Concept explainers
Interpretation:
The normality of sodium carbonate solution of volume 51.89 mL titrated with 5.038 g of HC7H5O2 is to be calculated.
Concept introduction:
An acid base reaction is one of the types of reaction which is known as neutralization reaction. Normality is used to express concentration of any solution. The n factor of any species is defined in different way depending upon the type of reaction. In an acid base reaction n-factor is defined as the number of replaceable H+ of an acid and for a base it is the total charge present on a species after removing all metal ions. Equation which equates the equivalents of an acid and a base is given below.
n1M1V1︸equivalents of acid =n2M2V2︸equivalents of base
Answer to Problem 105E
The normality of sodium carbonate of volume 51.89 mL which is titrated with 5.038 g of HC7H5O2 is 0.7952 N.
Explanation of Solution
The equation which is used to calculate the normality of sodium carbonate is given below.
n1M1V1︸equivalents of acid =n2M2V2︸equivalents of base …(1)
The molarity is given by the formula shown below.
M1=w(g)Molar mass of acid(g/mol)×Volume of acid(V in mL)M1V1=w (g)Molar mass of acid (g/mol) …(2)
Substitute relation (2) in (1) as shown below.
n1M1V1︸equivalents of acid =n2M2V2︸equivalents of base n1×w(g)Mo(g/mol)︸equivalents of acid HC7H5O2=n2M2V2︸equivalents of Sodium Carbonate …(3)
Where,
• n1 is the n-factor of benzoic acid(HC7H5O2).
• n2 is the n-factor of sodium carbonate base.
• w is the mass used of benzoic acid.
• M2 is the molarity of sodium carbonate base.
• Mo is the molar mass of acid used.
• V2 is the volume of sodium carbonate base used.
Benzoic acid has 1H+. So, n-factor of benzoic acid is 1 and sodium carbonate has n-factor 2.
The given value n1 is 1.
The given value of w is 5.038 g.
The given value of M∘ is 122.12 g/mol.
The given value of n2 is 2.
The given value of V2 is 51.89 mL.
Substitute all the values in the equation (2) as shown below.
1×5.038 g122.12 g/mol=2×M2×51.89 mLM2=1×5.038g122.12 g/mol×2×51.89 mLM2=0.0003976 mol/mL×(1000 mL1 L)=0.3976 mol/mL
Relation between molarity and normality is written below.
N=n×M
Where,
• N is the normality of the solution.
The calculation of normality of sodium carbonate is shown below.
N=n×MN2=2×0.3976 M=0.7952 N
Therefore, normality of sodium carbonate of volume 51.89 mL which is titrated with 5.038 g of HC7H5O2 is 0.7952 N.
The normality of sodium carbonate of volume 51.89 mL titrated with 5.038 g of HC7H5O2 is calculated as 0.7952 N.
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