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 log
_{10}of the hydrogen ion concentration in mol/dm^{3}

Ok! Let’s start with a little bit of simple mole concept first. Imagine that a container contains 1dm^{3} of 0.1mol/dm^{3} HCl. What volume of 0.1 mol/dm^{3} NaOH is needed to neutralize it completely?

A quick calculation should tell you that the volume of NaOH needed is also 1dm^{3}. 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 1dm^{3}

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 1dm^{3}. So, the half-way point of neutralization occurs when 0.5dm^{3} 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 0.5dm^{3} 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 = 1.5dm^{3}. 1dm^{3} originally from the acid, 0.5dm^{3} 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 0.999dm^{3} 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/dm^{3} 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/dm^{3}!

I hope this helps to clear the mystery and alleviate the misery!

It actually explains the shape of potentiometric titration curve too, there’s also a logarithm in the Nernst’s equation.

Great article!