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# Volumetric Analysis - Law with Solved problem

"Equal volume of equinormal solutions exactly neutralise the other solution having same concentration and volume". V1 N1 = V2 N2

Volumetric Analysis

An important method for determining the amount of a particular substance is based on measuring the volume of reactant solution. Suppose substance A in solution reacts with substance B. If you know the volume and concentration of a solution of B that just reacts with substance A in a sample, you can determine the amount of A.

Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution of A with known concentration of B untill the reaction of A and B is just completed. Volumetric analysis is a method of analysis based on titrations.

Law

"Equal volume of equinormal solutions exactly neutralise the other solution having same concentration and volume".

V1 N1  =  V2 N2

V1, V2  -  Volume of solutions.

N1, N2  -  Strength of solutions.

Solved problem

Calculating the volume of reactant solution needed

1. What volume of 6M HCl and 2M HCl should be mixed to get one litre of 3M HCl?

Solution

Suppose the volume of 6M HCl required to obtain 1L of 3M HCl = XL

Volume of 2M HCl required  =  (1-x)L

Applying the molarity equation

M1V1  +   +   M2V2  =   =   M3V3

6MHCl+   2MHCl=   3MHCl

6x+2(1-x)      =  3x1

6x+2-2x  =  3

4x  =  1

x  =  0.25L

hence, volume of 6M HCl required  =  0.25L

Volume of 2M HCl required      =  0.75L

2. How much volume of 10M HCl should be diluted with water to prepare 2.00L of 5M HCl.

Solution

N1V1  =   N2V2

10N HCl  =  5N HCl

10xV1  =  5 x 2.00

V1  =  (5 x 2.00 )/ 10

=  1.00L

Problems for Practice

1. NiSO4 reacts with Na3PO4 to give a yellow green precipitate of Ni3(PO4)2 and a solution of Na2SO4.

3NiSO4(aq) + 2Na3PO4(aq) Ni3(PO4)2 (s) + 3Na2SO4(aq)

How many mL of 0.375M NiSO4 will react with 45.7 mL of 0.265M Na3PO4?

2. What volume of 0.250M HNO3 reacts with 42.4 mL of 0.150M Na2CO3 in the following reaction?

2HNO3(aq)  +  Na2CO3(aq) 2NaNO3(aq)  +  H2O(aq)+CO2(g)

3. A flask contains 53.1 mL of 0.150M Ca(OH)2 solution. How many mL of 0.350M Na2CO3 are required to react completely with Ca(OH)2 in the following reaction.

Na2CO3(aq) +  Ca(OH)2(aq) CaCO3(aq)  + 2NaOH(aq)

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