## Titration

A titration is a lab procedure where a precise amount of solution (the titrant) in a burette is added to a measured volume of sample solution. Its purpose is to determine the amount of a specified chemical in the sample.

The addition of the titrant continues until the equivalence point is reached, which is when the acid and base completely react with each other. In other words, both are the limiting reactant. In a real titration, we stop adding titrant when a sharp change in colour occurs—this is called the endpoint, and ideally it happens at the same time as the equivalence point. In practice, though, it is always slightly off.

The equivalence point occurs at the middle of the sharp increase or decrease in pH on the titration curve. The best indicator for any given titration is the one whose colour change pH range is centred around the pH of the equivalence point. In the graph below, an indicator that changes colour at around 7 would be ideal, but we can usually get away with phenolphthalein (it changes colour on the pH interval 8–10).

The pH of the equivalence point depends on the type of titration:

• strong acid with strong base: "pH" = 7
• weak acid with strong base: "pH" > 7
• weak base with strong acid: "pH" < 7

After completing a titration, we can calculate the missing piece of information using

V_"a"c_"a" = V_"b"c_"b",

where the known quantities are the concentration of titrant, the volume of titrant added, and the original volume of sample. The unknown variable is usually the concentration of the sample.

### Example

In a titration, 10.00 mL of 0.200 mol/L HCl(aq) is titrated with standardized 0.250 mol/L NaOH(aq). Find the amount of unreacted HCl(aq) (in moles) and the pH of the solution

• before beginning the titration;
• after 2.00 mL of NaOH(aq) has been added.

In the beginning, the amount of HCl(aq) is given by

n = Vc = (0.0100\ "L")(0.200\ "mol/L") = 2.00\ "mmol",

and we can calculate pH with

"pH" = -log["H"^+] = -log(0.200\ "mol/L") = 0.6990.

When 2.00 mL of NaOH(aq) has been added, there is

n = Vc = (0.00200\ "L")(0.250\ "mol/L") = 0.500\ "mmol"

of NaOH(aq), and since each hydroxide ion reacts with each hydrogen ion, the amount of H+(aq) that remains is

2.00\ "mmol" - 0.500\ "mmol" = 1.50\ "mmol",

and the pH of the solution at this point is

"pH" = -log["H"^+] = -log((1.50\ "mmol")/(10.00\ "mL" + 2.00\ "mL")) = 0.9031.