[tanjinjack]

Hi, I am Tan from Malaysia.
I am currently doing my pre-u program and I came across one question that is somehow to be quite confusing.

What is the effect on pH when vinegar is added with water?
Here's my attempt.
CH₃COOH + H₂O ⇌ CH₃COO^- + H₃O⁺.
Alright, I can explain by the Le Chatelier's Principle.
When water is added, the equilibrium moves towards the right and more hydronium ions are produced and hence the pH will go down.
However, another concern is when more hydronium ions are produced, there are more water added into the solution and the volume will be different, thus affecting the concentration.
Even if there are more hydronium ions, but the volume also increases and the concentration will change. In the end, the pH may not go down but instead goes up.
By common sense, the pH should go up as we are diluting the acid.

As I ask my lecturer, she seemed to stick to the Le Chaletier's Principle but I haven't proposed to her with the change in volume yet.

So, how will the concentration be and eventually affecting the pH? Or, there are other factors in deciding the concentration? I am a Pre-U student and feel free to tell me what you think that I may know!

Thanks in advance!

Best regards from Malaysia,
-jin jack-

[Borek]

However, another concern is when more hydronium ions are produced, there are more water added into the solution and the volume will be different, thus affecting the concentration.

I don't see where the water comes from.

[tanjinjack]

I mean at the time we add more water, we have more hydronium ions, and we have a bigger volume.
If we say
number of moles = concentration x volume
We now have more hydronium ions, number of mols increases, but at the same time, volume also increases, so, how will the concentration behave?

[tanjinjack]

Hmm,
After some readings on the acidity constant, here's my answer. The pH will remain the same.
As more water is added, but the K_a is not affected. As K_a is not affected, the concentration of hydrogen ion is the same, and hence no change in pH.
So, am I correct?

[Borek]

Sorry but I am not able to understand the situation you are describing.

What do we start with? Pure water?

What do we add? Acetic acid? Of what concentration?

[tanjinjack]

Alright, we have acetic acid dissociated in water, and hence we got the equation
CH₃COOH + H₂O ⇌ CH₃COO- + H₃O+.

So, when we add pure water into the system, what will happen to the system?

[Astrokel]

Is water in the equilibrium constant?

[Hunt]

Hi, I am Tan from Malaysia.
I am currently doing my pre-u program and I came across one question that is somehow to be quite confusing.

What is the effect on pH when vinegar is added with water?
Here's my attempt.
CH₃COOH + H₂O ⇌ CH₃COO^- + H₃O⁺.
Alright, I can explain by the Le Chatelier's Principle.
When water is added, the equilibrium moves towards the right and more hydronium ions are produced and hence the pH will go down.
However, another concern is when more hydronium ions are produced, there are more water added into the solution and the volume will be different, thus affecting the concentration.
Even if there are more hydronium ions, but the volume also increases and the concentration will change. In the end, the pH may not go down but instead goes up.
By common sense, the pH should go up as we are diluting the acid.

As I ask my lecturer, she seemed to stick to the Le Chaletier's Principle but I haven't proposed to her with the change in volume yet.

So, how will the concentration be and eventually affecting the pH? Or, there are other factors in deciding the concentration? I am a Pre-U student and feel free to tell me what you think that I may know!

Thanks in advance!

Best regards from Malaysia,
-jin jack-

I think for all practical reasons the pH decreases unless the volume added is huge compared to the initial volume and the concentrations of the species are infinitely small, which would then lead do an increase in pH.

In your case assume that initially there exists C_o mol/L of H₃O⁺ ions. If a relatively small amount of water is added, the new concentration becomes : C_o f where f ( dilution factor ) = initial solution volume / total solution volume

According to Le Chatelier's principle the reaction advances to the right, so the concentration of H₃O⁺ at equilibrium is : C_o f + x

Ofcourse the new pH = - log (C_o f + x)

f is close to 1 , so the pH should increase but whether this change is negligible or not depends on x which is a function of Ka, the concentration of hydronium ions, , the concentration of the acid, concentration of its conjugate base, and the dilution factor f

Now if the volume of water added is very large so that the weak acid completely dissociates then [ H₃O⁺ ] = C_acid f which should be greater than the initial concentration of hydronium ions. The pH then decreases too.

C_acid is the initial concentration of the acid.

[Borek]

So, when we add pure water into the system, what will happen to the system?

OK, so bascially your question is: what will happen to the solution of acid, when we dilute it with pure water. pH will go up to 7.

If you will start with highly concentrated acid situation at the beginning will be more complicated due to the changing activities, and honestly I don't know how to describe the system. But let's say we start with about 1M solution:

C (mol/L)	pH
1	2.38
0.1	2.88
0.01	3.39
0.001	3.91
0.0001	4.47
0.00001	5.15
0.000001	6.02
0.0000001	6.79
0.00000001	6.98

[tanjinjack]

Thanks for the help.
After giving some thought, I guess I do not need to use Le Chatelier's in this situation.
When we add more water, we are not changing the concentration of water, and hence no change.
But, when we add water, will there be a change to the concentration of the weak acid? My answer is no as well as the acidity constant for weak acid, usually we do not include the concentration of water as it is insignificant.
If my assumption is correct, that means adding water will have NO change throughout, but that does not make sense.

I know when we add water to strong acid, the pH will go up, but do we change the concentration of a weak acid if we add water to it?

Confused,
-jin jack-

[Borek]

But, when we add water, will there be a change to the concentration of the weak acid? My answer is no

OK, so you started with 1L of 0.1M acid - it contained 0.1 mole of acid.

You added 1L of water - final volume is 2L. You think that concentration of acid has not changed - that means concentration of acid is still 0.1M. So now you have 2L of 0.1M solution - it contains 0.2 mole of acid.

Where did the additional 0.1 mole come from?

[tanjinjack]

Borek, the dilution is always true for a strong acid, but how about for the weak acid?

The concentration of the H⁺ ion is related to the acidity constant, given as
K_a=[H⁺][A^-]/[HA]
Generally adding water will not change this acidity constant as the concentration of water is not significant to the equilibrium. So, will adding water change the concentration of H+ ion?

[Borek]

Borek, the dilution is always true for a strong acid, but how about for the weak acid?

It gets diluted as well.

The concentration of the H⁺ ion is related to the acidity constant, given as
K_a=[H⁺][A^-]/[HA]

It is related to Ka AND acid concentration.

Generally adding water will not change this acidity constant as the concentration of water is not significant to the equilibrium.

Ka won't change, but total concentration [A^-]+[HA] goes down.

So, will adding water change the concentration of H+ ion?

Asking the same question for the umpteenth time won't change the answer. I have listed calculated values of pH for different concentrations of acetic acid two posts earlier, just look at them.

[tanjinjack]

Correct me if wrong.

Le Chatelier says that when we have more number of molecules of water, the water molecules will react and convert to hydronium ion and the carboxylate ion. So, there will be more hydronium ion and carboxylate ion in NUMBERS, and at the same time less acid molecules now.

Should we say now concentration of acid molecules decreases now as number of mols decreases, volume increases.

By K_a,
K_a=[H⁺][A^-]/[HA]
[HA] decreases, [H⁺] should decrease as [H⁺]=[A^-]

So, another problem I see, now number of hydronium ion also increases, but does the increase in volume as well eventually decreases the concentration of hydronium ion?

[Borek]

You can't mix intensive and extensive properties. Even if number of molecules grows volume grows too, these two effects cancel out. What is important is concentration. Stick to it.