NEUTRALIZATION From the previous lesson, you have learned that the Arrhenius theory defines neutralization as the reaction between an acid and a base to form a neutral solution. To produce a neutral solution, the acid and the base should have comparable strengths. Strictly, speaking, neutralization occurs when an equivalent amount of an acid reacts with an equivalent amount of a base. Nacl + H20 Neutral solution HCI + NaOH

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Chapter1: Chemical Foundations
Section: Chapter Questions
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A sample consisting of Na2CO3, NaHCO3 and inert matter weighs 1.179 grams. It is titrated with 0.1 N HCl with phenolphthalein as the indicator, and the solution became colorless after the addition of 24.00 mL. Another duplicate sample was titrated with HCl using methyl orange as indicator. It required 50.25 mL of the acid for the color change from yellow to red. What are the percentage of Na2CO3 and NaHCO3 in the sample? Note [EXAMPLE IN THE IMAGE]
NEUTRALIZATION
From the previous lesson, you have learned that the Arrhenius theory defines
neutralization as the reaction between an acid and a base to form a neutral solution.
To produce a neutral solution, the acid and the base should have comparable strengths.
Strictly, speaking, neutralization occurs when an equivalent amount of an acid reacts
with an equivalent amount of a base.
HCI + NaOH
Nacl + H20
Neutral solution
In a neutralization process, the equivalence point of a reaction is the point where the
acid or base has been added in an amount equivalent to the other. This is accomplished
approximately by means of titration, the progressive addition of a solution (acid or
base) of known concentration to a known amount of an acid or base of unknown
concentration. The solution whose concentration is known is called a standard solution.
The equivalence point is closely related to the end point of a titration. The end point
is the point in titration where the indicator used undergoes a color change.
Since
8/18
no.of equivalents
N =
Therefore, the number of equivalents of acid or base is calculated by multiplying the
volume by its normality.
No. of equivalents = NV
In neutralization,
No. of equivalents of acid = no. of equivalents of base
Thus,
NaVa = NoVb
Where,
Na = normality of acid
Va = volume of acid in litres
No = normality of base
Vo = volume of base in litres
Exercises
1. By experiment, the normality of H;SO, was found to be 0.517 2. If 39.65 mL of
this acid neutralized 21.74 mL of NAOH, what is the normality of the base?
2. 0.200 g of pure NazCO; requires 35.5 mL of vinegar for titration. Calculate the
normality of vinegar.
Transcribed Image Text:NEUTRALIZATION From the previous lesson, you have learned that the Arrhenius theory defines neutralization as the reaction between an acid and a base to form a neutral solution. To produce a neutral solution, the acid and the base should have comparable strengths. Strictly, speaking, neutralization occurs when an equivalent amount of an acid reacts with an equivalent amount of a base. HCI + NaOH Nacl + H20 Neutral solution In a neutralization process, the equivalence point of a reaction is the point where the acid or base has been added in an amount equivalent to the other. This is accomplished approximately by means of titration, the progressive addition of a solution (acid or base) of known concentration to a known amount of an acid or base of unknown concentration. The solution whose concentration is known is called a standard solution. The equivalence point is closely related to the end point of a titration. The end point is the point in titration where the indicator used undergoes a color change. Since 8/18 no.of equivalents N = Therefore, the number of equivalents of acid or base is calculated by multiplying the volume by its normality. No. of equivalents = NV In neutralization, No. of equivalents of acid = no. of equivalents of base Thus, NaVa = NoVb Where, Na = normality of acid Va = volume of acid in litres No = normality of base Vo = volume of base in litres Exercises 1. By experiment, the normality of H;SO, was found to be 0.517 2. If 39.65 mL of this acid neutralized 21.74 mL of NAOH, what is the normality of the base? 2. 0.200 g of pure NazCO; requires 35.5 mL of vinegar for titration. Calculate the normality of vinegar.
Solutions:
1. Given:
Na = 0.517 2 N
Va = 0.039 65 L
Vp = 0.021 74 L
Find:
No = ?
N.Va = N,Vo
NaVa
No
%3!
(0.517 2 N)(0.039 65 L)
0.021 74 L
= 0.943 3 N
2. Given:
3.
Mass = 0.200 g
NazCO3
Va = 0.035 5 L
Find:
Na = ?
No. of equivalent =
eq wt
0.200 g
No. of equivalents of base = 10
8/g-eq wt
No. of equivalents of base = 0.003 77
No. of equivalents of acid = no. of equivalents of base
N.Va =0.003 77
0.003 77 eq
Na
0.035 5 L
= 0.106 N
Transcribed Image Text:Solutions: 1. Given: Na = 0.517 2 N Va = 0.039 65 L Vp = 0.021 74 L Find: No = ? N.Va = N,Vo NaVa No %3! (0.517 2 N)(0.039 65 L) 0.021 74 L = 0.943 3 N 2. Given: 3. Mass = 0.200 g NazCO3 Va = 0.035 5 L Find: Na = ? No. of equivalent = eq wt 0.200 g No. of equivalents of base = 10 8/g-eq wt No. of equivalents of base = 0.003 77 No. of equivalents of acid = no. of equivalents of base N.Va =0.003 77 0.003 77 eq Na 0.035 5 L = 0.106 N
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