A titration can define the volume of one solution required to react correctly with an identified volume of a different solution. Titration commonly comprises of reactions such as redox reactions and reactions involving precipitations, different than acid-base reactions.

“Titration can discern the volume of one solution required to react exactly with a known volume of a different solution.”

The equation for Titration Formula is articulated as:





1000 = factor relating mg to grams

W = mass of sample

N = normality of titrant

V = volume of titrant

Eq.wt = equivalent weight of acid


Though commonly the Titration Formula is articulated as:




V1 = Volume of titrant

N = Normality of titrant

V2 = Volume of sample

Eq.wt = equivalent weight of predominant acid


Solved Examples

Problem 1: Compute the titratable acidity if 17.5ml of 0.085N NaOH is required to titrate a 15ml sample of juice, the sum titratable acidity of that juice, articulated as percent citric acid. (molecular weight = 192; equivalent weight = 64)


Now we use the equation:



% of acid = 0.085×17.5×64/15×10 = 0.635%

Notice that the equivalent weight of anhydrous citric acid always is used in calculating and reporting the results of titration.


Problem 2: Contemplate the reaction of sulfuric acid H2SO4 with sodium hydroxide NaOH.

H2SO4 + 2NaOH à 2H2O + Na2SO4

How many millimeters of 0.250M NaOH must be added to react completely with the sulfuric acid? Assuming that a beaker contains 35.0mL of 0.175M H2SO4.


The calculation is as follows.



                                                                                        = 4.90 ×× 10-2 L NaOH

precisely 49.0 mL of 0.250M sodium hydroxide solution reacts with 35.0 mL of 0.175M sulfuric acid solution.

Practise This Question

An electric dipole consisting of two opposite charges of magnitude 2×106C  each, separated by a distance of  3 cm is placed in an electric field of 2×105N/C. The maximum torque on the dipole will be