5.3.1 neutralization reactions 5.3.2 titration reactions

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Unit 5: Acids and Bases 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

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Page 1: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Unit 5: Acids and Bases

5.3.1 Neutralization reactions5.3.2 Titration Reactions

Page 2: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

5.3.1 Neutralization Reactions What happens when an acid such as

HCl is mixed with a base such as NaOH:

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

When an acid and a base are combined, water and a salt are the products.

Page 3: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Salts are ionic compounds containing a positive ion other than H+ and a negative ion other than the hydroxide ion, OH-.

Double displacement reactions of this type are called neutralization reactions.

Page 4: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

We can write an expanded version of this equation, with aqueous substances written in their longer form:

H+(aq) + Cl-(aq) + Na+

(aq) + OH-(aq) → Na+

(aq) + Cl-(aq) + H2O(l)

Removing the spectator ions we get the net ionic equation:

H+(aq) + OH-

(aq) → H2O(l)

Page 5: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

When a strong acid and a strong base are combined in the proper amounts ([H+] equals [OH-]) - a neutral solution results (pH = 7).

But, salt solutions do not always have a pH of 7. Through a process known as hydrolysis, the

ions produced may react with water molecules to produce a solution that is slightly acidic or basic.

Generally if a strong acid is mixed with a weak base there the resulting solution will be slightly acidic; if a strong base is mixed with a weak acid the solution will be slightly basic.

Page 6: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

5.3.2 Acid-Base Titrations

Acid-base titrations are lab procedures used to determine the concentration of a solution

Page 7: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

During an acid-base titration, an acid with a known concentration (a standard solution) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base.

Page 8: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

The indicator will change colour, when the base has been neutralized (when [H+] = [OH-]).

At that point - called the equivalence point or end point - the titration is stopped. By knowing the volumes of acid and base used, and the concentration of the standard solution, calculations allow us to determine the concentration of the other solution.

Page 9: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

It is important to accurately measure volumes when doing titrations. The instrument you would use is called a burette (or buret).

Page 10: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions
Page 11: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Titration Procedure

1.Rinse the burettes several times with the acid and base solutions, then fill them with the appropriate solution.

2.Using the burette, accurately measure a volume of the base into an Erlenmeyer flask. How much you use is not important, as long as you know exactly how much is in the flask.

3.Add a suitable indicator such as

Page 12: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

4. The acid (with the known concentration - the titrant) is then added to the base. Release the acid from the burette quickly at first, then more slowly as the endpoint is neared.

How do you know when you are reaching the endpoint? The indicator will begin to show a change in colour. Swirling the flask will cause the colour to disappear.

ENDPOINT IS REACHED AS SOON AS THE COLOUR CHANGE IS PERMANENT. ONE DROP WILL DO IT - once the colour change has occurred, stop adding acid.

If a pH meter is used instead of an indicator, endpoint will be reached when there is a sudden change in pH.

Page 13: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

5. Record the volume of acid added to reach endpoint. Calculations will allow us to calculate the concentration of the base.

6. The titration should be repeated several times and the results averaged. Often the first trial is done quickly to get a general idea of the volume of titrant required. Subsequent trials are done more slowly.

Page 14: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Titration Calculations

Use titration information to calculate the concentration of the unknown solution, you must know the following information.

The concentration of one of the solutions, the acid for example (MA)

The volume of acid used for the titration (VA) The volume of base used for the titration (VB) What you will calculate: The concentration of the other solution, the

base for example (MB)MAVA = MBVB

Page 15: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Example 1

During a titration 75.8 mL of a 0.100 standard solution of HCl is titrated to end point with 100.0 mL of a NaOH solution with an unknown concentration. What is the concentration of the NaOH solution?

Page 16: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Solution

MA= 0.100 M MB = MB

VA =75.8 mL VB =100.0 mLSubstitute in known values and solve for the

unknown:MAVA = MBVB

(0.100)(75.8) = MB(100.0)7.58 /100.0 = MB

0.0758 = MB

Answer: The concentration of the NaOH solution is 0.0758 M or 7.58 ×10-2M

Page 17: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Example 2

A 20.0 mL solution of strontium hydroxide, Sr(OH)2, is placed in a flask and a drop of indicator is added. The solution turns colour after 25.0 mL of a standard 0.0500M HCl solution is added. What was the original concentration of the Sr(OH)2 solution?

Page 18: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Solution Write a balanced equation for the neutralization reaction:2 HCl(aq) + Sr(OH)2 (aq) → SrCl2 (aq) + H2O(l)

Notice that 2 moles of the acid HCl are required to neutralize 1 mole of the base Sr(OH)2.

Define the variables to be used and the known values: MA = 0.050 M  MB = MB

VA = 25.0 mL  VB = 20.0 mL

To balance this out we need to modify our formula:MAVA = 2 MBVB

Notice that the "2" is on the base side of our formula even though it was in front of the acid side in the balanced equation - it "switches places".

Page 19: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Substitute in known values for the equation and solve for the unknown:

MAVA = 2MBVB

(0.050)(25.0) = 2MB(20.0)1.25 = 40.0MB

1.25/ 40.0 = MB

0.0312 = MB

Answer: The concentration of the Sr(OH)2,solution is 3.12 ×10-2M.

Page 20: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Indicator Selection for Titrations

(You will not be required to know this section for a test.)

What indicator is used depends on what type of titration you are performing. An indicator should be chosen that will change color when enough of one substance (acid or base) has been added to exactly use up the other substance.

The three main types of acid-base titrations, and suggested indicators, are:

Page 21: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Titration between . . . Indicator Explanation

1. strong acid anyand strong base

2. strong acid methyl changes color and weak base orange in the acidic

range(3.2-4.4)

3. weak acid phenolphthalein changescolor and strong basein the basic range (8.2 - 10.6)

Page 22: 5.3.1 Neutralization reactions 5.3.2 Titration Reactions

Lab 5.3.2- titrations Assignment 5.3.2