 ml of water to remove excess HCL, and then I dried the wet product with a paper towel. When I weighed the fumaric acid, it weighed 2.021 g.

1. What would be the %yield of the weighed fumaric acid?

I tested the following properties of each isomer and for solubility so I mixed 0.501g of maleic acid and 0.504 g of fumaric acid and put them in separate test tubes with 5 ml of water and the maleic acid dissolved faster than the Fumaric acid. After this step I dissolved 0.102g of Maleic acid and 0.102g of fumaric acid in separate test tubes with 20 ml of water. I measured the PH which was 1.6 for the maleic acid and 1.9 for the fumaric acid.

2. How can I use the molarity of each isomer in each test tube, the [H^+] from the pH readings and estimate the Ka for each isomer from equation Ka≈[H^+]^2/ [isomer]?

3. Based on the Ka, which acid would be stronger?

4. Assume that the equilibrium between the maleic acid and fumaric acid occurred when the product was filtered Maleic acid <=====> Fumaric acid (K2 is above the equilibrium sign), so the last question is How can I calculate the amount of unreacted maleic acid (in grams) and determine the gram ratio of fumaric acid/maleic acid to estimate K2 at room temperature?(Why is it NOT necessary to use molarities in this calculation?)

Answer 1

Answer 1

As maleic acid and fumaric acid are isomers, so, they have equal molar mass and for complete conversion, whole of maleic acid will get converted to fumaric i.e. amount of fumaric formed should be equal to amount of maleic taken.

Now, as per question, mass of maleic acid taken = 3.010 g

Expected mass of fumaric acid formed = 3.010 g

Mass of fumaric actually formed = 2.021 g

% yield = (2.021/3.010)*100 = 67.14%

Answer 2

Equation to be used, Ka = [H⁺]²/[isomer]

Here, [isomer] is the undissociated isomer

maleic acid

So, for Maleic acid, initial moles = 0.102 / 116 = 0.00087

Initial molarity =(0.00087/20)*1000 = 0.0439 moles/liter

As pH = 1.6, [H⁺] = 10^-1.6 = 0.025 M

So, undissociated maleic acid = 0.0439 - 0.025 = 0.0189

So, Ka = (0.025)²/0.0189 = 0.033

Fumaric acid

Initial concentration = 0.0439 M

pH = 1.9, [H⁺] = 10^-1.9 = 0.0126

So, undissociated acid = 0.0439 - 0.0126 = 0.0313

Ka = (0.0126)²/0.0313 = 0.0051

Answer 3

Based on Ka, maleic acid is stronger than fumaric acid because more is the Ka, more is the acidic strength.

Answer 4

If equilibrium would have been there, mass of unreacted maleic acid = 3.010 - 2.021 = 0.989 g

So, equilibrium constant will be

K2 = [fumaric acid formed]/ [Unreacted maleic acid]

K2 = (2.021)/(0.989) = 2.043

Molarities are not required as both the acids are present in same solution, so these will have same volume of solution.

Also, these are isomers, so molar mass is same, hence on calculating K2, all these will be canceled.