Ch 21.1

Neutralization Reactions

Acid-Base Reactions

• Acid + Base Water + Salt

• Properties related to every day:– antacids depend on neutralization– farmers use it to control soil pH– formation of cave stalactites

– human body kidney stones from insoluble salts

Acid-Base Reactions

• Neutralization Reaction - a reaction in which an acid and a base react in an aqueous solution to produce a salt and water:

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2 H2O(l)

– Table 21.1, page 614 lists some salts

Titration

• Titration is the process of adding a known amount of solution of known concentration to determine the concentration of another solution

• Remember? - a balanced equation has a mole ratio

Titrations

BuretBuret - volumetric glassware used for titrations.It allows you to add a known amount of your titrant to the solution you are testing.

If a pH meter is used, the equivalence point can be measured.

An indicator will give you the endpoint.

Titrations

• Note the color change which indicates that the ‘endpoint’ has been reached.

Start End

- Page 614

Titration

• The concentration of acid (or base) in solution can be determined by performing a neutralization reaction

– An indicator is used to show when neutralization has occurred

– Often we use phenolphthalein- because it is colorless in neutral and acid; turns pink in base

Steps - Neutralization reaction

#1. A measured volume of acid of unknown concentration is added to a flask

#2. Several drops of indicator added

#3. A base of known concentration is slowly added, until the indicator changes color; measure the volume– Figure 21.4, page 617– ASAP Sim26

Neutralization

• The solution of known concentration is called the standard solution– added by using a buret

• Continue adding until the indicator changes color– called the “end point” of the titration– Sample Problem 21.2, page 618

Equivalents

• An equivalent is the amount of acid or base that will give 1 mole of hydrogen ions, or in the case of a base, hydroxide ions ( remember not all acids and bases are the same strength and so, a 1M solution of an acid may or may not give 1 mole of hydrogen ions)

Example; 1 mole of HCl is 1 equivalent and will neutralize 1 mole of NaOH, which is also 1 equivalent

BUT, 1 mole of H2SO4 has 2 equivalents and will neutralize 2 moles of NaOH

Equivalents

• The mass of one equivalent of a substance is called the gram equivalent mass

• The gram equivalent mass for HCl is 36.5 g/mol

• The gram equivalent mass for H2SO4 is 49g/mol (1 mole H2SO4 is 2 equivalents, so 1 equivalent is ½ mole)

Equivalent Problem

• What is the mass of 1 equivalent of Ca(OH)2?

Molar mass is 74g/mol

1mole Ca(OH)2 has 2 equivalents

74g/mol = 37g/eq

2 eq./mol

Equivalent problem

• How many equivalents are in 14.6 g of sulfuric acid (H2SO4)

1mole H2SO4 has 2 equivalents

Molar mass is 98.1g/mol

So… 98.1g/2eq = 49.1g/eq

14.6 g x 1eq/49.1g=.297 eq

Normality

• Normality (N)=equivalent/L• IF 1 mole of acid or base gives 1

equiv. then Molarity (M) will equal Normality (N)

• If 1 mole of acid =2 equiv,, then N=2M; if 1 mole of acid =3 equiv, then N=3M

• If you know the normality you can find the # of equiv. by.

Equiv= V (in L) x N

Normality problem

• How many equiv. are in 2.5L of 0.60N H2SO4?

Equiv=V x N

Equiv=2.5L x 0.60N=1.5 equiv.

Normality & Titrations

• If you are diluting a solution of known normality, or in the case of titration you will find the following formula useful:

V1N1=V2N2

In the case of titrations

VaNa=VbNb

Problem Using V1N1=V2N2

• You need to make 250mL of 0.10N sodium hydroxide from a stock solution of NaOH that is 2.0N. How many mL of the stock solution do you need?

V1N1=V2N2

250mLx 0.10N=V2x2.0N

250mLx 0.10N = V2

2.0N

13mL=V2

• If a 35.0mL of a 0.20N HCl are needed to neutralize 25.0mL of an unknown base, what is the normality of the base?

VaNa=VbNb

35.0mL x 0.20N =25.0mLxNb

35.0mL x 0.20N=Nb

25.0mL

0.28N=Nb

Problem Using V1N1=V2N2

• How many mL of 0.500N sulfuric acid are needed to neutralize 50.0mL of 0.200N potassium hydroxide?

VaNa=VbNb

Va x 0.500N = 50.0mL x 0.200N

Va= 50.0mL x 0.200N 0.500N

Va=20.0mL

Problem Using V1N1=V2N2