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Prepare M/10 NaOH solution (250 ml) and using this solution estimate the

molarity and strength of the given HCL solution.

Step I : Preparation of 250 ml of M/10 NaOH :

Molecular Formula of

NaOH

= NaOH

Molecular Mass = 40 g/mol

NaOH required to prepare

250 ml of M/20 NaOH

= Required molarity X Molecular

mass of

NaOH

X Required volume of NaOH in mL

1000

= 1/10 X 40X 250

1000

= 1 g

Weight of the weighing bottle + NaOH = 1.7 g

Weight of the weighing bottle = 0.7 g

Weight of NaOH = 1.0 g

Molarity of NaOH solution = Strength (g/L)

Molecular mass of NaOH

= 1 X 1000

40 250

= 1 X 4

40

= 0.1 M

Step II : Molecular equation and Ionic equations:

Molecular equation:

NaOH + HCl NaCl + H2O

Ionic Equation:

OH- + H

+ H2O

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Step III : Simple Procedure

a) Burette solution : HCl

b)Pipette Solution : M/10 NaOH

c)Medium : Aqueous

d)Temperature : Room temperaturee)Indicator : Phenolphthalein

f)End Point : Appearance of pink colourg)Molecular Mass of HCl : 36.5 g/mol

Step IV: Observation Table:

NaOH X HCl

Sl.No. Volume ofNaOH in mL

Burette readings in mL Volume ofHCl in mLInitial Final

1 20 0 19.2 19.22 20 0 19.1 19.1

3 20 0 19.1 19.1

Concordant Value: 19.1 mL

StepV: Calculation of Molarity of HCl

In the balanced equation,

1moles of NaOH reacts with 1 moles of HCl

V HCl X M HCl 1

----------------------------------- = ----------V NaOH X M NaOH 1

V HCl X M HCl 1

= ---------- X V NaOH X M NaOH1

1 1V HCl X M HCl = ---------- X 20 X ------1 10

1 1

19.1 X M HCl = ---------- X 20 X ------1 10

1 1 1M HCl = ---------- X 20 X ------ X -----

1 10 19.1

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1 X 20 X 1 X 1M HCl = -----------------------------------

1 X 10 X 19.1

M HCl = 0.104712 M

Therefore, Molarity of given HCl Solution = 0.104712 M

StepVI: Calculation of Strength of HCl Solution:

Strength of HCl Solution = Molarity of HCl X Molecular mass of HCl

= 0.104712 X 36.5 g/L

= 3.821988 g/L

Step VII: Result:

Molarity of given HCl = 0.1047 M

Strength of given HCl Solution = 3.821988 g/Litre.

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2.0000

0 .500

0.050

0.001