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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

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

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter15: Acids And Bases
Section: Chapter Questions
Problem 15.20QP: Strong Acids, Weak Acids, and pH Two 0.10-mol samples of the hypothetical monoprotic acids HA(aq)...
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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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