# Chemistry Help!

1-The concentration of water,solid substances and precipates should not appear in the equilibrium constant equation Kc , why?I don't understand.Please don't tell me that because their concentration remains constant whatever their quantities. because I have read these words tens of times and I don't really understand them.

2-according to my text book " if the reactants and the products are in the gasoeus state,the concentration is expressed by their partial pressure" What does this mean?

3-CH3COOH <----->CH3COO- + H3O+

let a be the no. of dissociated moles.So if we want to calculate the concentration of CH3COOH why should we say its remaining concentration=C-a,where C is the orginal concentration of CH3COOH before dissociation????.

4-H2O <---->H+ + OH-
Kc= [H+] [OH-]/[H2O]=10^-14
Kw=[H+] [OH-] = 10^-14
how could the two equations have the same value?why we are totally neglecting water?and what is the importance of calculating Kw?

Help me please I'm very confused!

Borek
Mentor
1-The concentration of water,solid substances and precipates should not appear in the equilibrium constant equation Kc , why?I don't understand.Please don't tell me that because their concentration remains constant whatever their quantities. because I have read these words tens of times and I don't really understand them.

Concentration of water is almost constant - it is about 55.5 moles/liter. Even if water reacts with some of the substances present, its concentration rarely changes by more than tenths of percent. We rarely know equilibrium constants with accuracy high enough for this change to make a difference. So in most cases we can safely assume water concentration to be constant - but it is not always true.

Solids are not dissolved, they are reacting only on the surface, it is enough that they are present to take part in the reaction. Concentration of the solid - which is separate from water - doesn't make much sense.

2-according to my text book " if the reactants and the products are in the gasoeus state,the concentration is expressed by their partial pressure" What does this mean?

Not concentration, but activity, in fact each time we talk about reaction quotient we use activities, not concentrations. As a first approximation concentration and activity have the same value.

Do you know what partial pressure is?

3-CH3COOH <----->CH3COO- + H3O+

let a be the no. of dissociated moles.So if we want to calculate the concentration of CH3COOH why should we say its remaining concentration=C-a,where C is the orginal concentration of CH3COOH before dissociation????.

Stoichiometry. Imagine you started with 1 mole of acetic acid and 0.1 moles dissociated - obviously 1-0.1=0.9 moles are left. Now imagine it happened in a known volume of solution - you start with C=1/V, concentration of dissociated acid is a=0.1/V, concentration of the left acid is (1-0.1)/V=C-a.

4-H2O <---->H+ + OH-
Kc= [H+] [OH-]/[H2O]=10^-14
Kw=[H+] [OH-] = 10^-14
how could the two equations have the same value?why we are totally neglecting water?and what is the importance of calculating Kw?

They don't have the same value. See water ion product page.

1-
Concentration of water is almost constant
Yes.but if what is the problem if it is constant? why don't we substitute with the concentration of water as 55.5 moles/liter ?and what is the relation between calculation of Kw and assuming water concentration constant?

2-
Not concentration, but activity, in fact each time we talk about reaction quotient we use activities, not concentrations. As a first approximation concentration and activity have the same value.

Do you know what partial pressure is?
yes i know.Do u mean by "activity" the rate of chemical reaction ?
As a first approximation concentration and activity have the same value.
how?can u give me an example?

3-
Stoichiometry. Imagine you started with 1 mole of acetic acid and 0.1 moles dissociated - obviously 1-0.1=0.9 moles are left. Now imagine it happened in a known volume of solution - you start with C=1/V, concentration of dissociated acid is a=0.1/V, concentration of the left acid is (1-0.1)/V=C-a.
got it now!

4-
They don't have the same value. See water ion product page.
as usual..a mistake in my book.

Thanks very much.

Borek
Mentor
Yes.but if what is the problem if it is constant? why don't we substitute with the concentration of water as 55.5 moles/liter ?and what is the relation between calculation of Kw and assuming water concentration constant?

Not sure what you are asking about. We ignore water concentration to make calculations easier. It doesn't mean it is not there - just like in the case of water ion product, we move it to the equilibrium constant. This way we have one number less to worry about, it makes calculations easier.

yes i know.Do u mean by "activity" the rate of chemical reaction ?

No, rate is rate, activity is activity. For example

$$a_{H^+} = f_{H^+}[H^+]$$

activity of H+ equals its concentration times activity coefficient. For water ion product

$$K_w = a_{H^+}a_{OH^-}$$

For diluted solutions activity coefficients equal 1, for not too concentrated solutions (say below 0.1M) they are identical for ions of the same charge (so fH+=fOH-) and are less than 1. See ionic strength and activity coefficients. Please note that this is just an approximation of what is happening in reality - better one than using just concentrations, but failing in more concentrated solutions. In fact despite over 100 years of efforts we still don't have a reasonable theory allowing calculation of activities of ions in more concentrated solutions.

another question:
In Ostwald law, sometimes we assume (a) as no. of dissociated moles, sometimes we assume (a) as degree of dissociation which equals no. of dissociated moles/the total no. of moles of a substance before dissociation
http://www.pinkmonkey.com/studyguides/subjects/chem/chap12/c1212701.asp
so, what is (a) ?

There is some problem with formatting on the page, seems to me like whenever they use "a" they in fact mean α (degree of dissociation).

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