A titration can determine the volume of one solution required to react exactly with a known volume of another solution. Titration frequently involve the reaction other than acid-base reactions, such as redox reactions and reactions involving precipitations.

Titration is a method of finding out the strength of a solution either in the terms of molarity or normality or molality or acidity or alkalinity or precipitatability.

Titration can also be used to find out the presence of elements either in their elemental state or in their compound state. Some substances get adsorbed on the surface of other substances and this can also be estimated by the process of titration.

The equation for

- N = normality of titrant
- V = volume of titrant
- Eq.wt = equivalent weight of acid
- W = mass of sample
- 1000 = factor relating mg to grams

- N = normality of titrant
- V
_{1}= volume of titrant - Eq.wt = equivalent weight of predominant acid
- V
_{2}= volume of sample

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Solved problems based on titration are given below. ### Solved Examples

**Question 1: **Calculate the titratable acidity if it takes 17.5ml of 0.085N NaOH to titrate a 15ml sample of a juice, the total titratable acidity of that juice, expressed as percent citric acid. (molecular weight = 192; equivalent weight = 64)

** Solution: **

Now we use the equation

% of acid = $\frac{0.085\times17.5\times64}{15\times10}$ = 0.635%

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

**Question 2: **Consider the reaction of sulfuric acid H_{2}SO_{4} with sodium hydroxide NaOH.

H_{2}SO_{4} + 2NaOH $\rightarrow$ 2H_{2}O + Na_{2}SO_{4}

suppose a beaker contains 35.0mL of 0.175M H_{2}SO_{4}. How many millimeters of 0.250M NaOH must be added to react completely with the sulfuric acid?

** Solution: **

The calculation is as follows.

35.0 $\times$ 10^{-3} L H_{2}SO_{4} $\times$ $\frac{0.175\ mol\ H_{2}SO_{4}}{1L\ H_{2}SO_{4}}$ $\times$ $\frac{2mol\ NaOH}{1mol\ H_{2}SO_{4}}$ $\times$ $\frac{1 L\ NaOH}{0.250mol\ NaOH}$

= 4.90 $\times$ 10^{-2} L NaOH

Thus 35.0 mL of 0.175M sulfuric acidsolution reacts with exactly 49.0 mL of 0.250M sodium hydroxide solution.

Now we use the equation

% of acid = $\frac{0.085\times17.5\times64}{15\times10}$ = 0.635%

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

H

suppose a beaker contains 35.0mL of 0.175M H

The calculation is as follows.

35.0 $\times$ 10

= 4.90 $\times$ 10

Thus 35.0 mL of 0.175M sulfuric acidsolution reacts with exactly 49.0 mL of 0.250M sodium hydroxide solution.

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