CHAPTER 15

Acid-Base Titration and pH

15.1

Aqueous Solutions and the Concept of pH

Ionization of Water

• Self-ionization of water: two water molecules produce a hydronium ion and hydroxide ion by transfer of a proton

• Concentrations are represented by the molecule’s name enclosed in brackets.– Example: [H3O+]

• Concentrations of hydronium and hydroxyl ions are inversely proportional - as one increases, the other decreases.

Ionization Constant

• Ionization constant of water is Kw

• It’s equal to the concentration of hydronium ions times hydroxyl ions, which equals 1.0 x 10-14.

– [H3O+] x [OH-] = 1.0 x 10-14

• Ionization of water increases as temperature increases.

Solutions

• Neutral Solution

– [H3O+] = [OH-]

• Acidic Solution

– [H3O+] > [OH-]

• Basic Solution

– [H3O+] < [OH-]

• Calculating Hydronium and Hydroxyl ion concentrations– Do dissociation

problem to determine number of moles of specific ion made via one mole of solute

– Use strong acids and bases; they completely ionize

Strong Acids and Bases

Example

• A 1.0 x 10-4 M solution of nitric acid has been prepared in a lab. Calculate the hydronium ion concentration and the hydroxyl ion concentration.

Solution

• HNO3 + H2O H30+ + NO3-

• 1.0 x 10-4 M HNO3 =

(1.0 x 10-4 mol / 1 L) x (1 mol H30+ / 1 mol HNO3) = 1.0 x 10-4 M H30+

• [H3O+] x [OH-] = 1.0 x 10-14

– [1.0 x 10-4] x [OH-] = 1.0 x 10-14

• [OH-] = 1.0 x 10-10 M

Example

• Barium Hydroxide has a hydronium ion concentration of 1.0 x 10-11. What is the hydroxyl concentration? What is the molarity of solution?

Solution

• Ba(OH)2 –> Ba2+ + 2OH-

• [H3O+] x [OH-] = 1.0 x 10-14

– [1.0 x 10-11] x [OH-] = 1.0 x 10-14

– [OH-] = 1.0 x 10-3 M

• 1.0 x 10-3 M OH = 1.0 x 10-3 mol / 1L

• 1.0 x 10-3 mol OH x (1mol Ba(OH)2 / 2 mol OH) = 5.0 x 10-4 M Ba(OH)2

pH Scale

pH Scale

• pH: negative of the logarithm of hydronium ion concentration

• pH = -log[H30+]

• Example:

[H30+] = 1x10-7 M pH = -log[1x10-7] = 7

• pOH: negative logarithm of hydroxide ion concentration

• pOH = -log[OH-]

pH and pOH

• Recall that Kw = 1.0 x 10-14 M

– Therefore, pH + pOH = 14

• pH of solutions– Neutral: pH = 7.0; pH = pOH– Basic: pH > 7.0; pH > pOH– Acidic: pH < 7.0; pH < pOH

Calculating pH

• If starting from hydronium concentration, just take negative log

• If starting from hydroxyl concentration, find hydronium concentration, then take negative log

• Calculate pOH similarly– take negative log for hydroxyl concentrations – for hydronium concentrations, find hydroxyl

concentration and then take negative log

Examples• What is the pH of a 1.0 x 10-3 M NaOH solution?

– [OH-] = 1.0 x 10-3 and [H30+] = 1.0 x 10-11

– pH = -log(1.0 x 10-11) = 11

• What is the pOH of a 1.0 x 10-8 M NaOH solution?– pOH = -log(1.0 x 10-8) = 8

• What is the pH of a solution if the [H30+] is 2.7 x 10-3 M? – pH = -log(27 x 10-3) = 2.6

Common pH Ranges

Calculating Concentrations

• Find concentrations from pH in reverse order– Hydronium concentration = 10-pH

• Example– Determine the hydronium concentration of an

aqueous solution that has a pH of 4.0

• [H30+] = 10-pH = 10-4 = 1.0 x 10-4 M

Another Example

– A shampoo has a pH of 8.7. What are the hydronium and hydroxyl ion concentrations?

• [H30+] = 10-pH = 10-8.7 = 2.0 x 10-9 M

• [H30+][OH-] = 1.0 x 10-14

• [OH-] = 1.0 x 10-14 / 2.0 x 10-9 = 5.0 x 10-6 M

15.2Determining pH and

Titrations

Indicators

• Acid-Base Indicator: compounds whose colors are sensitive to pH

• Indicators change colors because they are either weak acids or weak bases

• Indicators come in different colors and work over a variety of ranges

Different Color Ranges of Various Indicators

How Indicators Work

• Equilibrium indicator eq: HIn <–> H+ + In-

• An indicator’s colors result from the fact that HIn and In- are different colors.

• Acidic solutions - In- acts as a base and accepts acid protons. Indicator is then present in largely unionized form, Hin.

• Basic solutions - H+ ions combine with the base’s OH- ions. The indicator further ionizes since H+ ions have been lost. Indicator is then largely present in the form of In- .

Indicators cont’d and pH Meters

• Transition Interval: pH range over which an indicator changes color

• The lower the pH that an indicator changes colors means the stronger the acid of the indicator.

• pH Meter: determines pH of solution by measuring the voltage between two electrodes placed in the solution

Titrations

• Definition: controlled addition and measurement of amount of solution of known concentration required to react completely with measured amount of solution of unknown concentration

More Titration

• Equivalence point: point at which two solutions used in titration are present in chemically equivalent amounts

• End point: point in titration where indicator changes color

• If we know the concentration of one solution, we can find the concentration of the other in a titration from the chemically equivalent volumes.

Titration Equivalence Points

Titration Solutions

• Standard solution: solution that contains a precisely known concentration of a solute

• Compare our known solution concentrations with a solution of a primary standard

• Primary standard: highly purified solid compound used to check concentration of known solution in titration

Steps to Solve a Titration Problem

1) Start with a balanced equation for neutralization reaction and determine chemically equivalent amounts of acid and base

2) Determine moles of acid or base from known solution used during titration

3) Determine moles of solute of unknown solution used during titration

4) Determine molarity of unknown solution

Example

• In a titration, 27.4 mL of .0154 M Ba(OH)2 are added to a 20.0 mL sample of HCl solution of unknown concentration until the equivalence point is reached. What is the molarity of the acid solution?

Solution

• Ba(OH)2 + 2HCl 2H2O + BaCl2

– Ba(OH)2 : 27.4 mL, 0.0154 M– HCl: 20.0 mL, ?M

• 0.0154 M = ? Mol Ba(OH)2 / 0.0274 L

? = 4.22 x 10-4 mol Ba(OH)2

• 4.22 x 10-4 mol Ba(OH)2 x (2HCl / 1 Ba(OH)2) = 8.44 x 10-4 mol HCl

• 8.44 x 10-4 mol HCl / 0.0200 L = 0.0422 M

Example

• You have a vinegar solution you believe to be 0.83 M. You are going to titrate 20.00 mL of it with a NaOH solution that you know to be 0.519 M. At what volume of added NaOH solution would you expect to see an end point?

Solution

• HC2H3O2 + NaOH H2O + NaC2H3O2

– HC2H3O2: 0.83 M, 0.02000L– NaOH: 0.519 M, ? L

• 0.83 M HC2H3O2 = ? Mol / 0.02000 L

? = 0.0166 mol HC2H3O2

• 0.0166 mol HC2H3O2 x (1 mol NaOH / 1 mol HC2H3O2) = 0.0166 mol NaOH

• 0.519 M NaOH = 0.0166 mol / ? L– ? = 0.03198 L = 32 mL

THE END

Created by:

Savannah Sisk