Why do liquids and solids have constant concentrations?

[scilearner]

Hello everyone,

 

Why is that liquids and solids have constant concentrations and are eliminated from equilibrium constants. I don't get it why can't their concentrations change. More importantly does this mean that if water (liquid) is a reactant in a reaction and if we add more water to the system the system would not move in the forward reaction to get rid of excess water. Help would be appreciated. Thanks

 

Example of my reaction 2H20 <> H30 + OH

[insane_alien]

they are not dispersed in solution. the density doesn't change(except in extreme circumstances where you'd be using a different model anyway) therefore the concentration doesn't change.

 

for the water dissociation the water just has the same concentration. so the concentration of H3O and OH are the same. but if you take into account the volume you have more H3Oand OH than before.

[scilearner]

they are not dispersed in solution. the density doesn't change(except in extreme circumstances where you'd be using a different model anyway) therefore the concentration doesn't change.

 

for the water dissociation the water just has the same concentration. so the concentration of H3O and OH are the same. but if you take into account the volume you have more H3Oand OH than before.

 

Thank you for your help:-). But I didn't understand the last sentence. How can you have more H30 and OH if you have more volume?

[insane_alien]

The total amount of a substance in solution is its concentration multiplied by the volume of solution.

 

If volume increases and concentration stays the same then the total amount must also increase.

[scilearner]

The total amount of a substance in solution is its concentration multiplied by the volume of solution.

 

If volume increases and concentration stays the same then the total amount must also increase.

 

Thanks once again for your help . Are you saying concentrations stay the same because of particles in both sides being equal or water that has a constant concentration is added. This is where I'm most confused. If you add more water do you increase the moles of water. If so wouldn't there be a faster reaction with more moles of reactants?

[insane_alien]

You do increase the number of molecules but the volume also increases but a proportional amount. The end result is that the concentration stays constant as opposed to when you dissolve a solid, for instance salt, in the solvent, water.

 

Reaction rates depend on the concentration rather than the total number of molecules, so unless the concentration rises, the reaction rate will not.

 

I know this is a tricky bit as i had problems getting it at first but once you understand it, it looks so obvious. So, if i've not explained anything very well please tell me and i'll have another go at it.

[scilearner]

You do increase the number of molecules but the volume also increases but a proportional amount. The end result is that the concentration stays constant as opposed to when you dissolve a solid, for instance salt, in the solvent, water.

 

Reaction rates depend on the concentration rather than the total number of molecules, so unless the concentration rises, the reaction rate will not.

 

I know this is a tricky bit as i had problems getting it at first but once you understand it, it looks so obvious. So, if i've not explained anything very well please tell me and i'll have another go at it.

 

Thanks once again for the help and understanding . Or is it that even though the moles are increased the volume is also increased at the same time. So the probababily of fruitful collisons are still the same. I think I got it . So in equilibrium if you add solid or liquid does that mean there would be no effect. Still I'm bit confused with having more moles of H30 and OH. If the concentration of water stays the same with added water wouldn't the concentrations of H30 and 0H... Oh I get it so if this didn't happen in water and if we added more solid the concentrations of the products and the moles would stay the same. Basically no effect. Am I right?

 

EDIT: Actually now I think about it. When an acid and a base react water is produced. If there is no limiting or excess in this case. So wouldn't this water produced create a dilution. Making H30 and OH gone down. According to Le chatelir the reaction would try to move to more particle side. But both sides have same number. So isn't the concentration going down.

Edited July 18, 2008 by scilearner

[insane_alien]

Or is it that even though the moles are increased the volume is also increased at the same time. So the probababily of fruitful collisons are still the same. I think I got it .

 

Yep, thats about it.

 

So in equilibrium if you add solid or liquid does that mean there would be no effect.

 

In the dissolving of salt adding more water will dilute the solution meaning the concentration of dissolved salt will decrease and more salt will dissolve but the new equilibrium will be achieved at the same concentration of salt in solution.

 

So wouldn't this water produced create a dilution. Making H30 and OH gone down

 

yes, but remember the reaction going the otherway takes a water molecule and splits it. making the concentration go up. At equilibrium the concentration would be constant as both reactions proceed at the same rate. As with the salt solution, the equilibrium will happen at identical concentrations as before.

[scilearner]

Yep, thats about it.

 

 

 

In the dissolving of salt adding more water will dilute the solution meaning the concentration of dissolved salt will decrease and more salt will dissolve but the new equilibrium will be achieved at the same concentration of salt in solution.

 

 

 

yes, but remember the reaction going the otherway takes a water molecule and splits it. making the concentration go up. At equilibrium the concentration would be constant as both reactions proceed at the same rate. As with the salt solution, the equilibrium will happen at identical concentrations as before.

 

Thanks a lot for your help Insane Alien . I had a good think about the moles and worked it out. When we add water there is more water and concentration is decreased. If moles in increases you can get closer to the intital level.

 

What you are saying should be true because if the concentrations do not change to their initial level. The PH of water would change. This can only happen at temperature.

 

I think why I'm confused is that I have never seen what Le chatelier principle state when both product concentrations are decreased at the same time. Le chatelier only talks about one reactant or product or when the whole concentration of products and reactant are decreased. So according to Le chatelier principle if the concentration of H30, OH would go down at the same time if water is added. Concentration of water wouldn't change because of liquid. Then how can the concentration come to back to intitial level if the system only partially counteracts the change. But what you are saying must be right because or else Kw would change. Looking forward to your reply

[insane_alien]

yep, you've pretty much got it.

 

though i think the water example has a major flaw in it. the water added also contains an equal concentration of H30 and OH ions so it will not have to reach equilibrium again.

 

although, i suppose you could use it as an idealised example where you have a sample of water not at equilibrium added.

[scilearner]

yep, you've pretty much got it.

 

though i think the water example has a major flaw in it. the water added also contains an equal concentration of H30 and OH ions so it will not have to reach equilibrium again.

 

although, i suppose you could use it as an idealised example where you have a sample of water not at equilibrium added.

 

Yeah I made up the water example and I asked one of my teachers and she said acidity decreases . lol it's not her fault I think I did a bad job at explaining my question . Anyway I have one more question

 

If 0.10 M Mg(OH)2 calculate the PH

 

I know how to do this. Why can we use mole ratios to work out concentrations. I thought it was only for moles. So when mgoh2 is dissolved do I have to think of concentration of OH going up or concentration of H30 going down at equilibrium. When working out the concentration of OH do we have to assume that the concentration we work out is the concentration of OH when the equilibrium position has moved to the left. Thanks a lot for your help so far You are the only one who has helped so far