“Titration is the slow addition of one solution of a known concentration to a known volume of another solution of unknown concentration until the reaction reaches neutralization, which is often indicated by a color change.”
Basically, titration is the use of one chemical that you know the concentration of, to find the concentration of another chemical that you know the volume of. If you use an acid and a base, then you would add acid to the volume of base until they neutralized, which an acid/base indicator like phenolphthphthalphein (which is not recommended for use) could help show you.
From there, you could use the equation M1V1 = M2V2, which states that a substance of a known molarity and a known volume can be used to find the volume or molarity of that same substance when diluted, when you know the molar ratio in which the two chemicals react.
For example, suppose you had an antacid tablet–that nauseating, chalky magnesium bicarbonate stuff you take when you’re real sick. Normally, this shouldn’t dissolve; but let’s suppose it does, and you put 25mL of a solution into a flask. If you added a pH indicator like BTB to the tablet solution until it turned blue (and indicated a base), then dripped 12mL of nitric acid of a .75 Molar concentration into it until it turned greenish (and indicated the mixture was neutral), then you could find the concentration of the antacid solution. But you would also need the balanced equation for your work, which looks like this:
Mg(HCO3)2(aq) + 2HNO3(aq) –> Mg(NO3)2(aq) + 2H2O(l) + 2CO2(g). This tells you the antacid and acid mix in a 1:2 ratio.
First, we work with the amount of acid: (.75mol HNO3/1L) * (.012L) = .009mol HNO3
Next comes the molar ratio: (.009mol HNO3) * (1mol Mg(HCO3)2/2mol HNO3) = .0045mol Mg(HCO3)2
Finally, finding the concentration: (.0045mol Mg(HCO3)2) ÷ (.025L) = .18 Molar concentration. This means that there are .18 moles of magnesium bicarbonate in every one liter of the original solution.
Titration also helps when you’re adding a base to an acid, such as to determine the amount of citric acid in fruit juice. Suppose you start by taking 250mL of the fruit juice and like three drops of BTB, and assuming that everything but the citric acid is absolutely neutral. Then, drip in NaOH of 1.25 Molarity until the mixture changes from acidic yellow to neutral green, and find the amount of used–say, 589 mL. Then, balanced equation:
HC6H7O7(aq) + NaOH(aq) –> NaC6H7O7(aq) + H2O(l) (and assume citric acid has a single ionizable hydrogen atom)
So now, (.589L) * (1.25mol NaOH/1L) = .73625 mol NaOH
Then, (.73625mol NaOH) * (1mol HC6H7O7/1mol NaOH) = .73625 mol HC6H7O7
Last, (.73625 mol HC6H7O7) ÷ (.250mL) = 2.945 Molarity for the citric acid.
Or, if you wanted the mass of citric acid present, (.73625 mol HC6H7O7) * (192.124gHC6H7O7/molHC6H7O7) = 141 grams of citric acid in that 250 milliliter sample.
In conclusion, as you can see, titration is very useful.