# Why do we say that the concentration of solids and liquids are constant?

• Chemistry

## Homework Statement:

Concentration of solids and liquids are constant, but it seems to me non-intuitive.

## Homework Equations:

concentration = density/ molar mass
I was introduced to the concepts like molarity, molality , mole fraction, formality, normality etc. during general chemistry and it was said that these concepts are used to measure the concentration of solutions. But as the thing called Chemical Equilibrium came I was surprised when I read "In heterogenous chemical reaction the concentration of solids and liquids are taken to be constant in calculating $K$"
I did some research and found that people are explaining it like this :-
Let's assume that you have a liquid with density $\rho$ and molar mass $m$ and since units of $\rho$ is $\frac{g}{L}$ and that of $m$ is \frac{g}{mol} therefore $\frac{\rho}{m}$ will have units as$\frac{mol}{L}$ and hence $concentration = \frac{\rho}{m}$ and $\rho= \frac{mass}{volume}$. Therefore, $conc. = \frac{mass\cdot molar mass}{volume}$ and as you can see all three quantities are constant(volume is constant for solids and liquids) and hence the concentration of solids and liquids are constant.
I agree with that mathematical explanation but what does it mean when we say "that bottle have concentrated NaOH and that bottle have dilute NaOH" isn't it contrary to what has been just described above? I mean if we mix more $CaO$ in water we will get a highly concentrated solution then if we mix just a pinch of $CaO$ in water.
Any help will be much appreciated. I know that I'm having some misconception and therefore I request you all to please help me.
Thank you.

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Borek
Mentor
I mean if we mix more $CaO$ in water we will get a highly concentrated solution then if we mix just a pinch of $CaO$ in water.
That's not a solution, that's a suspension - you have two separate phases that react only the at the phase boundary.

(Well, CaO easily reacts with water, so the situation is much more complicated, think in terms of sand in water to make things conceptually easier).

That's not a solution, that's a suspension - you have two separate phases that react only the at the phase boundary.

(Well, CaO easily reacts with water, so the situation is much more complicated, think in terms of sand in water to make things conceptually easier).
I can mix one spoon of sand in a glass of water and will get a solution of say conc. $X$.
But when I will mix five spoons of sand in same quantity of water then according to my knowledge (till now) I should get a solution of concentration more than that of $X$.

Borek
Mentor
Solution is a homogeneous mixture. Is the mixture of sand and water homogeneous?

Solution is a homogeneous mixture. Is the mixture of sand and water homogeneous?
No. From personal experience I can say that sand will settle down. Hence composition is not uniform all over.

Borek
Mentor
So it is not a solution and you can't speak about the concentration of sand in water.

So it is not a solution and you can't speak about the concentration of sand in water.
Yes I think I'm getting you. But if we mix 500 ml of $NaOH$ in 1 litre of water and in a different bottle we mix 900 ml of $NaOH$ in 1 litre of water then what will be the situation? I think that the latter case will be a more concentrated base.

So it is not a solution and you can't speak about the concentration of sand in water.
I must express my feelings, you have taught such a great lesson which I missed to notice the significance of. Really I didn't ever notice that concentration is defined only for homogenous mixtures (well it makes sense too, it would be senseless to talk about concentration when concentration itself varies over the solvent).
I just really want to thank you. Thank you so much.
I apologize for this off-topic remark.

Borek
Mentor
Yes I think I'm getting you. But if we mix 500 ml of $NaOH$ in 1 litre of water and in a different bottle we mix 900 ml of $NaOH$ in 1 litre of water then what will be the situation? I think that the latter case will be a more concentrated base.
Depends on how much will dissolve. At some point you will get a saturated solution with some solid left, then adding more NaOH will change the amount of solid, but not the concentration of the solution.

I just really want to thank you. Thank you so much.
The pleasure is all mine

Depends on how much will dissolve. At some point you will get a saturated solution with some solid left, then adding more NaOH will change the amount of solid, but not the concentration of solution
So sir, what does it actually mean by "the concentration of liquid and solids are constant" . I'm not getting it.
I mean before saturation is attained we can have differently concentrated solutions of NaOH and water.

Ygggdrasil
Gold Member
Here's how I think about the issue. For a reaction between a an aqueous substance and a solid (say the dissolution of solid NaOH(s) into Na+(aq) and OH-(aq) ions), the reaction can occur only at the interface between the solid NaOH and the aqueous solution.

Because equilibrium doesn't depend on the total amount of substance, let's just consider a fixed area of solid NaOH in contact with water. Changing the concentration of Na+ or OH- ions in solution will change the number of these ions that can collide with and react with the surface. However, (by definition) the number of solid NaOH units in contact with the water over that fixed area is fixed and cannot be changed.

Adding more solid NaOH will provide more surface area over which the reaction can occur (speeding up the reaction), but it won't do anything to change the relative abundance of solid NaOH units to Na+ and OH- ions at the interface. Thus, the eqiuilibrium does not change when you change the amount of solid NaOH.

Here's how I think about the issue. For a reaction between a an aqueous substance and a solid (say the dissolution of solid NaOH(s) into Na+(aq) and OH-(aq) ions), the reaction can occur only at the interface between the solid NaOH and the aqueous solution.

Because equilibrium doesn't depend on the total amount of substance, let's just consider a fixed area of solid NaOH in contact with water. Changing the concentration of Na+ or OH- ions in solution will change the number of these ions that can collide with and react with the surface. However, (by definition) the number of solid NaOH units in contact with the water over that fixed area is fixed and cannot be changed.

Adding more solid NaOH will provide more surface area over which the reaction can occur (speeding up the reaction), but it won't do anything to change the relative abundance of solid NaOH units to Na+ and OH- ions at the interface. Thus, the eqiuilibrium does not change when you change the amount of solid NaOH.
First of all thank you so much for replying me. I almost believed that Borek has not replied to me because he has thought that I can't ever understand it but your answer has really given me ample amount of pleasure.
Your answer is very nice, it has cleared all my doubts. I'm really grateful to you for such a nice answer.