effect of dilution on pH (weak vs. strong acid)

[chocolatecake2206]

Hey everyone

please help with a question from VCAA 2008 unit 4 exam!

Acid I: pH=1.0
Acid II: pH=2.1

Both acids are diluted by a factor of 10.  Which acid has a greater change in pH units?

The answer is acid II 'because it is a weak acid' but this doesn't really explain it for me.

Anyone able to help?

thank you!!

[vox nihili]

Hey everyone

please help with a question from VCAA 2008 unit 4 exam!

Acid I: pH=1.0
Acid II: pH=2.1

Both acids are diluted by a factor of 10.  Which acid has a greater change in pH units?

The answer is acid II 'because it is a weak acid' but this doesn't really explain it for me.

Anyone able to help?

thank you!!

Do the maths for both and see

[H⁺]=-log₁₀pH

[Bruzzix]

Do the maths for both and see

[H⁺]=-log₁₀pH

Ummm I think you mean pH=-log₁₀[H⁺] :p

[chocolatecake2206]

ah yes the math does make it clear, thanks

I'm just thrown by the comments in the assessment report:

'a change in concentration by a factor of 10 corresponds to a one unit change in pH - While this is true for strong acids, it is certainly not true for weak acids.'

^not quite sure why it is true for strong acids but not for weak acids 

[IndefatigableLover]

ah yes the math does make it clear, thanks

I'm just thrown by the comments in the assessment report:

'a change in concentration by a factor of 10 corresponds to a one unit change in pH - While this is true for strong acids, it is certainly not true for weak acids.'

^not quite sure why it is true for strong acids but not for weak acids

Strong Acids completely ionise in water whilst weak acids do not ionise completely. With strong acids, we are able to use the formula that Bruzzix has stated above as one mole of Acid equals one mole of H+ (for monoprotic acids).
On the other hand, we can't assume this to happen for weak acids as they do not completely ionise so normally (since the backwards reaction is favoured) when you're calculating pH of weak acids, you'll typically use Ka values from your data booklet in order to do so!

[vox nihili]

Ummm I think you mean pH=-log₁₀[H⁺] :p

Just keeping you all on your toes




*dies inside*

[chocolatecake2206]

Strong Acids completely ionise in water whilst weak acids do not ionise completely. With strong acids, we are able to use the formula that Bruzzix has stated above as one mole of Acid equals one mole of H+ (for monoprotic acids).
On the other hand, we can't assume this to happen for weak acids as they do not completely ionise so normally (since the backwards reaction is favoured) when you're calculating pH of weak acids, you'll typically use Ka values from your data booklet in order to do so!

thanks!

[kim21]

I had thought the strong acid has a greater change in pH since dilution would push the weak acid forward creating more H+ and resiting change whereas since the strong acid has nearly completely ionised 

[cosine]

What does it mean when 'strong acids completely ionise in water'? Strong acids donate protons to the bases, hence they're ionising, but what does it mean to completely ionise? Does it refer to donating all the available protons or?

[kim21]

What does it mean when 'strong acids completely ionise in water'? Strong acids donate protons to the bases, hence they're ionising, but what does it mean to completely ionise? Does it refer to donating all the available protons or?

It means theres a hell of a lot more of its conjugate base than the acid.
HCl + H2O --> H3O+ + Cl-
theres almost no HCL ions in the solution meaning it completely ionises therefor more H+ and lower pH

[cosine]

It means theres a hell of a lot more of its conjugate base than the acid.
HCl + H2O --> H3O+ + Cl-
theres almost no HCL ions in the solution meaning it completely ionises therefor more H+ and lower pH

So would you say that it's a non-reversable reaction and so all the Cl- ions remain as they are? But whereas a weak acid would be reversible so the Cl- ions are capable of going back into HCl?

[kim21]

So would you say that it's a non-reversable reaction and so all the Cl- ions remain as they are? But whereas a weak acid would be reversible so the Cl- ions are capable of going back into HCl?

Weak acids will still have plenty of its acid compared to the base
CH3COOH + H2O --> CH3COO- + H3O+ unlike strong acids, only a small amount is converted to H3O+ therefore a high pH

I think the idea is that the addition of water will increase the pH overall but with weak acids the water will push the reaction forward creating more H3O+ thus slightly opposing the change whereas a strong acid has completely ionised so the addition of water wont create anymore H3O+ since theres no acid left

[vox nihili]

So would you say that it's a non-reversable reaction and so all the Cl- ions remain as they are? But whereas a weak acid would be reversible so the Cl- ions are capable of going back into HCl?

Every reaction is reversible under the right conditions. In the case of strong acids, the equilibrium is shifted towards the right (i.e. towards dissociation), whereas in weak acids, it's roughly equal or even towards the left (i.e. association).

[tommyl97]

Ah, I did this question from Checkpoints; as the others stated, the change in pH is greater for Acid I (pH = 1.0) than Acid IV (pH = 2.1).

Here is their (Cambridge's) explanation:

The change in pH is greater for acid I than for acid IV.
Acid I is a strong acid ([acid] = [H+]) and is fully ionised. When diluted by a factor of 10 the resulting pH will be 2. However acid IV is a weak acid. It is only partially ionised. When diluted its percentage ionisation will increase (or adding water pushes the position of equilibrium to the right and more H+ is formed). Thus the resulting pH will not be 3.1, but between 2.1 and 3.1.

Here's my attempt in explaining it further/more in detail:

We can already establish that acid IV is a weak acid. This is because of the previous question from the VCAA 2008 Unit 4 Exam (Question 3c) which asks to calculate the percentage ionisation of Acid IV using the given concentration and pH (0.10M and 2.1 respectively).


As acid I is a strong acid (pH = 1.0, 0.1M, 100% ionisation) we can confirm that its pH will change to 2 when we dilute by a factor of 10. Assuming an initial volume of 1L:



However, acid IV is a weak acid so the above does not apply. Lets pretend that acid IV is ethanoic acid and write out its ionisation equation in water.



Remember than water is amphiprotic, that is, it can act as either an acid or a base. In this particular instance, it is acting as a base. Adding water will cause the OH- ions from the water to react with the H+ ions in the solution of ethanoic acid. The concentration of H+ ions will decrease. Subsequently, the system will try to oppose this change by producing more H+ to make up for the lost H+. But this effect is only partial (Le Chatelier's Principle) and therefore the pH will increase because there is less H+ than what we had before adding the water. However this pH won't jump to 3.1 as we'd expect for a strong acid, but be situated somewhere between 2.1 and 3.1.

[chocolatecake2206]

The answer is acid II 'because it is a weak acid' but this doesn't really explain it for me.

apologies to anyone puzzled by the mistake in my question - the answer is I as the others have pointed out, not II. just realized I must have overlooked it as I was too focused on the theory!