**Acid base titration determines the exact strength of acids and bases.** If the strength of acid is known the strength of the base can be estimated and vice versa by this acid base titration process.**The exact point of complete neutralization is called the equivalence point or end point where the pH stands at 7. This is true if both acid and base are stronger.** If the base is weak and acid stronger the pH will be slightly less than 7 and if the base is stronger and the acid weaker the pH is slightly more than 7. A large variety of indicators are known and a proper indicator is to be used for a proper range of pH.

Generally the indicators change their color in a significant manner in a pH range of less than 2 units. This range is called the color change interval of that particular indicator. A suitable indicator must be selected for a particular acid base titration which will indicate the color change at the nearest point of the neutralization or equivalence.

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When a weak acid is titrated with a strong base the curve is quite different in two important ways.

The titration curve of a weak base has an initial point at which the system is a weak base solution followed by a buffer region, then as equivalence point at which the system is a weak acid solution, and finally a region in which strong acid is in excess.

A titration curve is a plot of pH versus the amount of acid or base added. It displays graphically the change in pH as acid or base is added to a solution and shows how pH changes during the course of the titration.

As the base is added the acid is slowly neutralized. At first the change in pH is minimal. This resistance is due to the fact that the flask has a much greater number of H_{3}O^{+} ions than the OH^{-} ions available from the added titrant. As more and more base is added, more OH^{-} ions are added and thus more H_{3}O^{+} ions gets neutralized.

In this case the titration curve will be as follows.

In the titration of a weak acid and a weak base a titration curve is like

The change in pH at the end point is gradual and indicators will change color gradually. No indicator will give a sharp end point.

The way out of this difficulty is to titrate the weak acid against a strong base and the weak base against a strong acid. One of the important application of titration is the acid base titrations. It involves either the estimation of the strength of an acid or a base of unknown strength by titrating it against a base or an acid of known strength respectively. Suitable indicator is used to ascertain the correct end point of the titration.

There are several steps for the Acid base titration calculations. Since the mineral acids are in liquid form it is not possible to make a standard solution which is also called the primary standard solution with them. It is always required to standardize these acids with another titration.

Hence, these acids form secondary standards, with which the main titration of finding the strength of a given base is done. In the same way the alkalis like NaOH, KOH are hygroscopic solids and will not give the proper result. These are again unfit to make primary standards. They will be standardized by a primary standard made from organic acid like oxalic acid which is in crystalline form.

Example : Finding the strength of given hydrochloric acid (HCl) using oxalic acid [(COOH)_{2}.2H_{2}O] and NaOH.

**1. Calculations involved in making of Primary standard of Oxalic Acid**

Oxalic acid Eq.mass = Mol. mass /2. (126/2 = 63) = (a – b) / 63 N

**2. Calculations involved in making of Secondary standard of Sodium hydroxide solution**

**3. Calculations involved in finding the strength of Hydrochloric Acid**

Below you could see problems### Solved Examples

**Question 1: **If 0.5gram of a mixture of K_{2}CO_{3 }and Li_{2}CO_{3} requires 30ml of 0.25M acid solution for neutralization what is the percentage composition of the mixture.

** Solution: **

Let x = gram of K_{2}CO_{3} y = gram of Li_{2}CO3

Then x + y = 0.5

No. of meq of K_{2}CO_{3} present = $\frac{x}{K_2CO_3/2000}$ = $\frac{x}{0.060}$

No. of meq of Li_{2}CO_{3} = $\frac{y}{Li_2CO_3/2000}$ = $\frac{y}{0.0375}$

No. of meq of acid required = 30 x 0.25

Hence, x/0.69 + y/0.0375 = 30 x 0.25

Solving the above simultaneous equations we have

x = 0.247 and y = 0.253

Hence, percentage 100x/0.5 = 49.4% K_{2}CO_{3}

and 100y/0.5 = 50.6% Li_{2}CO_{3}

**Question 2: **A 1.2gram sample of a mixture of (Na_{2}CO_{3} + NaHCO_{3}) is dissolved and titrated with 0.5N HCl. With phenolphthalein the end point is at 15ml while after further addition of methyl orange a second end point is at 22ml. Calculate the percentage composition of the mixture.

** Solution: **

15 + 15 = 30ml acid is necessary to neutralize Na_{2}CO_{3} completely. Total volume needed = 15 + 22 = 37ml that is (37-30) = 7ml acid is needed for neutralizing NaHCO_{3}

Therefore, Na2CO3 composition (%) is $\frac{30 \times 0.5 \times 0.053}{1.2}$ $\times $100 = 66.25%

$\frac{7 \times 0.5 \times 0.042 \times 100}{1.2}$ = 24.50% NaHCO3

- Once the addition of strong base begins the solution is buffered before the equivalence point.
- The solution is basic at the equivalence point because a salt of a weak acid and a strong base undergoes hydrolysis to give a basic solution.

- Before any base is added the pH depends on the weak acid alone.
- After some base has been added but before the equivalence point a series of weak acid/ salt buffer solutions determines the pH.
- At the equivalence point, hydrolysis of the anion of the weak acid determines the pH.
- Beyond the equivalence point, excess strong base determines the pH.

In this case the titration curve will be as follows.

The titration curve of a weak base has an initial point at which the system is a weak base solution followed by a buffer region, then as equivalence point at which the system is a weak acid solution, and finally a region in which strong acid is in excess.

A titration curve is a plot of pH versus the amount of acid or base added. It displays graphically the change in pH as acid or base is added to a solution and shows how pH changes during the course of the titration.

As the base is added the acid is slowly neutralized. At first the change in pH is minimal. This resistance is due to the fact that the flask has a much greater number of H

In this case the titration curve will be as follows.

In the titration of a weak acid and a weak base a titration curve is like

The change in pH at the end point is gradual and indicators will change color gradually. No indicator will give a sharp end point.

The way out of this difficulty is to titrate the weak acid against a strong base and the weak base against a strong acid. One of the important application of titration is the acid base titrations. It involves either the estimation of the strength of an acid or a base of unknown strength by titrating it against a base or an acid of known strength respectively. Suitable indicator is used to ascertain the correct end point of the titration.

There are several steps for the Acid base titration calculations. Since the mineral acids are in liquid form it is not possible to make a standard solution which is also called the primary standard solution with them. It is always required to standardize these acids with another titration.

Hence, these acids form secondary standards, with which the main titration of finding the strength of a given base is done. In the same way the alkalis like NaOH, KOH are hygroscopic solids and will not give the proper result. These are again unfit to make primary standards. They will be standardized by a primary standard made from organic acid like oxalic acid which is in crystalline form.

Example : Finding the strength of given hydrochloric acid (HCl) using oxalic acid [(COOH)

- Weight of the dish with Oxalic acid before transferring it in to standard flask = 'a' grams
- Weight of the dish with oxalic acid after transferring it in to standard flask = 'b' gms.
- Exact weight of Oxalic acid = (a – b) gms.

Oxalic acid Eq.mass = Mol. mass /2. (126/2 = 63) = (a – b) / 63 N

- Volume of Oxalic acid solution taken ( Pipette reading) = V
_{ox}mL. - Normality of Oxalic Acid = (a – b) / 63 = N
_{ox} - Volume of NaOH solution ( Burette reading) = V
_{NaOH}_{ }mL - Normality of NaOH = N
_{NaOH }= V_{ox }X N_{ox }/ V_{NaOH}

- Volume of NaOH solution ( Burette reading) = V
_{NaOH}_{ }mL - Normality of NaOH = N
_{NaOH } - Volume of HCl taken ( Pipette reading ) = V
_{HCl} - Normality of HCl = N
_{ HCl}= V_{NaOH }X N NaOH_{ }/ V_{HCl} - Strength of HCl = N X Eq.mass ( For HCl Eq.mass= Mol.mass) = N
_{HCl}X 36.5 g in 1 L.

Below you could see problems

Let x = gram of K

Then x + y = 0.5

No. of meq of K

No. of meq of Li

No. of meq of acid required = 30 x 0.25

Hence, x/0.69 + y/0.0375 = 30 x 0.25

Solving the above simultaneous equations we have

x = 0.247 and y = 0.253

Hence, percentage 100x/0.5 = 49.4% K

and 100y/0.5 = 50.6% Li

15 + 15 = 30ml acid is necessary to neutralize Na

Therefore, Na2CO3 composition (%) is $\frac{30 \times 0.5 \times 0.053}{1.2}$ $\times $100 = 66.25%

$\frac{7 \times 0.5 \times 0.042 \times 100}{1.2}$ = 24.50% NaHCO3

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