Commenting on enthalpy change of neutralisation.

The definition of the enthalpy change of neutralisation is the heat evolved when one more of water is formed between an acid and an alkali. Essentially, it is the reaction between 1 mole of H⁺ and 1 mole of OH^- to give 1 mole of H₂O.

Typically, enthalpy change of neutralisation between a strong acid and a strong base is -57 kJ mol^-1. While, if it is between a weak acid and a strong base, or a strong acid and a weak base, the enthalpy change of neutralisation will be slightly less exothermic than -57 kJ mol^-1. This is largely because some amount of heat is taken in my the molecule to allow for complete acid (or base) dissociation.

1. Interesting question:

However, when we react 1 mole of H₂SO₄ with 2 moles of NaOH, do we get a more exothermic enthalpy change of neutralisation? The answer is NO.

This is because the reaction between sulfuric acid and sodium hydroxide produces 2 moles of water. This implies that the enthalpy change for this reaction is twice of that of enthalpy change of neutralisation; since the latter is defined to be per mole of water formed.

Therefore, when we obtain -114 kJ mol^-1 (which is more exothermic than -57 kJ mol^-1) for the reaction between 1 mole of sulfuric acid and 2 moles of sodium hydroxide, this value refers is the enthalpy change of reaction. Hence, when this number is divided by two because two moles of water is obtained, we will get enthalpy change of neutralisation.

2. Application - in terms of planning an experiment:

In conclusion, we often use calorimetry method to typical to determine the enthalpy change of neutralisation of an acid and a base. However, in using energetics in determining strength of acid, we are essentially trying to determine enthalpy change of reaction.

For example, when the reaction between sulfuric acid and sodium hydroxide is compare it with ethanoic acid and sodium hydroxide (and both acids are equimolar), it is important to ensure that the amount of hydroxide used is in excess. It is also good to ensure the volumes of hydroxide used are the same - this facilitates the comparison.

The former will produce an enthalpy change of reaction that is slightly more than twice of that of the latter; resulting in the former to have a temperature change that is twice as much as the latter.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

• ^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
• ^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Ionic equilibrium - deciding on end point

Carbonic acid, H₂CO₃, is an example of a weak dibasic acid. As such, carbonic acid has two end points when titrated with NaOH using the a pH meter. Hence, it can be slightly confusing when deciding on which equilibrium to use to calculate the pH of the first end point.

1. Question:

Do we take equilibrium B, which shows us HCO₃^- dissociating into H⁺ and CO₃^2-? Or do we take equilibrium C, which shows us HCO₃^- functioning as a conjugate base?

The choice is simple, in equilibrium C, the equilibrium constant, K_b is larger than the equilibrium constant for equilibrium B. Therefore, it is likely that HCO₃^- will function as a base rather than an acid. Hence, the pH of the end point A, will give us an alkaline pH.

However, when we add more NaOH, such that we obtain point B. At that point, we have only HCO₃^- and CO₃^2-, thus to calculate the pH, we are essentially making use of equilibrium B. At that point, pH = -lg(4.8 x 10^-11) > 9. In addition, the addition of NaOH will inhibit the forward reaction of equilibrium C. Hence, only equilibrium B is occuring.

2. Application to similar patterns:

Therefore, when we are dealing with a species that has two possible equilibria, (i.e it can function as an acid or a conjugate base), we will need to see which equilibrium (the acid dissociation or the base dissociation) has the higher equilibrium constant. The equilibrium which has the higher equilibrium constant will be the one that we choose to calculate the pH at that equivalent point.

In conclusion, do you notice that I mentioned we will be able to see two equivalent points when we use a pH meter. Why do you think it is not suitable to use phenolphathelein to detect the two end points?

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

• ^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
• ^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

ionic equilibrium titration curves

Titration curves are informative. They are able to give us information about the bascity of the acid, the volume of base required for neutralisation. the pH of the solution and the type of indicator which should be used There are a few situations to take note when we sketch a titration curve.

1. General points to take note.

In drawing the titration curve, we will need to take note what solution would be in the conical flask and what will be in the burette. If an acid is in the conical and the base is in burette, the pH curve will show an increasing pH as the solution in the burette is added into the conical flask.

In sketching the titration curves, the following are points to take note.

• The shape of the curves (e.g is it moving from low pH to high? the number of equivalence points?)
• The volumes where neutralisation has occurred (i.e. the verticals of the graph).
• The volumes and the pH of the effective buffer region. This requires you to check which point the number of moles of weak acid is the same as its conjugate base (or vice versa).
• For the final point, do not assume that the effective buffer capacity is before neutralisation.

2. Examples of weak dibasic acid and weak diacidic base.

Organic bases and acids are generally examples of weak base and weak acid respective. For example CH₃NH₂ and CH₃COOH respectively. The former is a base because N has a lone pair of electrons to form a dative bond with H⁺, hence making it a proton acceptor (thus a base). While the latter has a -COOH which enables a H⁺ (thus an acid) to be donated. The above are examples of organic moles which are dibasic acid and diacidic base.

Using dibasic acids as an example, these substances are able to donate two H⁺. However, the protons are not donated at the same time. The acid (malic acid = HOOCCH₂CH(OH)COOH) will dissociation once first to form HOOCCH₂CH(OH)COO^-, then the acid will dissociate a second time (from HOOCCH₂CH(OH)COO^-) to give ^-OOCCH₂CH(OH)COO^- and H⁺.

Each dissociation has its own acid dissociation constant with the first acid dissociation constant generally larger than the second.  The easiest reason is that the second acid dissociation is the removal of a H⁺ from an anion; which will be more difficult than removing a proton from a neutral acid molecule.  

3. Titrating a weak monobasic acid against a strong base

When CH₃COOH placed in a conical flask and NaOH is added to the acid. The pH increases as more NaOH is added and that is due to decreasing amount of the weak acid CH₃COOH and increasing amount of conjugate base CH₃COO^-. At the point where n(CH₃COOH) = n(CH₃COO^-), that is the maximum buffer capacity. At 2A cm³ of NaOH added, all the acid is used up and only the conjugate base is left, hence the pH is greater than 7.

4. Titrating a weak diacidic base against a strong acid.

When titrating a weak diacidic base against a strong monobasic acid, with both having the same concentration. The volume required for the monobasic acid to completely react with the two basic groups in the diacidic base is twice the volume of the base used. In addition, the effective buffer capacity will be 14 - pK_b values.

However, when we titration against a strong dibasic acid, the volume required by the dibasic acid is the same as the volume of diacidic base used. This is when both acid and base have the same concentration. The pH of the effective buffers remain the same.

5. Titrating a weak dibasic acid against a strong base.

When titrating a weak dibasic acid against a strong monoacidic base, with both having the same concentration. The volume required for the monoacidic base to completely react with the two acidic groups in the dibasic acid is twice the volume of acid used. In addition, the effective buffer capacity will be the pK_a values.

However, when we titration against a strong diacidic base, the volume required by the base is the same as the volume of acid used. This is when both acid and base have the same concentration. Note that the pH of the effective buffers remain the same. 

6.  Conclusion

Hence, when sketching the titration curves, it is good advice to think about how the titration is done.  Do not rush through with fixed model, e.g. the point where maximum buffer capacity exists is always between the starting point and the equivalence points.  Always fist ask yourself what is in the solution at any particular significant point on the curve.

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^{Article written by Kwok YL 2010. (edited in 2011)}

^{Disclaimer and remarks:}

• ^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
• ^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}