ACID-BASE TITRATION

There are two ways to make a solution of known molarity. The first and most convenient way to make the solution, by dissolving exactly 1 mole of solute in a litre (dm3) of solution. The second way is to make up a solution quickly, using an estimated amount of solute and an estimated amount of solution, and then determine the solution's exact molarity by titration.

Titration is the chemical process by which we can determine the concentration of unknown solution, that reacts with a standard solution, whose concentration is known.

Titrations play a major role in determining, amounts of solutes present in a solution. Titration process is an important tool of the analytical chemist.

9.6.1 Molarity (M) and Molar Solution:

The most generally useful means of expressing a concentration of solution is molarity.

It is defined as the number of moles of solute dissolved per 1 litre (1dm3) of a solution; it is denoted by (M).

Thus 1 mole of H2SO4 (i.e. its gram formula mass) 98g dissolved in one litre (1dm3) of solution is said to be 1 molar (1M) solution, if only one-half the mole i.e. 49g of H2SO4 are dissolved in one litre of solution, the solution is said to be one-half molar (i.e. 0.5 M)

The molarity of any solution is found by dividing the number of moles of solute by the number of moles/litres of the solution.

Molarity (M)   =       Moles of solute/Litres of solution

We known that,

(i)       Number of moles      =       Mass of solute/ Gram formula mass

(ii)      Litres of solution       =       Volume of solution in cm3 (ml)/1000

Molarity (M) = Mass of solute/ Gram formula mass x 1000/ Volume of solution in cm3 (ml)   

For example:

(i) Calculate the molarity of a solution, containing 1.5g of NaOH in 250 cm3 of solution?

Solution

Molarity (M) = Mass of solute/ Gram formula mass x 1000/ Volume of solution in cm3 (ml)

Data  

(i)       Molarity                                               =       M       = ?

(ii)      Mass of solute (NaOH)                            =       l.5g

(iii)     Gram formula mass of solute (23+16+1) =       40g

(iv)     Volume of solution = 250 cm3

M = 1.5g x 10000/40 x 250 = 0.15

Result:   Molarity of NaOH solution is = M = 0.15M

Example. 2

i) What mass of (NaOH) must be dissolved in 500cm3 of solution to make 1.5 M-solution?

Solution

M       =       Mass of solute/ Gram formula mass x 1000/cm3 of solution

By cross-multiplication, we get,

Mass of solute =    M x G.F. mass x cm3 of solution/1000         

(i)       Mass of solute (NaOH)                  =       ?

(ii)      Molarity                                      =      M       =       1.5 M

(iii)     G.F mass of solute (23+16+1)       =       40 g

(vi)     Cm3 of solution                            =       500 cm3

Mass of solute = 1.5gx 40 x 500cm3 1000 cm3/1000cm3

Result mass of solute i.e. NaoH = 30g

9.6.2 Preparation of Solution of Known Molarity:

For example, how to prepare 1 M-solution of NaOH

To prepare 1M solution of NaOH you first weigh put one mole of NaOH i.e. 23 +16 + 1 - 40g of NaOH, place 40g of NaOH in a 1 litre (1 dm3) volumetric flask, and add some water to bring the level of solution to calibration mark as shown in figure given below, finally the solution is shaken until uniform

9.6.3   Standard Solution:

A solution whose molarity or strength is known is called standard solution. For example 1M.KOH solution, contains 1mole of (KOH) i.e. formula mass expressed in grams =39+16+1 = 56g of KOH are dissolved in 1 litre of solution is said to be 1 molar (1M). If we use only one-half mole i.e. 28g of KOH in 1-litre of solution our solution would be one-half molar i.e. 0.5 M.

9.6.4 Acid-Base Titration:

In acid-base titration, a solution of known concentration (say base) is added gradually to a solution of unknown concentration (say an acid) so as to determine the concentration of unknown solution. The point at which the reaction is completed is called the end point.       

Steps for Carrying out Titration:

1.    Fill the burette with the given solution of NaOH; read and record the initial reading of the burette (Remember to read the lower meniscus) if the burette is not full refill it with NaOH solution. Then let the burette drain into beaker down to the zero mark before using it. The solution in the burette is called "Titre".

2.    Pipette out 10 cm of HCl in a conical flask, (titration flask) and add one or two drops phenolphthalein indicator. The solution in titration flask is called "Titrant". A suitable acid-base indicator, such a methyl orange, litmus or phenolphthalein is used to detect the end point.

S.No.

Indicator

Colour in acids         

pH-range

Colour in bases

l.

Methyl orange

Red

3 - 5

Yellow

2.

Litmus

Red

6-8

Blue

3.

Phenolphthalein

Colour less

8- 10

Pink (red)

 

3.    Add slowly the NaOH solution, from the burette into the conical flask (titration flask) with-constant shaking. Stop adding the NaOH solution, when the mixture in the titration flask becomes light pink. Record the final burette reading. If you have been adding the reactant rapidly and think you have over run the end point, repeat the entire titrations slowing down when you think you are nearing the end point. Rinse out the flask with water before the next sample is titrated.

4.    Record the initial and final burette readings for each experiment.

To see how titration works, let us consider that we have (HCL) solution acid) whose concentration we want to find by allowing it to react with a BI>   lard base such as NaOH.

We begin- the titration by measuring out a known volume of HCL in a conical flask (titration flask) and adding few drops of an indicator, such as phenolphthalein (colourless liquid in acidic solutions but turns pink colour in basic solutions).

Next, we fill a burette with the(NaOH) standard solution (of known concentration) and we slowly add the NaOH to the HCL, until the phenolphthalein just begins to turn pink, indicating that all the HCL has been reacted and that the solution is starting to become basic. Then by recording the reading from the burette, to find the volume of the NaOH (standard solution) that has reacted with a known volume of HCL, we can calculate the concentration (molarity) of the HCL.

Let us assume for example that we take 10 mls of HCl-solution and find that we have to add 30 mls of 0.1 M NaOH from the burette to obtain complete reaction. Using the following formula we can calculate the molarity of unknown solution of HCl?

HCl (aq) + NaOH (aq) ------à NaCl (aq) + H2O (l) 

The balanced equation shows that:

1 mole of HCL reacts with 1 mole of NaOH

1-HCl   ------à  1-NaOH

Burette         =       NaOH solution molarity        = M    =       0.1 M.

Pipette           =       HCl solution molarity = M    =       ?

Indicator        =       Phenolphthalein.

Colour change = Colourless to pink.

V1 - Volume of HCL taken              =       10 mls

M1 - Molarity of HCL taken             =       ?

V2 - Volume of NaOH solution         =       30 mls

M2 - Molarity of NaOH solution        =       0.1M

n1 - Number of moles of HCL         =       1

n2 - Number of moles of NaOH       =        1

V1XM1/n1 = V2XM2/n2

10x?/1 = 30x0.1/1 = 30x0.1/10 =3/10

M1 = .3M

 Molarity of HCL = 0.3M

We can also find out the amount of HCL dissolved per 1-litre = (1dm3) of a solution by the following formula.

Amount of HCl per dm3 = molarity x F. mass in grams

Amount of HCl per dm3 = 0.3 x 36.5 = 10.95g

(As F. mass of HCl = 1+35.5 = 36.5g)

Result:   Molarity of HCL = 0.3M

Amount of HCl/dm3     = 10.95g

Example.

A flask contain 30 ml of NaOH solution, it require 50ml of 0.15M-H2SO4, to complete the reaction. Calculate the molarity of NaOH and how many grams of NaOH were in the flask?

Solution.

The balance equation for the reaction is:

2NaOH (aq) + H2SO4 (aq) ------à  Na2SO4 (aq) + 2H2O (l)

2 mole          1 mole

Since 2 mole of NaOH _____ 1mole of H2SO4we need twice the amount of NaOH to react completely with H2SO4 solution as needed for the same concentration of HCl solution.

M1xV1/n1 = M2xV2/n2

Where:

V1 = Volume of H2SO4 = 50 ml

M1 = Molarity of H2SO4 = 0.15M

n1 = Number of moles of H2SO4 = 1

V2 = Volume of NaOH = 30 ml

M2 = Molarity of NaOH = ?

n2 = Number of moles of NaOH = 2

 

M1xV1/n1 = M2xV2/n2        0.15x50/1 =M2x30/2

M2 = 0.15x50x2/30 = 0.5M

Amount of NaOH per 30 ml = MxG.F.mass x volume of solution/1000

0.5x40x30/1000 = 0.6g

Result: Molarity of NaOH = 0.5M

Amount of NaOH in the flask = 0.6g

 

 
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