So! Earlier today.. Somebody asked this question using someone else’s phone.. And then to add to my bemusement, I realized that there were (at least) three people involved not two.. So strange.. But nvm.. Back to the topic at hand..
Summary for the super-fast people:
- If you understand that logarithm results in a compressed scale (i.e. lg(10) = 1, lg(100)=2 etc) you should be able to deduce the shape of the titration curve pretty quickly if you know the definition of pH
For the rest of us plebeians, there are a couple of things to bear in mind while you read this post:
- There is some simple math involved so have a calculator beside you so that you work out the values with me as you go along
- Remember that the pH of a solution is defined as the negative log10 of the hydrogen ion concentration in mol/dm3
Ok! Let’s start with a little bit of simple mole concept first. Imagine that a container contains 1dm3 of 0.1mol/dm3 HCl. What volume of 0.1 mol/dm3 NaOH is needed to neutralize it completely?
A quick calculation should tell you that the volume of NaOH needed is also 1dm3. This value is important for understanding what happens before, during and after the point of complete neutralization!
The table below summarizes the 6 key plot points in the titration curve
Plot point 1 – What is the initial pH of the solution?
At first, no NaOH is added, thus total volume in container comes from the acid which is 1dm3
Mole of H+ ions at first = conc X volume = 0.1 X 1 = 0.1
Concentration of H+ ions in the container = 0.1 /1 = 0.1
pH of solution = – lg(0.1) = 1
Plot point 2 – What is the pH at the half-way stage of neutralization?
Earlier, we calculated that the volume of NaOH needed to completely neutralize the HCl was 1dm3. So, the half-way point of neutralization occurs when 0.5dm3 of NaOH has been added. A common misconception that many students have is that the solution now has a pH of 4. How did they get this value? (A completely neutralized solution has pH of 7 while the initial pH was 1, so at the half way mark, the pH is 4). See graph below.
This is WRONG. The reason is due to the math involved in calculating pH.
After adding 0.5dm3 of NaOH, Mole of H+ ions left = 0.05 (as half the acid has been neutralized)
Concentration of H+ ions in the container= 0.1 /1.5 = 0.0333
(Note: Total volume of solution in container = 1.5dm3. 1dm3 originally from the acid, 0.5dm3 from the NaOH added)
pH of solution = – lg(0.0333) = 1.48
Notice that even at the half-way point, the pH is still quite low and not at pH 4 that you expected!
Plot point 3 – occurs when three-quarters of the acid has been used up. You can work the math out on your own
Plot point 4 – occurs just before the point of complete neutralization
When 0.999dm3 of NaOH has been added, the solution is almost completely neutralized. Work out the pH value on your own and you should realize that the pH is still quite low at 3.3
So, plotting the first 4 points on the graph, you should get the following graph below! See how the curve only increases gradually?
Plot point 5 – point of complete neutralization
At point of complete neutralization, there are no more H+ ions left in the solution, and the concentration of the H+ ions is zero.
pH of solution = – lg(0) = ERROR 2
So why is it we say that the pH of the neutral solution is 7?
This is because, water, which is a polar molecule is able to auto-ionize very slightly on its own to give H+ and an OH– ions. Empirically it has been found that for every dm3 of water, there are around 0.0000001 moles of H+ and 0.0000001moles of OH– ions. Hence, the concentration of H+ ions is 0.0000001mol/dm3 and the pH of the solution which is calculated by – lg(0.0000001) = 7. (Note that I have ignored the OH–. Because of the definition of pH, I am only interested in the H+)
Plot point 6 – When NaOH is in excess
The simple way of understanding this portion is that, since NaOH is in excess, and since NaOH is a strong alkali, the pH will increase to around 12 or 13.
The excess NaOH reacts with the H+ ions formed by the auto-ionization of water to further decrease the amount of H+ present in the solution. BUT the amount of H+ is never 0 because water will always auto-ionize, so no matter what happens there will be a tiny bit of H+ ions present in the solution. To give you an indication of how tiny, a pH of 13 means that the concentration of H+ ions in the solution is 0.0000000000001mol/dm3!
I hope this helps to clear the mystery and alleviate the misery!
Curiosity is a wonderful thing and should be encouraged. For instance, questions such as why is aluminium chloride considered to be a covalent compound when it is made up of a metal and non-metal, deserve to be addressed even if they are not part of the requirements of the ‘O’ Level syllaus.
However, due to time constraints, I am often unable to answer these types of questions in class. Furthermore, not everybody wants to know about about these (wonderful) things especially since it won’t affect the grade. They are the ones who.. you know.. still believe that curiosity killed the cat..
But if you ask me.. They’re just pussies.
Anyways.. After much consideration, I’ve come to the conclusion that even dinosaurs must move into the digital age and hence this abomination of a blog. So I hope you guys read, enjoy and ask me some (chem-related) questions!
Begin by scrolling your mouse over the menu/topics header to view the list of subjects covered!
PS: A note to JC students who have stumbled onto this page accidentally or otherwise. As this page is primarily targeted at O level students who wish to learn more, I have intentionally reduced the usage of chem jargon and (slightly) simplified some concepts. Do not be alarmed if the explanations here differ slightly from what your JC lecturers preach. Use their version in your answers!
3. Doc Brown
Under the “O” level syllabus, all salts are considered to be neutral unless otherwise stated. However, in reality, there are actually acidic, basic salts as well as neutral salts.
Let me explain basic salts first, as it builds on topics I’ve covered in previous posts.
What are basic salts?
A basic salt is a salt that dissolves in water to produce a solution with a pH of more than 7. In other words, there must be more OH– ions than H+ ions present in the solution
How are they formed?
When a strong base reacts with a weak acid.
For instance, when sodium hydroxide (strong alkali) reacts with ethanoic acid, the salt formed, sodium ethanoate, is a basic salt.
Why is it basic?
When sodium ethanoate is dissolved in water, it dissociates to form 2 ions, Na+ and CH3COO–.
Recall what you learned in the previous post about the behaviour of the Na+ ion in water. Due to its low charge density of the Na+, it is unable to hydrolyse (break-up) the water molecule, hence will produce few if any H+ ions.
Recall what you learned in the previous post about the behaviour of the ethanoate ion (CH3COO-) in water. It forms an alkaline solution with plenty of OH– ions.
Thus, there will be more OH– ions than H+ ions in a solution containing dissolved sodium ethanoate and the solution is said to be basic/alkaline.
How about acidic salts?
An acidic salt is a salt that dissolves in water to produce a solution with pH less than 7. In other words, there must be more H+ ions than OH– ions in the solution
How are they formed?
- When a strong acid reacts with a weak base. (E.g. HCl + NH4OH, the salt formed, NH4Cl will be an acidic salt)
- When the salt contains a cation with high charge density. (E.g. AlCl3)
Why is it acidic?
Let’s talk about scenario 1 first!
HCl is a strong acid that dissociates completely in water to produce H+ and Cl– ions. This implies that the Cl– ion has relatively weak attraction to the H+ as it is willing to give it up easily. Hence, when dissolved in water, the Cl– ion, is unable to hydrolyse the water molecule to produce any OH– ions.
Put another way, the Cl– ion which is negative is attracted to the partial positive H atom of the water molecule. It is unlikely to react with water as it has relatively low charge density. Even if it reacts with water to produce HCl and OH–, since the HCl formed is a strong acid, it dissociates immediately to form back H+ and Cl–. The H+ then reacts with the OH– ion to form H2O. In other words, no net OH– ions are produced.
On the other hand, when the ammonium ion is dissolved in water, it reacts with the water to form H3O+ (which is commonly written as H+ in most textbooks*).
NH4+ + H2O ⇌ NH3 + H3O+
Hence, there will be more H+ than OH– ions in a solution containing dissolved ammonium chloride and the solution is said to be acidic.
Onwards to scenario 2!
When the metal ion has high charge density (E.g. Al3+), it is able to hydrolyse the water molecule to produce H+ ions. The Al3+ which has high charge density is attracted to the partial negative end (O) of the water molecule. It hydrolyses the water molecule to form the aluminate ion (Al(OH)4–) and 4 H+ ions**. The key point to note is that the OH– ions are covalently bonded to the Al3+ to form one large complex ion but there are 4 H+ ions. See diagram below.
However, the Cl– ion which has relatively lower charge density is unable to react with water to produce OH– ions as discussed above. Hence, there are more H+ ions than OH– ions in a solution containing dissolved AlCl3 and the solution is said to be acidic.
Likewise, if the anion has high charge density while the cation has low charge density, the resulting solution is most likely to be basic.
Ah ha! I hear voices again.. (not that its good thing mind you.. ) But.. What if the charge density of the cation and anion are both high? In that case, the ionic bonds between the ions will be very strong and it will probably not dissolve in water as it requires too much energy to break. (See previous post about why ionic compounds are soluble in water)
As the name suggests, neutral salts dissolve in water to produce a solution with pH 7. This means that there are equal amounts of H+ and OH– ions in the solution.
How are they formed?
They are formed when a strong acid reacts with a strong alkali. (E.g. HCl + NaH, the salt formed, NaCl will be neutral).
Tell me why! (Ain’t nothing but..)
As discussed in the earlier sections, due to its low charge density of the Na+, it is unable to hydrolyse (break-up) the water molecule, hence will produce few if any H+ ions. Likewise, the Cl– ion which also has relatively low charge density is unable to react with water to produce OH– ions. Hence the solution is neutral!
How about the salts formed by a weak acid and a weak base? Are they acidic/basic/neutral?
The solution can be acidic, basic or neutral depending on the relative strengths of the cation and the anion since both the cation and anion are able to hydrolyse the water molecule to produce H+ and OH– ions respectively.
*Note: The true form of H+ is H3O+. That is because when an acid dissociates in water, it is actually forming bonds with water, hence H3O+. But to simplify things, most textbooks will just put H+.
*For A level students, I have deliberately removed some jargon like conjugate acid/base to simplify the discussion. But you should know that the conjugate base of a strong acid is weak and vice versa, etc.
**For A level students, you should know that this is not strictly speaking true. It should form hydrated aluminium ions (hexaqua complex) first then form H+ ions when the complex is deprotonated. It need not necessarily form the aluminate ion. This is just a simplification to make it easier for O level students to understand.
As you know, when ethanoic acid (CH3COOH) is dissolved in water, the solution formed will be acidic due to the presence of the dissociated H+ ions. But what happens if you dissolve a CH3COO– ion by itself into water. Will the solution be acidic/alkaline/or neutral?
Tell me why! (Ain’t nothing but…)
Again, we need to make some reference to your existing knowledge. As you know, ethanoic acid is a weak acid and dissociates partially in water to produce H+ ions and CH3COO– ions.
The fact that it dissociates only partially implies that the CH3COOH is very unwilling to give up its H+ yes? Put another way, the CH3COO– ion has relatively strong attraction to the H+ and is unwilling to give it up.
So, when you dissolve the CH3COO– ion by itself into water it is attracted to the H atom of the water (Remember that water is a polar molecule with permanent partial positive charges on the H and partial negative charge on the O).
It is then able to hydrolyse (break apart) the water molecule, stealing the hydrogen for itself to form a CH3COOH molecule, leaving behind an OH- ion causing the solution to be alkaline!
In equation form:
CH3COO– (aq) + H2O(l) ⇌ CH3COOH(aq) + OH– (aq)
Ah ha! I know what you guys are thinking… Since I form CH3COOH, the solution will be acidic no? Imagine this situation…
I dissolve 100 CH3COO- ions into water.
According to the equation above, I should get 100 CH3COOH molecules and 100 OH- ions. (Stage1)
As you know, CH3COOH is a weak acid, so it dissociates partially in water. So out of 100 CH3COOH molecules, maybe 3 will dissociate to give me 3 CH3COO– and 3H+. (Stage2)
So, now, the solution contains 3 H+, 3 CH3COO–, 97 CH3COOH molecules (from stage 2) and 100 OH– ions (from stage 1). Since there are more OH– ions than H+ ions, the solution is alkaline! Mind reading complete!
Based on feedback from T1, I realize that perhaps, the table form isn’t the best presentation.. Perhaps the diagram below would help you understand better?
Thank you Ms S (Sec3, 2013) for asking such an interesting question & T1 for her feedback!
As you guys know, the key idea in solubility is that “like dissolves like”. In other words, compounds with the same type of bonding tend to be soluble/miscible with one another. For instance iodine, a simple covalent molecule would be soluble in organic solvents like hexane (another simple covalent molecule).
However, why are ionic compounds soluble in water?
That is because water is a polar molecule. The positive end (H) of the water molecule is attracted to the anions on the surface of the solid ionic compound while the negative end (O) is attracted towards the cations, forming bonds with the anions and cations respectively. (Fancy A level term: ion-dipole interactions). See diagram below.
Bond formation results in the release of energy. This energy is transferred to the cations and anions making them vibrate faster eventually breaking the ionic bonds between the cations and anions causing them to break free from the ionic lattice and enter the solution! See diagram below. Note the ions breaking away from the lattice!
However, not all ionic compounds are soluble in water. This is because in certain cases, the energy released from the ion-dipole interactions is insufficient to break the strong ionic bonds between the cations and anions.
This explains why compounds like MgO are insoluble in water (energy released from ion-dipole interaction is insufficient to break the “double ionic bond” between Mg2+ and O2-).
At ‘O’ levels, we accept that aluminium oxide is amphoteric because it is able to behave as a base in the presence of an acid and behave as an acid in the presence of an alkali.
But, we know that metal oxides are basic. So, what gives rise to this special property?
In order to answer this question, we first need to understand why metal oxides are basic.
Key idea to bear in mind throughout:
- Charge density of the ion: Ions with high charge density are able to cause the water molecules to break apart (hydrolyse) into H+ and OH– ions.
Why are metal oxides basic?
The simple answer is that they react with water to produce an alkali. For example:
When Na2O is dissolved in water, there are two ions formed, Na+ and O2-. Na+ has a relatively low charge density, hence only forms ion-dipole interactions with water but it does not hydrolyse the water molecule.
On the other hand, O2- ion has high negative charge density. Hence it is strongly attracted to the partially positive H atom of the water, such that it is able to break the H – O bond in water to form 2 OH– ions!
Why is Al2O3 amphoteric?
As discussed above, the presence of the O2- ion makes Al2O3 basic (i.e. able to react with acid). However, the Al3+ ion has high positive charge density, hence is strongly attracted to and reacts with the electron rich OH– ions present in alkalis to form aluminate ions [Al(OH)4–]. See below.
Why is the OH – ion “electron-rich”? Recall the bonding in sodium hydroxide:
There is a shared pair of electrons between the hydrogen and oxygen atoms in the hydroxide ion. Since oxygen is more electronegative than hydrogen, the shared pair of electrons within the ion are pulled closer to the oxygen atom, making the oxygen atom “electron rich”.
The aluminium ion having high positive charge density attracts and forms dative covalent bonds with hydroxide ions to form the aluminate ion! See diagram below