• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Mod 6 neutralisation multiple choice (1 Viewer)

NexusRich

Member
Joined
Jul 6, 2019
Messages
87
Gender
Male
HSC
2021
In this question, I thoguht that since acetic acid is a weak acid, it partially ionises in water and will not produce 0.100M of <H+>, so that it would require less than 23.4ml of NaOH to neutralise. But the answer turns out to be B. Could someone pls explain? Thanks in advance.
 

Attachments

Pikapizza

Member
Joined
Oct 8, 2020
Messages
26
Gender
Male
HSC
2020
I believe the keyword here is "completely react". Yes, acetic acid is weak and in solution will partially ionise, however in neutralisation with a base it fully reacts to form a salt + water.

Not sure if this is the right way to think about it (haven't touched chem in a while), but possibly considering it like: as the neutralisation reaction proceeds, H+ concentration decreases as it is being used up; hence the acetic acid will keep proceeding with its ionisation reaction all until the H+ is fully neutralised - this means eventually the acetic acid will end up producing its 0.1M of H+ (since monoprotic) and hence why 23.4ml of NaOH is required for the right amount of OH- to react.

I do apologise for any mistakes but I do hope it explains a little while you wait for the more experienced helpers!
 

TheShy

Active Member
Joined
Nov 12, 2018
Messages
130
Gender
Male
HSC
2021
might not be the best explanation, but you can think of it in terms of mole ratio. With NaOH and HCl, ittl be a 1:1 reaction, and with acetic acid and NaOH, will also be a 1:1 ratio. So it doesn't matter really matter that its a weak acid, but since the molar ratios are the same, the same amount of volume can be used, as long as the [NaOH] is the same.
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
Pragmatic / Strategic answer... if it wasn't B, that would mean the equivalence point in a titration depended on both the stoichiometry and the strengths of the acids and bases, in which case titration would only be accurate for strong acid + strong base. Since titration and volumetric analysis are highly accurate, this can't be true and the equivalence point must depend solely on stoichiometry.

For a more theoretical answer, you could use @Pikapizza's approach and say that the reaction occurring is all hydroxide plus hydronium and the ionisation of the weak will be drawn right by Le Chatelier's Principle until the concentration of the unionised form reaches practically zero.

Another approach is to consider that the hydroxide can react with hydronium from after the weak acid ionises or with the unionised weak acid, and will continue until both are essentially all consumed. That is, we have reaction continuing until

H3O+ + OH- ---> 2 H2O

and

CH3COOH + OH- -----> CH3CO2- + H2O

are complete, and so all of the weak acid reacts - some directly, some indirectly by ionising and the hydronium thereby produced reacting.
 

idkkdi

Well-Known Member
Joined
Aug 2, 2019
Messages
2,566
Gender
Male
HSC
2021
Pragmatic / Strategic answer... if it wasn't B, that would mean the equivalence point in a titration depended on both the stoichiometry and the strengths of the acids and bases, in which case titration would only be accurate for strong acid + strong base. Since titration and volumetric analysis are highly accurate, this can't be true and the equivalence point must depend solely on stoichiometry.

For a more theoretical answer, you could use @Pikapizza's approach and say that the reaction occurring is all hydroxide plus hydronium and the ionisation of the weak will be drawn right by Le Chatelier's Principle until the concentration of the unionised form reaches practically zero.

Another approach is to consider that the hydroxide can react with hydronium from after the weak acid ionises or with the unionised weak acid, and will continue until both are essentially all consumed. That is, we have reaction continuing until

H3O+ + OH- ---> 2 H2O

and

CH3COOH + OH- -----> CH3CO2- + H2O

are complete, and so all of the weak acid reacts - some directly, some indirectly by ionising and the hydronium thereby produced reacting.
does ch3cooh have to be dissociated/in ion form to react?
aren't there acid/base reactions where there aren't two aqueous things?

just wondering about this lol, might have missed something in mod 6.
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
does ch3cooh have to be dissociated/in ion form to react?
aren't there acid/base reactions where there aren't two aqueous things?

just wondering about this lol, might have missed something in mod 6.
Acetic acid is a weak acid in water and so in aqueous environments exists in equilibrium between its ionised and unionised forms. As I wrote above, it can react in either form. For the purposes of stoichiometry, the equilibrium makes no difference so for a titration calculation we'd usually write

CH3COOH + NaOH ----> CH3CO2Na + H2O​

but the reaction is actually occurring both when hydroxide ions encounter unionised acetic acid molecules and when hydroxide ions encounter hydronium ions.

In fact, there is a third mechanism for the reaction that involves the indicator, but the mechanisms don't matter to the stoichiometry as the overall net equation that describes the equivalence point can be summarised as the above.

Glacial acetic acid is so concentrated that it exists as almost exclusively unionised acetic acid molecules. Even pure acetic acid acid will possess species other than the unionised acetic acid molecule due to autoionisation, but that can be ignored for most purposes... though a difficult MCQ can be written involving conjugate pairs where it is presented with its conjugate acid.
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
Actually, acetic acid is one of the comparatively uncommon substances that has two different conjugate acids.
 

idkkdi

Well-Known Member
Joined
Aug 2, 2019
Messages
2,566
Gender
Male
HSC
2021
Acetic acid is a weak acid in water and so in aqueous environments exists in equilibrium between its ionised and unionised forms. As I wrote above, it can react in either form. For the purposes of stoichiometry, the equilibrium makes no difference so for a titration calculation we'd usually write

CH3COOH + NaOH ----> CH3CO2Na + H2O​

but the reaction is actually occurring both when hydroxide ions encounter unionised acetic acid molecules and when hydroxide ions encounter hydronium ions.

In fact, there is a third mechanism for the reaction that involves the indicator, but the mechanisms don't matter to the stoichiometry as the overall net equation that describes the equivalence point can be summarised as the above.

Glacial acetic acid is so concentrated that it exists as almost exclusively unionised acetic acid molecules. Even pure acetic acid acid will possess species other than the unionised acetic acid molecule due to autoionisation, but that can be ignored for most purposes... though a difficult MCQ can be written involving conjugate pairs where it is presented with its conjugate acid.
in prior yrs, we define acid + base -> salt + water.
is an acid-base reaction now defined irrespective of products and only based on reagents?

"For the purposes of stoichiometry, the equilibrium makes no difference" - when does the equilibrium make a difference?

Actually, acetic acid is one of the comparatively uncommon substances that has two different conjugate acids.
huh?
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
in prior yrs, we define acid + base -> salt + water.
is an acid-base reaction now defined irrespective of products and only based on reagents?
There are different models that are used, chosen on which is most suitable.

I can look at a reaction like

CH3COOH + NaOH ---> CH3CO2Na + H2O​

using the model

acid + base ---> salt + water​

which I could describe as an Arrhenius theory interpretation, but I can also look at it from a Lowry-Bronsted perspective as

CH3COOH + OH- ---> CH3CO2- + H2O​

which fits with a model

acid 1 + base 1 ---> base 2 = conjugate base of acid 1 + acid 2 = conjugate acid of base 1​

Each of these is reasonable but has different emphasis and is suited to explaining different aspects of the process.

"For the purposes of stoichiometry, the equilibrium makes no difference" - when does the equilibrium make a difference?
If I put 25.00 mL of a monoprotic acid HA into a conical flask and add 0.1000 mol L-1 of sodium hydroxide from the burette and find 25.00 mL of the NaOH is needed to reach the end point, then the concentration of HA will be 0.1000 mol L-1 from stoichiometric calculations irregardless of whether HA is strong or weak. However, the pH at the equivalence point will depend on the nature of the acid (and thus on its equilibrium properties).

CM_Tutor said:
Actually, acetic acid is one of the comparatively uncommon substances that has two different conjugate acids.
huh?
Just like water, acetic acid undergoes autoionisation, reacting with itself to ionise:

2 CH3COOH <---> CH3CO2- + CH3COOH2+

and so acetic acid has a conjugate acid with formula CH3COOH2+.

However, there are two possible forms for this species:

1. CH3COOH can be protonated at the O atom of the hydroxyl group, so that the conjugate acid has a condensed structural formula CH3-C(=O)-OH2+, in which case the +ve charge is located on the O atom with one O-C and two O-H single bonds.

2. CH3COOH can be protonated at the O atom of the carbonyl group, so that the conjugate acid has a condensed structural formula CH3-C(=OH)+-OH, in which case the +ve charge is located on the O atom with one O=C double bond and one O-C single bond. Oxygen being much more electronegative than carbon, this will rearrange to CH3-C(-OH)2+, a carbocation where the C atom of the (former) carboxyl group has three single bonds (one C-C, two C-O) and a +ve charge.

So, there are two forms for the conjugate acid of acetic acid, each with the formula C2H5O2+ but with two distinct structures CH3-C(=O)-O+(-H)2 and CH3-C+(-OH)2, which will exist in equilibrium.
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
is an acid-base reaction now defined irrespective of products and only based on reagents?
How an acid-base reaction is defined really depends on what model is used for the definition. If I look online, I will find that the pKa for methane is about 50. If I do the calculation as you have learned it, I can calculate that a 0.0100 mol L-1 of methane has a pH of 26. This result is meaningless in the theory that you have covered which points to the model used to find the result is flawed in this situation... but it doesn't invalidate the calculation that a 0.100 mol L-1 solution of acetic acid (pKa = 4.76) has a pH of 2.88 at 298 K.

In some ways, the topic of acids and bases is an excellent example of an important aspect of models, extending way beyond chemistry.

In both science and mathematics, much early learning concerns learning about methods of problem solving and situations with a single answer that represents (in some sense) "truth". Mathematics helps us to learn that an answer can be found in multiple ways and thus that there are significant differences between methods and outcome.

Take a simple example like the equation .

I can square root to get , reject the possibility of being negative, square root again, and reach a solution . If I am dealing solely in cardinals, I have the single answer .

OR, I can factorise , recognise that this must be zero from the starting equation, and thus have solutions where , , and . I get the same result on rejecting being negative.

OR, from the factorisation, I can form two quadratic equations, and , each of which I can solve using the quadratic equation:



and


I can see my four solutions are and over or, over , the two solutions are , as was found above.

Of course, the earlier approaches will also find the solutions if the rejection of an is negative case is recognised as reasonable over but not over .

Which of these is "correct" or "true" depends on whether the outcome is considered over or over , but the answers are the same. Which of the methods is preferable has no universal answer as they differ in efficiency but not in accuracy.

In other words, one thing Maths is helping students to learn is that there are different ways to approach a problem which might be more or less simple / efficient but which still lead to the same outcome. This insight is just as important for any field of problem solving, and even one where there is no clear cut standard of accuracy / "truth" by which to determine an outcome. An essay on the exploration of a theme in an English set text can be thought of as a problem allowing different methods and different outcomes that can be assessed for the quality of the argument, persuasiveness of the evidence, and significance of the outcome in response to the problem.

Acids and bases allows students to explore some of these ideas in a scientific context. In learning about historical conceptions, Arrhenius theory, Lowry-Bronsted theory, and even Lewis theory, what is developed is NOT a progression from less accurate to more accurate theories. It is an exploration of theories that are more or less suitable to understanding and interpreting evidence through different contextual lenses. Some students may think that Lowry-Bronsted theory is "better" or more accurate / true than Arrhenius theory, but more able students will hopefully realise that each is useful for different contexts and so can be used selectively depending on contexts. There are plenty of contexts in which each is applicable and will yield comparable outcomes making each useful for that context without making either superior than the other. That a theory is inapplicable to situation A does not make its validity / utility to situation B any different, it just makes the theory not useful for A.

This idea arises all over the place. If I want to determine how high a projectile thrown at 10 m s-1 on Earth will rise, simple two dimensional motion will give a reasonable answer. The same problem with an initial speed of 10 km s-1 will need a model that includes other factors. If I want to launch a rocket to Mars, yet more factors are needed. Does this make the rocket model better for the simple thrown rock example? No - because the difference is in the applicability of the models and the reasonableness of their underlying assumptions. I am not going to model a car travelling at 60 km h-1 by rejecting Newtonian notions in favour of quantum mechanics because the differences in such a case will be trivial, no matter that quantum mechanics is arguable more "true" as it considers more factors. Similarly, I need not solve a problem where Arrhenius theory is perfectly applicable by using Lewis theory even if Lewis theory is a more broadly applicable theoretical framework.

Models are used as tools in a myriad of human contexts. Much early learning presents this as accessing "truth" without really studying limitations, usefulness, and the idea of what truth actually might be. The acids and bases topic can be treated as requiring guessing about which model is sought in a given circumstance, but is much more significantly trying to help students to recognise that models are not about truth nor about which is "correct" in a given circumstance. By nature, models are about usefulness, applicability and limitations, and judgement about when they can be used or whether what they produce is reasonable.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top