Titration
Titration is an important laboratory procedure that is used in performing chemical analyses.  The apparatus that is employed is usually an Erlenmeyer flask and a burette.   The burette is along glass tube and it is marked in volumes, usually in increments of 0.10 mL.  In a typical analysis, a solution containing an unknown amount of substance to be analyzed is placed in the receiving flask.  Precisely measured volumes of a solution of known concentration - a standard solution - are then added from the burette.  This addition is continued until some visual effect, such as a colour change, signals that the two reactants have been combined in just the right ratio to give a complete reaction.
The valve at the bottom of the burette is called a stopcock, and it permits the analyst to control the amount of titrant (the solution in the burette) that is delivered to the receiving flask.  This permits the addition of the titrant to be stopped once the reaction is complete.
The titration procedure is applicable to many different kinds of reactions.  We will however focus our attention on acid-base titrations in which the reaction is neutralization.  When an acid-base neutralization is performed the analyst adds a drop or two of an acid-base indicator solution to the solution in the receiving flask prior to the start of the titration.
An acid-base indicator is a chemical dye that is capable of being one colour at a low pH and a different colour at a higher pH.  Two common types are litmus, which is red in acidic solutions and blue in basic solutions and phenolphthalein which is colourless in acids and pinkish-violet in basic solutions. 
In performing the titration, the analyst looks for a change in colour that signals that the solution has changed from acidic to basic (or basic to acidic, depending upon the nature of the reactants in the burette and flask.)    Usually this colour change is very abrupt --- as the end of the reaction is approached, the change in colour normally takes place with the addition of only one drop of the titrant.  Then the colour change finally occurs, delivery of the titrant is stopped -- the end point has been reached -- and the total volume of the titrant that was added to the receiving flask is recorded.
Sample Calculations
We will use the acid-base titration equation:      Ma * Va * Ca = Mb * Vb * Cb

where M is the molarity of the acid and base respectively.   V is the volume of the acid and base.  Ca and Cb  is the coefficient of the acid and base.  The coefficient of the acid and base is basically the number of acid hydrogens and hydroxides that are present.
 

For example:
Acids with a coefficient of "1"
HCl, HBr, HCH3COO, HNO3, HI, HClO3, HClO4

Acids with a coefficient of "2"
H2SO4, H2SO3, H2CO3

Acids with a coefficent of "3"
H3PO4

It should be very obvious that the coefficient and the number of acid hydrogens are the same.
 

Bases with a coefficient of "1"
LiOH, NaOH, KOH, RbOH

Bases with a coefficient of "2"
Mg(OH)2, Ca(OH)2, Sr(OH)2

Bases with a coefficient of "3"
Al(OH)3

It should be very obvious that the coefficient and the number of hydroxides are the same.
 

Titration Example #1
In a titration, a sample of H2SO4 solution having a volume of 15.00 mL required 36.42 mL of 0.147 M NaOH solution for complete neutralization.  What is the molarity of the H2SO4 solution?
Solution
Ma = ?                            Mb = 0.147 M
Va = 15.00 mL                Vb = 36.42 mL
Ca = 2                             Cb = 1
Ma = MbVbCb / VaCa
      =  (0.147 M * 36.42 mL * 1 ) / (15.00 mL * 2)
      =  5.354 / 30 M
      =   0.178 M

The sulphuric acid has a molarity of 0.178 M
 

Titration Example #2
In a titration involving phosphoric acid, H3PO4, solution having a molarity of 0.345 M and a volume of 20.00 mL is titrated against a sample of  Ca(OH)2.   If the reaction requires  22.25 mL of base to reach neutrality what is the molarity of the calcium hydroxide?
Solution
Ma = 0.345 M                Mb = ? M
Va = 20.00 mL                Vb = 22.25 mL
Ca = 3                             Cb = 2
Mb = MaVaCa / VbCb
      = (0.345 M * 20.00 mL * 3) / (22.25 mL * 2)
      = 20.70 / 44.5 M
      = 0.465 M

The phosphoric acid has a molarity of 0.465 M

Go to the Titration Worksheet
Go to the Solution Stoichiometry Review Worksheet