Acidimetry

ACIDIMETRY – STANDARIZATION OF ACIDS

Neutralimetry which includes all acid base titrations can be divided into two types.

1.The determination of the amount of acid in a sample by titration with a standard alkali,called acidimetry

2.The determination of the amount of base in a sample by titration with a standard acid solution called alkalimetry

Principle

            The principle of alkalimetry is used (ie) the determination of the strength of an acid by titrating it against standard solution of a base which may be a primary or secondary standard.

Equal volumes of all acids / bases containing their gram equivalent weights per litre of the solution react completely with each other.

Na­2CO3  + 2 HCl           -->      2 NaCl + H2O + CO2

Na­2CO3  + 2  HNO3       -->     2 NaNO3 + H2O + CO2

Na­2CO3 + H2SO4       -->       Na2 SO4 + H2O + CO2

 In other words, the product of volume (V1) and normality (N1) of one solution (acid) is equal to the product of volume (V2) and normality (N2) of another solution (alkali / base)

V1N1 = V2N2 (Law of chemical equivalence or Law of Volumetric Analysis)

Example:         Preparation of secondary standard solution of acid

ie., 0.1 N HCl of 250 mL

• Molecular weight of HCl=36.45 g

• Equivalent weight of HCl=Molecular weight/valency=36.45/1=36.45 g

• Therefore, one equivalent weight of HCl=36.45 g

• Specific gravity of concentrated HCl=1.18

• Per cent purity of concentrated HCl=35%

• One litre of 0.1 N solution of HCl contains 3.645 g HCl.

First calculate the volume of conc. HCl that contains this amount of HCl.  By proportion, for 35.00 g of HCl , 100g of HCl solution is required For 3.645 g HCl=(100/35) x 3.645 = 10.4 g HCl solution      

To convert this weight of acid to volume, it must be divided by the specific gravity of the acid (1.18), we then have

10.4 / 1.18 = 8.8 mL

 For 1000 mL of 0.1 N HCl, 8.8 mL of conc. HCl is required. 

Then for 250 mL (8.8 /1000)  250 = 2.2 mL

(Similarly calculate for HNO3 and H2SO4, the details of specific gravity and per cent purity can be referred from the respective containers).

Reagents

1.Concentrated HCl  

2.0.1 N Na2CO3 and

3.Methyl orange indicator.

Apparatus                      

1. 400 mL beaker

2. 10 mL measuring cylinder

3. 50 mL burette

4. 10 mL pipette 

5. 250 mL conical flasks and 

6. 250 mL measuring cylinder.

Procedure

• Measure out 250 mL of distilled water into a 400 mL beaker using a 250 mL measuring cylinder.

•  Add 2.2 mL (roughly 2.5 mL) of conc. HCl through a 10 mL measuring cylinder to the beaker containing water and thoroughly mix it using a glass rod. This is approximately a 0.1 N solution of HCl.

•  Now, fill up a burette with 0.1 N Na2CO3 (primary standard) and mount it in the burette stand.

• Pipette 10 mL of the approximately 0.1 N solution of HCl into a 250 mL conical flask and add a few drops of methyl orange indicator. Note down the initial reading of the burette.

• Titrate the contents of conical flask against 0.1 N Na2CO3. End point is the change of colour from red to pale yellow.

• abulate the burette readings and find out the volume of 0.1 N Na2CO3   used in the titration.

• Repeat the titration until you get concordant titre values.

Now, using the relationship, V1 N1 = V2 N2 (Law of volumetric    analysis), find out the normality of the HCl. If the Na2CO3 consumed is exactly 10 mL then according to the above relationship the normality of the prepared HCl solution is 0.1 N. If not, the normality of the prepared HCl solution may be higher or lower than 0.1 N.

Observation table

0.1 N Na2CO3 Vs approximate 0.1 N HCl

Sl.No.   Aliquot taken (mL)    Burette readings(mL)          Volume of titrant used (mL)            Indicator               End point

      i            ii                              Initial           Final                                                iv-iii = v                 vi                              vii

                                                   iii                     iv

1.             10 

2.              20

Calculation

Volume of 0.1 N Na2CO3   consumed in the titration=V1=X mL

Volume of 0.1 N  HCl taken=V2=10 mL

Normality of Na2CO3 =N1= 0.1N

Normality of HCl=N2= Y(N)

V1 N1=V2  N2

X  x  0.1=10 x Y\ N2=Y

  Calculate the volume of water to be added in case the normality is more than 0.1N or the volume of concentrated HCl to be added in case the normality is less than 0.1 N.

If the prepared solution is > 0.1 N

Volume of 0.1 N HCl taken (V2)=10 mL

Volume of 0.1 N Na2CO3   used (V1)=X mL

Amount of water to be added for 10 mL    =(X – 10) mL

Therefore, for (250-A) mL=Where A = volume of the prepared HCl solution already used for rinsing the pipette and or for titration.

If the prepared solution is < 0.1 N

Volume of 0.1 N HCl taken (V2)=10 mL

Volume of 0.1 N Na2CO3   used (V1)=X mL

Normality of Na2CO3 (N1)=0.1N 

                                                        = V1 N1 = V2 N2        

N1 – N2  = 0.1 – Y   =  y (N)

• For the present 250 mL preparation of 0.1 N we have taken 2.5 mL

• Add either the volume of water or the volume of acid as the case may be to the solution in 400 mL beaker and mix it thoroughly by stirring.

• Repeat the titration and calculation processes until the normality of the HCl prepared reaches 0.1 N.