weak acids and ph

[Martoman]

So when you have a weak acid, or any acid, if you dilute it the pH will increase. Similarly, the percentage ionisation will increase because the concentration of the acid will decrease? I'm going off the fact that for the same amount of ionisation the concentration of the acid ionising will be less, meaning in effect, a higher percentage ionisation. For the more mathy, a smaller denominator means a bigger result in the % ionisation fraction. Is this right?

[kyzoo]

xD I had the same problem understanding this as well until I saw the solutions to a particular exam (forget which one)

Let's take the ionisation reaction of ethanoic acid

CH3COOH + H2O <--> CH3COO- + H3O+

Ok now if you add water (which is a reactant), you shift the equilibrium to the right. Hence
n(CH3COO-) increases, thus increasing the percentage ionisation.

However overall concentration of the ion will overall be decreased, because by Le Chatelier's principle the equilibrium system only partially opposes the decrease in concentration induced by the addition of water.

Hope that makes sense =D

[bomb]

So when you have a weak acid, or any acid, if you dilute it the pH will increase. Similarly, the percentage ionisation will increase because the concentration of the acid will decrease? I'm going off the fact that for the same amount of ionisation the concentration of the acid ionising will be less, meaning in effect, a higher percentage ionisation. For the more mathy, a smaller denominator means a bigger result in the % ionisation fraction. Is this right?

Only with weak acids, strong acids fully ionise so don't experience Le Chatelier effects when diluted. Strong acids have the same  %Ionisation.
As with weak acids, yes, when you dilute it, [H3O+] decreases and hence system shifts to the right (increasing %Ionisation) to partially oppose the decrease in [H3O+] concentration.

[fady_22]

For a strong acid, the percentage ionisation doesn't really change (as it will always remain at ~100%, due to being ionised completely).

For a weak acid, however, diluting the acid will increase the percentage ionisation. This is because, as the solution is diluted, there will be a net forward reaction to partially oppose the dilution. This increases the amount of hydronium and conjugate base, and so the percentage ionisation increases.

[kyzoo]

So when you have a weak acid, or any acid, if you dilute it the pH will increase. Similarly, the percentage ionisation will increase because the concentration of the acid will decrease? I'm going off the fact that for the same amount of ionisation the concentration of the acid ionising will be less, meaning in effect, a higher percentage ionisation. For the more mathy, a smaller denominator means a bigger result in the % ionisation fraction. Is this right?

Only with weak acids, strong acids fully ionise so don't experience Le Chatelier effects when diluted. Strong acids have the same  %Ionisation.
As with weak acids, yes, when you dilute it, [H3O+] decreases and hence system shifts to the right (increasing %Ionisation) to partially oppose the decrease in [H3O+] concentration.

But then what about the decrease in the concentration of the weak acid (e.g. CH3COOH) which is on the left hand-side of the equation?

[Martoman]

just F off Le chandelier for the moment

What i'm saying is that when you put a weak acid in water it will ionise to a small extent. If you dilute this, this will decrease the concnetration of the acid. What this means effectively is that there is more ionisiation per unit concentration of the acid. Hence the % increases. I'm looking for intuitive explanation, not Mr Chandelier, who explains nothing about this stuff

[bomb]

So when you have a weak acid, or any acid, if you dilute it the pH will increase. Similarly, the percentage ionisation will increase because the concentration of the acid will decrease? I'm going off the fact that for the same amount of ionisation the concentration of the acid ionising will be less, meaning in effect, a higher percentage ionisation. For the more mathy, a smaller denominator means a bigger result in the % ionisation fraction. Is this right?

Only with weak acids, strong acids fully ionise so don't experience Le Chatelier effects when diluted. Strong acids have the same  %Ionisation.
As with weak acids, yes, when you dilute it, [H3O+] decreases and hence system shifts to the right (increasing %Ionisation) to partially oppose the decrease in [H3O+] concentration.

But then what about the decrease in the concentration of the weak acid (e.g. CH3COOH) which is on the left hand-side of the equation?

H2O Is a reactant though, your adding reactant while decreasing all concentrations. Guess there's a few shifts then but the overall shift would be to the right. I actually asked that question to my teacher he said it shifts towards the right.

[fady_22]

just F off Le chandelier for the moment

What i'm saying is that when you put a weak acid in water it will ionise to a small extent. If you dilute this, this will decrease the concnetration of the acid. What this means effectively is that there is more ionisiation per unit concentration of the acid. Hence the % increases. I'm looking for intuitive explanation, not Mr Chandelier, who explains nothing about this stuff

This does not explain it, because although you are correct in stating that the concentration of acid decreases, you are forgetting that the concentration of hydronium also decreases (by the same factor), which would "cancel out" the decrease in concentration, as is the case in strong acids.
In weak acids, the only thing that explains the increase is Le Chatelier's.

[kyzoo]

just F off Le chandelier for the moment

What i'm saying is that when you put a weak acid in water it will ionise to a small extent. If you dilute this, this will decrease the concnetration of the acid. What this means effectively is that there is more ionisiation per unit concentration of the acid. Hence the % increases. I'm looking for intuitive explanation, not Mr Chandelier, who explains nothing about this stuff

Ok how about this.

There are 2 reasons why percentage ionisation decreases

1. The reason you described: decreased value of the denominator --> increased value of the overall fraction

2. Water is a reactant to the ionisation reaction. So if you add water to the equilibrium system of "CH3COOH + H2O <--> CH3COO- + H3O+", the rate of the forward reaction is increased (more reactant particles) whilst the rate of the backwards reaction remains the same. So that means that momentarily, until equilibrium is reestablished (forward reaction rate = backward reaction rate), the rate of the forward reaction > rate of the backward reaction. Thus  n(CH3COO-) is increased.

And yes lol, Le Chatelier's Principle isn't first principles, it's like using derived formulas for motion in Physics.

[Martoman]

just F off Le chandelier for the moment

What i'm saying is that when you put a weak acid in water it will ionise to a small extent. If you dilute this, this will decrease the concnetration of the acid. What this means effectively is that there is more ionisiation per unit concentration of the acid. Hence the % increases. I'm looking for intuitive explanation, not Mr Chandelier, who explains nothing about this stuff

Ok how about this.

There are 2 reasons why percentage ionisation decreases

1. The reason you described: decreased value of the denominator --> increased value of the overall fraction

2. Water is a reactant to the ionisation reaction. So if you add water to the equilibrium system of "CH3COOH + H2O <--> CH3COO- + H3O+", the rate of the forward reaction is increased (more reactant particles) whilst the rate of the backwards reaction remains the same. So that means that momentarily, until equilibrium is reestablished (forward reaction rate = backward reaction rate), the rate of the forward reaction > rate of the backward reaction. Thus  n(CH3COO-) is increased.

And yes lol, Le Chatelier's Principle isn't first principles, it's like using derived formulas for motion in Physics.

Yeah this is how i think, rate of reactions, i've never had the need to consult chandelier's principle.

[kenhung123]

Haha great old Mr Chandelier