CHEMISTRY CHAPTER 13

Titrations

13.1 TitrationsA. Titrations – is an experimental procedure in which a

standard solution is used to determine the concentration of an unknown solution.

1. Standard Solution – is one of known concentration. The titration involves the gradual addition of one solution to another until the solute in the first solution has completely reacted with the solute in the second solution.

a) This point is called the equivalence point (same as stoichiometry point).

b) The equivalence point is detected using an indicator.

c) The point at which the indicator changes color is called the end point of the

titration.

2. The most common titrations involve the reaction of an acid solution with a basic solution.

3. The reaction of the acid and base is termed a neutralization.

B. The three types of titration reactions.

1. An acid with a base to give a soluble salt and water

2. A soluble salt with a second soluble salt to give a precipitate..

3. An oxidizing material with a reducing material.

C. The reaction between a strong acid (e.g., HCl or HNO3) and a strong base (e.g., NaOH) gives salts (e.g., NaCl or

NaNO3)

Since these salts are products of strong acids and strong bases, the resulting solution is neutral.

D. For a strong acid / strong base titration, the pH at the equivalence point is 7; but only a small amounts

of reagent causes a major pH change.

E. The titration curve for the neutralization reaction is shown in Figure 25.1a.

1. Curves can be produced using a pH meter connected to chart recorder.

2. The indicator selected should change color in pH range from about 4 to 10.

3. Phenolphthalein is usually used since it is easy to detect visually a slight pink color from a

colorless liquid.

F. Reaction between a strong acid (HCl, HNO3, or H2SO4) and a weak base (NH4) also produces salts (NH4Cl, NH4NO3, or

(NH4)2SO4).

1. These salts hydrolize to form slightly acidic solutions.

2. The titration curve for this reaction is shown Figure 24.1b.

3. Methyl orange can be used as a indicator because of the low pH region in which it changes color.

G. The reaction between a weak acid (CH3COOH) and a strong base (NaOH) gives a salt (NaCH3COO).

H. The concentration of the acid and basic solution will change the position of the curve only slightly

(especially at the start and completion of the titration), in relation to pH.

I. Sample Problem #1

If 20cm3 of a 0.300M solution of NaOH is required to neutralize completely 30cm3 of sulfuric acid

solution, what is the molarity of the H2SO4 solution?

Solve solution on next page

Step A: First write a balanced equation.

2NaOH + H2SO4 Na2SO4 + 2H2O

Step B: Since the concentration of the base is given, determine the moles of NaOH.

Number of moles = 20 cm3 0.300 mol NaOH 1 dm3 = 1000 cm3

= 0.00600 mol NaOH

Step C: From the coefficient of the balanced equation, 2 moles of base are required for reaction with 1 mole of acid.

Number of moles = 0.00600 mol NaOH 1 mol H2SO4 = 2 mol NaOH

= 0.00300 mol H2SO4

Step D: Determine the molarity of the acid.

Molarity = 0.00300 mol H2SO4 1000 cm3 = 30 cm3 1dm3

= 0.100 mol H2SO4 / dm3

J. Sample Problem #2

What volume of 0.500M HNO3 is required to neutralize 25.0 cm3

of a

0.200M solution?

Solving Process:

Step A: First write a balanced equation

HNO3 + NaOH NaNO3 + H2O

Step B: Since the concentration of the base is given, determine the moles of NaOH.

mol NaOH = 25cm3 0.200 mol NaOH 1dm3 = 0.00500 1.00 dm3 solution 1000cm3 mol NaOH

Step C: From the coefficients of the balanced equation, 1 mole of acid will react completely with 1 mole of base.

0.00500 mol NaOH = 0.00500 mol HNO3

Step D: Therefore

volume = 0.00500 mol HNO3 1.00 dm3 solution 1000 cm3

0.500 mol HNO3 1dm3

= 10.0 cm3 solution

13.2 Titrations with Normal SolutionsA. Another quantitative expression for the concentration of solutions is normality. A 1N solution contains one

equivalent mass of solute per dm3 of solution.

Normality = number of equivalents of solute dm3 of solution

1. An equivalent is that quantity of a substance that provides 1-mole of charge. Consider these three substances.

NaNO3 Na + NO3

Na2SO4 2Na + SO4

Na3PO4 3Na + PO4

2. NaNO3 contains 1 equivalent per mole, Na2SO4 contains 2 equivalent mole. The equivalent

masses of these substances.

Examples on Next Page

NaNO3 = 85.0g 1 mol = 85.0 g / eq

1mol 1 eq

Na2SO4 = 142.1g 1 mol = 71.05 g / eq

1mol 2 eq

3. Titration problems can also be solved using the concentration expression normally used and substituting into a mathematical equation.

4. Consider a reaction between 1.00N HCl solution and a 1.00N NaOH solution.

a) It is necessary to add equal volumes of the acid and base to have a complete reaction.

b) An equal number of H3O+ ions and OH- ions react to produce water.

c) The other product, NaCl, is a salt of a strong acid and a strong base, which produces a neutral solution.

5. The equivalent mass of a substance is that mass of material that will produce 1-mole of charge. One equivalent mass of hydrogen ions, will

react with one equivalent mass of hydroxide ion.

a) Therefore, a equal number of equivalents of acid and base will react completely to form salt.

b) normality acid x volume acid will equal normality base x volume base

Na x Va = Nb x Vb

B. If any three values are known, the fourth value can be calculated.

1. For the factors to be equivalent, both volume terms must be in the same unit.

C. Acids and bases with one reacting hydrogen ion or hydroxide ion per formula unit contain 1 equivalent per

mole

D. Acids and bases with two reacting hydrogen ions or hydroxide ions contain 2 equivalents per mole.

one reacting ion HCl 2.00M HCl = 2.00N HCl

(1eq = 1mol) NaOH 0.300M NaOH = 0.300N NaOH

two reacting ions H2SO4 0.750M H2SO4 = 1.50N H2SO4

(2eq = 1mol)

Example Problem #1

If 20.0cm3 of a 3.00N solution of NaOH is required to neutralize 30.0 cm3 of a sulfuric acid solution. What is the normality of the H2SO4?

Solving Process:

equivalents of acid = equivalents of base or Na x Va = Nb x Vb

change cm3 to dm3 and then substitute:

Na = 3.00N 0.0200 dm3 = 2.00N

0.0300 dm3

F. Sample Problem #2

How many grams of KOH are required to neutralize completely 2.00.0 cm3 of a 4.00N solution of HNO3?

Solving Process:

For neutralization to be complete, the number of equivalents of acid must be equal to the number of equivalents of base. First, calculate the number of equivalents of acid. Then convert this answer to grams of KOH.

Equivalents HNO3 = 4.00 eq HNO3 0.200 cm3 = 0.800 eq

1 dm3

equivalents acid = equivalents base

mass KOH = 0.800 eq KOH 1mol KOH 56.1g KOH = 44g

1 eq KOH 1 mol KOH