Chemistry. Need to understand this.?

Question

Guys, for all the smart chemists out there, if they can please explain this in their own words, rather than words from book because I don't understand all these big chemistry words. Since you guys don't know the question asked and the big 1 page solution it would be hard for you guys to explain, but please try. Here is the Question.

Due to the addition of the acetate ion from the full dissociation of sodium acetate the dissociation of the weak acid(CH3COOH) is suppressed equilibrium is shifted to the left following Le Chatlier's principle, therefore the amount of CH3COOH that ionizes is negligible.

Since the ionization of the weak acid is negligible the full dissociation of sodium acetate is the main source of acetate ion in solution.

**WHAT IN THE WORLD DOES THAT MEAN?? Simpler words would help, best answer(VOTE) will be given out to best answers. Please help me.

Answer 1

let HOAc = acetic acid

let NaOAc = sodium acetate

HOAc <--> H^+1  +  OAc^-1

This is an equilibrium given by the constant, Ka

Ka = [H+][OAc-]/[HOAc]

Since Ka is constant. if you increase [OAc-] then [H+] must decrease (which will cause [HOAc] to increase). That is the only way Ka can remain constant.

Look at the equilibrium eequation. An increase in OAc^-1 will force the reaction to the left (more HOAc).

HOAc is very weakly ionized, so if you decrease the ionization even further (less H+ and more HOAc), the amount of AcO^-1 from the ionization becomes very small compared to the added AcO^-1 from NaOAc.

Answer 2

The equilibrium that they're talking about is this:

CH3COOH  =  CH3COO(-)  +  H(+)

The equilibrium constant for this reaction is very small, which means that if you have some acetic acid in solution, only a very tiny fraction of it is dissociated (broken apart) into the two ions shown on the right hand side of the equation.

In addition, they said that sodium acetate was added.  That looks like this:

CH3COONa =  Na(+) + CH3COO(-)

In this case, the equilibrium constant is very large, which means all of the sodium acetate is actually present in the form of the two ions on the right (the dissociated form).  

Notice that in both reactions you have CH3COO(-) on the right.  The large amount of CH3COO(-) that is formed in the second reaction actually creates kind of a "back pressure" in the first reaction, pushing it back toward the CH3COOH reactant.  That is why the amount of CH3COO(-) that is formed from the first reaction is considered negligible.  

This is an example of a principle known as the "common ion effect."  Two reactions are trying to create the same ion ... but the ion created in one reaction frustrates the other reaction from creating the same ion.  

I hope that helps. That's about the simplest language I can put that in.

Answer 3

Suppose a problem in which a 0.1 molar solution of acetic acid is also 0.1 molar in sodium acetate. What is the pH?

Let acetic acid be abbreviated HOAc and sodium acetate NaOAc

HOAc <===> H+ + OAc+

Ka = [H+][OAc-]/[HOAc] = 1.85 x 10^-5

pH = -Log[H+], so we seek the [H+]



Let x = [H+]. Then [HOAc] = (0.1 - x) and [OAc-] = (0.1 + x). Now here is the operative phrase! Compared with 0.1, x is very small. (You'll see.) So [HOAc] = 0.1 and [OAc-] = 0.1.

Ka = (x)(0.1)/(0.1) = 1.85 x 10^-5

x = 1.85 x 10^-5 M = [H+]

Log(1.85x10^-5) = -4.7

pH = +4.7

So you see that [H+] of 1.85 x 10^-5 is indeed minute compared with 0.1. That is what is meant by negligible.

Answer 4

I agree with "Fly on the wall".

CH3COOH is a weak acid, so it does not dissociate completely in water. The dissociation is shown as:

CH3COOH <--------> CH3COO(-)   +  H(+)  

(The 2 headed arrows in the center means that the reaction is in the equlibrium(the state of rest)... that means the reactants and the products like to remain in 50-50 state. Also remember that the equilibrium reactions are reversible to some extent and can go in any direction.

Let's say for simplicity that :

EQUILIBRIUM<----->means 50% reactants and 50% Products

If you add something on the "reactant" side, then lets say, we have 90% reactants and still 50% Product.

The equilibrium arrow is not going to like it, so addional reactant will react and make more "Product"......by moving the arrow towards right side(we can say that the equlibrium has moved to forward or to right hand side), giving you 70% Reactant and 70% Product.

If you add some common compound  to the product side then , then the excess Product will go towards reversible path giving you more Reactant. We say that the reaction has moved to left hand side. I don't want to re-write the whole reactions again as it is nicely explain by Mr. Fly on the wall.

Answer 5

On their own, CH3COOH ionises a bit, and CH3COONa ionises a lot.  In a mixture of the two, the CH3COO- ion provided by the CH3COONa stops the CH3COOH ionising.

Answer 6

we know the basic neutralisation reaction is:

acid+base ->salt+water.

here CH3COOH is the acid ,NaOH the base,the products are CH3COONa and H2O.

according to Le Chatelier's principle,which is similar in nature to the conservation of mass,a given amount of reactants will produce only the same amount of products.if the amount is disturbed on either side,the reaction will shift in that direction in which the mass is less so that the reaction comes back to it's normal rate and so that the same amount of product is produced.

here over time,Ch3COOH reacts with NaOH to produce a given amount of CH3COONa and remaining amount of water.

if the amount of product increases by a certain amount,the same amount of reactant has to decrease.

here NaOH is a strong base and Ch3COOH is a weak acid

Therefore the strong base will force greater dissociation of the weak acid and becasue the basic strrngth of the reactants is more than their acidic strength, the final product CH3COONa is also more basic than acidic

Since the srtong base and the weak acid combination causes the strong base to react with the weak acid more quickly and consequently the reaction time decreases if we were to increase the concentration of acetate ions the strength of CH3COOH which is present on the reactants side would go up.

As a result if we were to make the acidic strength of the reactants equal to the basic strength of the reactants, the end result would be a neutralisation reaction resulting in the formation of a salt and water and not the formation of a basic product and water as we have seen in the above reaction

The reaction would then stop and the only available concentration  CH3COONa would be from that avalable due to the full dissociation of CH3COONa as :

CHCOONa --> CH3COO- + Na+

This implies that the dissociation of Ch3COOH could be suppressed by adding more acetate ions and making it as equally as strong as the reactant base

This is what Le Chatlier's principle seeks to explain.