Question

Malic acid (H2C4H4O5) has a Ka1=3.48x10^-4 and Ka2=8.00x10^-7. Consider a 0.02 M aqueous malic acid solution:

A.) What is the pH of the solution (found using an I.C.E table)?
B.) What are the equilibrium concentrations (M) of the HC4H4O5^-1 ion and the C4H4O5^-2 ion in solution?
C.) What is the percent ionization of malic acid in the solution?

Answer 1

Would you know how to solve one of these with a single Ka Value

Answer 2

I will give you the steps to solve this problem

Answer 3

Just lost my entire answer give me a second to see if I can recover it

Answer 4

Here are the Ice tables and Ka1 equations you would have to set up to solve this problem please try to understand how I got all these values

Answer 5

I did the first half of the question for you because it might a little confusing if you are new to this stuff please please please read it over and try to make sense of how I arrived at my conclusion for both ICE tables and how I set up the formulas.

A side note water isnt in either the Ka formula or ICE table because it can pretty much just be assumed to be constant concentration since it is in such a large amount that any reacted is negligible.

1. Essentially you need to sub in the equilibrium values determined from the K1 ICE table into the Ka1 formula, then solve for the roots of that formula by setting the equation equal to 0 (you can use the quadratic formula), solve for x you should get two answers but only the positive one should be used as a negative concentration makes no physical sense.

check out example 2 on this web page to see what I mean
http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Le_Chatelier%27s_Principle/Ice_Tables

2. Once you have solved for x input it into the K2 ICE table then simply do the same thing you did in step 1 to solve for y.

check out example 2 on this web page to see what I mean
http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Le_Chatelier%27s_Principle/Ice_Tables

3. For A) pH = -log(H3O+)

4. For B) you have the ICE table solved for it should be evident what the answer is

5. For C) add up the concentration of both forms of ionized malic acid and divide it by the initial concentration of malic acid (0.2M) then multiply by 100

Answer 6

if you have questions feel free to ask them

Answer 7

Sorry for the delayed response, I had to step away from my computer for a bit. That makes sense, except the part for C. when you said add up the concentrations. which concentrations were you referring?

Answer 8

I made a mistake in the second ICE table

Answer 9

For HC4H4O5- in the second ICE table it should be x-y

Answer 10

You want to divide the ionized molecules of malic acid by the origianl amount of malic acid

Answer 11

to find the percent ionized

Answer 12

Oh okay that makes more sense now thank you :D