Chemistry 30

Unit 6: Acids and Bases

Teacher

General Info about Acids and BasesAcids:

React with bases

Taste Sour (don’t taste in chemistry)

React with certain active metals to produce hydrogen

Turn litmus paper red

Bases:

Reacts with acids

Bitter taste

Feel ‘soapy’

Turn litmus paper blue

Arrhenius TheoryAcids vs Bases

All acids produce hydrogen ions when they are dissolved in water. These ions are responsible for the acidic properties of these solutions.

All bases produce hydroxide ions when they are dissolved in water. These ions are responsible for the basic properties of these solutions.

Strong acids and bases have high K values (because the products are favoured) and weak acids and bases have low K values because they do not produce a lot of ions (reactants are favoured).

Since strong acids and bases have lots of ions in solution (favour products) they conduct electricity well.

General Idea

The acidic properties of aqueous solutions of HCl (hydrochloric acid) and HNO3(nitric acid) must be due to the hydrogen ions since they are the only ions common to both solutions.

The Arrhenius Model of Neutralization

When equal molar quantities of an acid and a base are mixed in water, a solution is obtained that does not act as an acid or a base; it is neutral in terms of its acid-base properties. This kind of chemical reaction is a neutralization reaction.

ACID + BASE SALT + WATER

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

acid + base salt + water

H+(aq) + Cl−

(aq) + Na+(aq) + OH−

(aq) Na+(aq) + Cl−

(aq) + H2O(l)

H+(aq) + OH−

(aq) H2O(l)

Problems with Arrhenius’ Theory

The Solvent Problem: The nature of the solvent is very important in determining the acidic behaviour of a substance.

HCl is a strong acid and when it dissolves in water it conducts electricity well because there are lots of H+ and Cl- ions, however when HCl dissolves in toluene it does not conduct electricity which means there are few ions present. If there are few H+ ions present it would no longer be a strong acid.

The Salt Problem: Salts solutions should not change the colour of litmus paper if they do not contain H+ ions or OH- ions because they would not be acids or bases (should be neutral).

Salt Effect on Litmus ConclusionNaCl None NeutralNa3PO4 Red turns blue BaseNa2CO3 Red turns blue BaseNaNO3 None NeutralNH4Cl Blue turns red AcidPb(NO3)2 Blue turns red Acid

*How can Pb(NO3)2 be an acid if it needs to produce H+ ions and how can Na2CO3 be a base if it needs to produce OH- ions?

The Brønsted-Lowry TheoryAcids vs Bases

An acid is a molecule or ion that can give up a hydrogen ion. Any ion containing H atoms is a potential acid.

A base is any molecule or ion that can react with (accept) a hydrogen ion. Any ion with a pair of valence electrons available for bonding is a potential base.

General Idea

An acid is a hydrogen-ion donor, and a base is a hydrogen-ion acceptor.

An acid can only act as an acid if a base is present and is willing/able to accept a hydrogen ion.

This theory was able to explain the role of the solvent as well as the existence of acidic and basic salt solutions.

Dissociation of Brønsted-Lowry Acids

Acids donate a proton (H+) which combines with water to form hydronium (H3O+). Often H+ is used for convenience but the meaning is H3O+.

Ex. HCl(g) + H2O(l) H3O+(aq) + Cl−

(aq)

Water functions as a hydrogen-ion acceptor, or base, in this reaction.

H3O+ is an acid itself and can release an H+ ion if the right base is present.

Ex. H3O+(aq) + H2O(l) H⇌ 2O(l) + H3O+

(aq)

Dissociation of Brønsted-Lowry Bases

Metal hydroxides dissolve in water and produces OH - ions directly which can accept an H+ ion.

Ex. NaOH(s) Na+(aq) + OH−

(aq)

Ba(OH)2(s) Ba2+(aq) + 2 OH−

(aq)

There are bases that do not contain OH- ions. With these bases H2O can act as an acid by releasing a hydrogen ion and leaving an OH- ready to react.

Ex. NH3(g) + H2O(l) NH4+

(aq) + OH−(aq)

Example: Liquid hydrogen perchlorate, HClO4, dissolves in water to form a solution of perchloric acid. Identify the hydrogen ion donor and acceptor, and write an equation for the solution process.

Remember, acids react with water to for H3O+

HClO4(l) + H2O(l) H3O+(aq) + ClO−

(aq)

donor acceptor

Example: A solution of gaseous methylamine, CH3NH2, turns red litmus blue. Write the balanced equation.

Blue litmus = base. Base releases OH- or reacts with water to release OH-.

CH3NH2(g) + H2O(l) CH3NH3+

(aq) + OH−(aq)

See Acids and Bases Assign

Weak and Strong Acids and Bases The strongest acid that can exist in water is the hydronium ion. The strongest base that can

exist in water is the hydroxide ion.

To differentiate acids that ionize completely from those that do not, we use the terms strong and weak. A strong acid is completely ionized in solution. A weak acid is only partially ionized

in solution. Hydrochloric acid is a strong acid even when it is dilute. Acetic acid is a weak acid even when it is concentrated.

Bases can also be weak or strong. Sodium hydroxide is an example of a strong base. It completely dissociates in water. Ammonia, however, is a weak base.

Using a table of acid-dissociation constants is the surest way to quantify relative strengths of weak acids, but you can often classify acids and bases qualitatively as strong or weak just from their formulas:

Strong acids. Two types of strong acids, that you should memorize are:

1. The hydrohalic acids HCl, HBr, and HI

2. Oxoacids in which the number of O atoms exceeds the number of ionizable protons by 2 or more, such as HNO3, H2SO4, HClO4; for example, in H2SO4, 4 O’s – 2 H’s = 2

Weak acids. There are many more weak acids than strong ones. Four types are:

1. The hydrohalic acid HF

2. Acids in which H is not bonded to O or to a halogen, such as HCN and H2S

3. Oxoacids in which the number of O atoms equals or exceeds by one the number of ionizable protons, such as HClO, HNO2, and H3PO4

4. Carboxylic acids (general formula RCOOH, with the ionizable proton shown in italics), such as CH3COOH and C6H5COOH

Strong bases. Water-soluble compounds containing O 2− or OH − ions are strong bases. The cations are usually those of the most active metals:

1. M2O or MOH, where M = Group IA (1) metal (Li, Na, K, Rb, Cs)

2. MO or M(OH)2, where M = Group IIA (2) metal (Ca, Sr, Ba)

Weak bases. Many compounds with an electron-rich nitrogen atom are weak bases. The common structural feature is an N atom with a lone pair (shown here in italics in the formulas):

1. Ammonia ( H3)

2. Amines (general formula R H2, R2 H, or R3 ), such as CH3CH2 H2, (CH3)2 H,

and (C3H7)3

Examples H2SeO4 strong acid oxoacid, more O than H by 2

(CH3)2CHCOOH weak acid carboxylic acid, −COOH present

KOH strong base Group 1 hydroxide

(CH3)2CHNH2 weak base amine, lone pair on N

Salts of Weak Acids and Bases

Ions of Neutral Salts

Cations

Na+ K+ Rb+ Cs+

Mg2+ Ca2+ Sr2+ Ba2+

Anions

Cl- Br- I-

ClO4- BrO4

- ClO3- NO3

-

Acidic Ions

NH4+ Al3+ Pb2+ Sn2+

HSO4- H2PO4

-

Basic Ions

F- C2H3O2- NO2

- HCO3-

CN- CO32- S2- SO4

2-

HPO42- PO4

3-

Example: Using an equation, show why a sodium nitrite solution is basic.

NaNO2(s) Na+(aq) + NO2

−(aq)

The sodium ions are neutral, but the basic nitrite ions will react with water:

NO2−

(aq) + H2O(l) HNO2(aq) + OH−(aq)

The formation of some hydroxide ions will render the solution basic.

Example: Account for the fact that sodium hydrogen carbonate produces a basic solution when it is dissolved in water.

NaHCO3(s) Na+(aq) + HCO3

−(aq)

The sodium ions are neutral, but the basic hydrogen carbonate ions will react with water:

HCO3−

(aq) + H2O(l) H2CO3(aq) + OH−(aq)

The formation of some hydroxide ions will render the solution basic.

Example: Account for the fact that sodium hydrogen sulfite produces an acidic solution when it is dissolved in water.

NaHSO3(s) Na+(aq) + HSO3

−(aq)

The sodium ions will remain as is, but the basic hydrogen sulfite ions will react with water:

HSO3−

(aq) + H2O(l) SO32−

(aq) + H3O+(aq)

The formation of some hydronium ions will render the solution acidic.

Example: Account for the fact that salts containing ammonium produce acidic solutions when they are dissolved in water.

Ammonium reacts with water (which acts as a base, accepting a hydrogen ion).

NH4+

(aq) + H2O(l) H3O+(aq) + NH3(aq)

Amphoteric Substances We`ve learned that water can act as an acid or a base:

o NH3 + H2O NH⇌ 4+ + OH- here water acts as a(n) acid.

o HCl + H2O → H3O+ +Cl- here water acts as a(n) base. Substances that can act as an acid in one reaction and a base in another are called amphoteric

or amphiprotic substances. Another amphoteric substance is HSO4

- (bisulphate ion).

Polyprotic Acids Acids that can give up more than one hydrogen ion per molecule are called polyprotic acids.

Diprotic acids, which can release two hydrogen ions, are much more common than triprotic acids, which can release three hydrogen ions.

Ex. Ionization of sulfuric acid:

H2SO4(aq) + H2O(l) H3O+(aq) + HSO4

−(aq)

HSO4−

(aq) + H2O(l) H3O+(aq) + SO4

2−(aq)

Ex. Ionization of Phosphoric acid

H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4

−(aq)

H2PO4−

(aq) + H2O(l) H3O+(aq) + HPO4

2−(aq)

HPO42−

(aq) + H2O(l) H3O+(aq) + PO4

3−(aq)

Example: Write equations for the ionization of sulfurous acid in water.

H2SO3(aq) + H2O(l) H3O+(aq) + HSO3

−(aq)

HSO3−

(aq) + H2O(l) H3O+(aq) + SO3

2−(aq)

See Acids and Bases Assignment

Conjugate Acid-Base Pairs Acid-base reactions involve the exchange of hydrogen ions. When an acid loses a hydrogen ion

it becomes a base, and when a base accepts a hydrogen ion it becomes an acid. Structures that differ in only one hydrogen ion are called conjugate acid-base pairs. The stronger the acid, the weaker will be its conjugate base, and vice versa. In these reactions, not only the reactants but also the products are acids and bases.

H2O(l) + NH3(aq) NH4+

(aq) + OH−(aq)

acid1 base2 acid2 base1

HNO3(aq) + H2O(l) H3O+(aq) + NO3

−(aq)

acid1 base2 acid2 base1

o Acid1 is the conjugate acid of base1, and base2 is the conjugate base of acid2.o Notice, as well, that the formula for the conjugate base always has one more

negative charge than the formula for the corresponding acid.

*Notice that water is acting as an acid in the first reaction, but as a base in the second one.*

Acid Base Conjugate Base Acid ConjugateHCl Cl- OH- H2OH2O OH- HSO4

- H2SO4H3O+ H2O NO3

- HNO3NH4

+ NH3 H2O H3O+

See Conjugate Acids and Bases Assignment

Dissociation of Water Write out the chemical equation for the dissociation of water:

H2O(l) + H2O(l) H3O+(aq) + OH−

(aq)

The equilibrium constant expression for the self-ionization of water can be written as follows:

Kw = [H3O+][ OH−] = 1.0 × 10-14 = ion product constant of water

Since Kw is so small, H2O is favored over H+/OH- production (ie. Not a lot of water dissociates)

Determine [H3O+] and [OH-] for pure water

Kw = [H3O+][ OH−] = 1.0 × 10-14

In pure water [H3O+]=[OH-]=x Kw = 1.0 x 10-14 = x2

x = 1.0 x 10-7 so in pure water [H3O+]=[OH-] = 1.0 x 10-7M

The ion product constant is valid for all aqueous solutions. If an acid is added to pure water, the hydronium ion concentration will increase. Since the value of Kw is constant, the concentration of the hydroxide ion will decrease.

[H3O+] does not have to equal [ OH−], they just need to multiply to give 1.0 × 10-14

Example: What are the hydronium ion and hydroxide ion concentrations in a 0.050 M aqueous solution of hydrogen chloride at 25°C?

Equation: HCl(g) + H2O(l) H3O+(aq) + Cl−

(aq)

HCl almost completely dissociates, [H3O+] =[H3O+](from water) + [H3O+](from HCl)= 1.0 x 10-7M + 0.050M

= 0.050M

Use Kw expression to calculate [OH-] Kw = [H3O+][ OH−] = 1.0 × 10-14

[ OH−] = = 2.0 × 10-13 mol/L

Example: What are the hydroxide ion and hydronium ion concentrations in an aqueous solution containing 0.010 M barium hydroxide?

Equation: Ba(OH)2(s) Ba2+(aq) + 2 OH−

(aq)

Since Ba(OH)2 is a strong base, it completely dissociates: [OH-](from Ba(OH)2)=2 x 0.010M = 0.020M

[OH-]final= [OH-](H2O) + [OH-](Ba(OH)2)

[OH-]final= 1.0 x 10-7M + 0.020M = 0.020M

Use Kw expression to determine [H3O+] Kw = [H3O+][ OH−] = 1.0 × 10-14

[H3O+] = = 5.0 × 10-13 mol/L

Ex. What is [H+] in an aqueous solution in which [OH-] = 1.0 x 10-3?

Kw = [H+][OH-]= 1.0x10-14

1.0 x 10-14 = [H+](1.0x10-3)

[H+] = 1.0 x 10-11

See Dissociation of Water Assignment

pH Scale and Indicators In 1909 the Danish biochemist, Søren P. Sørenson, introduced the pH scale as a way to

express the acidity and basicity of an aqueous solution (pH is a measurement that expresses how acidic or basic a solution is).

pH scale ranges from 0-14 when pH = 7 the solution is

neutral and moles of H3O+ = moles of OH-

if the pH < 7 the solution is acidic and [H3O+]>[OH-]

if the pH > 7 the solution is basic and [H3O+]<[OH-]

The pH of a solution is defined as the negative logarithm, to the base ten, of the hydronium ion concentration.

pH = −log10[H3O+]

In a neutral solution at 25°C, the hydronium ion and the hydroxide ion concentrations are both 1.0 × 10-7 mol/L. Thus, the pH of a neutral solution is 7.

pH = −log10(1.0 × 10-7) = 7.00

Example: What will be the pH of an aqueous solution containing 0.040 M sodium hydroxide?

Kw = [H3O+][ OH−] = 1.0 × 10-14

[H3O+] = = 2.5 × 10-13 mol/L

pH = −log10[H3O+] = −log10(2.5 × 10-13) = 12.60

Example: What is the hydronium ion concentration of a solution with a pH of 2.50?

pH = −log10[H3O+] = 2.50

log10[H3O+] = −2.50

[H3O+] = antilog (−2.50) = 3.2 × 10-3 mol/L

Determining pOH pOH is another way to express how acidic or basic a solution is. It’s basically just the opposite of pH.

pH + pOH = 14.00 pOH = −log10[OH−] pOH = 14.00 – pH

Example: A solution has a pH of 3.00, is it an acid or base? What would be the pOH of the same solution? Acid, pOH = 11.0

Example: An aqueous solution containing 0.040 M sodium hydroxide has the following pOH:

pOH = −log10[OH−] = −log10(0.040) = 1.40

pH + pOH = 14.00

pH = 14.00 – pOH = 14.00 – 1.40 = 12.60

See pH Assignment

Acid-Base Indicators Indicators are chemicals that appear different colours in acids and bases. Indicators are weak acids or bases Indicators change colours when the pH (or pOH) of a solution reaches a certain range, so they

can be used to determine the pH of solutions.

Acid-Base Titrations

Recall: neutralization refers to the double displacement reaction which occurs when an acid is mixed with a base, the products of neutralization are a salt and water.

To achieve neutralization we use a process called titration.

o One reactant is carefully added to another reactant until the two have combined in their exact stoichiometric proportions.

The objective of a titration is usually to find the number of moles or grams, the concentration, or the percentage of the analyte (the substance we are looking for) in a sample. This is usually done by measuring the precise volume and concentration of a titrant (the solution being added) needed to react completely with the analyte. Stoichiometric relationships are then used to determine the quantity of analyte present from the number of moles (volume × molarity) of titrant added.

o During titration, the solution reaches an equivalence point where the moles of acid equals the moles of the base (this does not necessarily mean pH = 7).

o When doing titrations we use a burette because it is really accurate at measuring volumes.

o When performing a titration you add your titrant to the analyte (with indicator) until there is ONE DROP that has a permanent colour change.

When you hear the term ‘neutralizatoin’ you may think that results in a final pH of 7, but this is not the case. The final pH depends on the strength of the acid and base being combined.

o Strong acid + Strong base pH = 7 (neutral) when neutralized.

o Strong acid + weak base pH < 7 (acidic) when neutralized.

o Weak acid + strong base pH > 7 (basic) when neutralized.

Titration Calculations: When neutralized, moles H+ = moles OH-, so we can use the formula:

o C (MA x VA ) = D (MB x VB), where C and D represent the coefficients from the balanced equation.

o Alternatively the formula could be MA x VA = MB x VB where MA represents the molarity of H+ and MB represents the molarity of OH+

Example: In a titration, a few drops of bromothymol blue indicator are added to 16.80 mL of an aqueous sodium hydroxide solution of unknown concentration and is neutralized by 25.00 mL of a 0.190 M solution of sulfuric acid. What is the concentration (initial) of the sodium hydroxide solution? Remember: when neutralized, moles H+ = moles OH-

H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l)

moles of H2SO4 = 0.02500 L × 0.190 mol/L

= 4.75 × 10-3 mol

moles of NaOH = 4.75 × 10-3 mol H2SO4

= 9.50 × 10-3 mol NaOH

c = [NaOH] = = 0.565 M

Example: A 20.0 mL aqueous solution of strontium hydroxide, has a drop of indicator added to it. The solution turns colour after 25.0 mL of a standard 0.0500 M HCl solution is added. What was the original concentration of the strontium hydroxide?

Va = volume of acid solution in litres

Vb = volume of basic solution in litres

Ma = molar concentration of the acid solution

Mb = molar concentration of the base solution

2 HCl + Sr(OH)2 SrCl2 + 2 H2O 2 mol 1 mol

moles of acid = 2(moles of base)

VaMa = 2VbMb

Mb = = =0.0312 M

See Titration Assignment

Primary Standards To perform a titration, the concentration of one of the solutions must be precisely and

accurately known. These substances are called primary standards. A primary standard must meet the following criteria:

1. It should be obtainable as a very pure solid at reasonable cost.

2. The substance should be air stable. That is, it should not react with any component in the air, such as oxygen, carbon dioxide, or water vapour.

3. The substance should be stable in solution for a reasonable length of time.

4. A substance with a high molar mass is preferred. This minimizes weighing errors.

Ionization Constants for Acids and Bases Since weak acids and bases partially dissolve in water and reach equilibrium, equilibrium

law expressions can be written.

HF(aq) + H2O(l) H3O+(aq) + F−

(aq)

Ka = = acid ionization constant

The larger the acid ionization constant, the stronger the acid (really strong acids almost completely dissociate so no equilibrium is reached).

The base ionization constant is the equilibrium constant expression for a weak base. Conjugate Acid-Base Pairs:

Ka x Kb = Kw here`s why:

HF(aq) + H2O(l) H3O+(aq) + F−

(aq) Ka =

F

−(aq) + H2O(l) HF(aq) + OH−

(aq) Kb =

Ka × Kb = × = [H3O+][OH−] = Kw

Ka × Kb = Kw = 1.0 × 10-14

We can calculate the value of an acid ionization constant if we know the acid concentration and the pH of the solution. Conversely, if we know the concentration and acid ionization constant, we can calculate the expected pH of the solution.

Example: Acetylsalicylic acid (aspirin) is a weak monoprotic acid. A 0.100 M solution of the acid has a pH of 2.24. Calculate the acid ionization constant of the acid (Ka).

HAsp(aq) + H2O(l) H3O+(aq) + Asp−

(aq)

Ka =

[H3O+] = antilog(−pH)

= antilog(−2.24)

= 0.0058 M

Ka = = 3.6 × 10-4

Example: Ascorbic acid (vitamin C) is a weak monoprotic acid. It has a K a = 8.0 × 10-5. Calculate the pH of a 0.100 M solution.

[HAsp] [H3O+] [Asp−]

Initial 0.100

Change −0.0058

+0.0058 +0.0058

Equilibrium 0.094 0.0058 0.0058

HAsc(aq) + H2O(l) H3O+(aq) + Asc−

(aq)

Ka =

8.0 × 10-5 = , approx. = = 1300 > 1000

8.0 × 10-5 =

x2 = (8.0 × 10-5)(0.100) = 8.0 × 10-6

x = 2.8 × 10-3

pH = −log(2.8 × 10-3) = 2.55

Example: Caffeine is a weak base that is related to ammonia. It has a Kb = 4.1 × 10-4. Calculate the pH of a 0.70 M solution.

cafN(aq) + H2O(l) cafNH+(aq) + OH−

(aq)

Kb =

4.1 × 10-4 = ,

approx. = = 1700 > 1000

x2 = (4.1 × 10-4)(0.70) = 2.9 × 10-4

x = 1.7 × 10-2

[HAsc] [H3O+] [Asc−]

Initial 0.100

Change − x + x + x

Equilibrium

0.100 − x x x

[cafN] [cafNH+] [OH−]

Initial 0.70

Change − x + x + x

Equilibrium 0.70 − x x x

[OH−] = 1.7 × 10-2 mol/L

pOH = −log(1.7 × 10-2) = 1.77

pH = 14.00 – 1.77 = 12.23

Buffers Buffers are mixtures of chemicals that

make a solution resist a change in pH. Solutions that have a resistance to changes in their pH because of the presence of buffers are called buffer solutions.

In general, buffers are made up of (1) a weak acid and one of its soluble salts, or (2) a weak base and one of its soluble salts.

Buffers of the first type buffer a solution in the acid range. Buffers of the second type buffer a solution in the base range. Both kinds of buffers make solutions in which an equilibrium exists that shifts in response to the addition of an acid or base so as to allow only small changes in the hydrogen-ion concentration.

Acids & Bases http://www.saskschools.ca/curr_content/chem30_05/5_acids_bases/labs/labs.htm

Acid-Base Titration

OVERVIEW

Often we want to determine the concentration of a solution. One way to do so is to carrying out an analytical procedure known as a titration. During a titration a carefully measured volume of the solution with the unknown concentration (called the analyte) is reacted with a second solution (the titrant) whose concentration is known (a standard solution). By knowing how much of the standard solution is required to react completely – no more, no less – with the solution with the unknown concentration we can calculate that solution’s concentration.

The point at which stoichiometrically equal amounts of the two solutions have been combined is called the equivalence point. When we neutralize an acid with a base, this will occur when [H+] = [OH-]. By using an appropriate indicator we can detect this point by noting when the indicator changes colour. This will be used to signify the end point of the titration. A balanced equation and simple calculations will then allow us to determine the concentration of the solution.

PURPOSE

To determine the concentration of a solution of NaOH by titration with a standard solution of HCl.

To determine the concentration of a sample of white vinegar by titration with a standard solution of NaOH

SAFETY

Acids and bases are corrosive substances. Safety goggles must be worn. Be sure to report any spills to your teacher so they may be cleaned up properly.

EQUIPMENT AND MATERIALS

two 50-mL buretsburet stand and clampsErlenmeyer flask, 125-mLErlenmeyer flask, 250-mLwash bottledistilled water10-mL graduated cylinder10-mL volumetric pipette (optional)

0.100M HCl standard solution

NaOH solution with unknown concentrationvinegar (acetic acid, HC2H3O2)phenolphthaleindistilled water

PROCEDURE

Part A. Titration of Base of Unknown Concentration

1. Wash two burets with detergent solution. Rinse them thoroughly.

2. With a grease marking pencil or tape identify which buret is to hold each solution, the acid or base.

Rinse each buret with about 10 mL of solution that it is to hold – rinse the acid-containing buret with the HCl solution and the base-containing buret with the NaOH solution. Allow the acid or base to run out of the buret tip to rinse them.

3. Fill each buret with the proper solution and allow the some of each solution to run out of the buret tip. Make sure no drop remains hanging on the buret. Be sure there are no air bubbles in the tips.

It is very important that you accurately read and record the initial and final volumes. It is not necessary that the burets be filled to the very top mark (0.0 mL) at the start of the titration, but it is important that the level never go below the bottom mark (50.0 mL). Be sure to read the bottom of the meniscus at eye level. You may find it helpful to hold a white card with a large black streak or rectangle behind the buret to make it easier to read.

4. Place the 125-mL Erlenmeyer flask beneath the acid buret. Add 10.0 mL of acid to the flask. Use your rinse bottle to make sure all drops make it to the bottom of the flask; rinse any drops that remain on the sides of the flask. Read the buret carefully and record both the initial and final volumes from the buret into your data table.

5. Add 10-mL of distilled water to the flask.

6. Add three drops of phenolphthalein to the flask, and swirl the flask to mix thoroughly.

7. Move the flask so it it beneath the base buret. Place the flask on a sheet of white paper so a colour change will be more readily observed.

8. After recording the initial volume of base in the buret, begin the titration by adding NaOH to the flask. For your initial trial you may want to add the base fairly quickly until you notice a pink colour appearing in the flask. Swirling the flask should make the pink colour disappear. At that point begin adding the NaOH more slowly, swirling the flask after each drop is added. As soon as a faint pink colour becomes permanent, stop the titration – the end point has been reached. Do NOT continue until a darker pink colour has been reached – if that happens you’ve gone past the end point.

If you do go past the end point, add a few drops of acid (be sure to record the new volume used), then add more base.

Record the final volume of base in the buret.

9. Repeat the titration, performing at least four trials. Be sure to rinse the Erlenmeyer flaks well between trials.

For your other trials add the base more slowly as you near the end point in order to get more accurate readings. You do not need to refill burets between trials.

Part B. Titration of Vinegar

1. Using the volumetric pipette (or another clean buret), add exactly 10.0 mL of vinegar to a clean 250-mL Erlenmeyer flask.

2. Add 100 mL of distilled water to the flask and three drops of phenolphthalein.

3. Titrate the vinegar with the NaOH solution used in Part A. If necessary add more NaOH to the buret before beginning the titration. Record the initial volume of base in the buret.

4. As before, the end point will be reached as soon as a permanent, pale pink colour appears in the flask. Record the final volume of base.

5. Repeat the titration at least two more times.

RESULTS

Copy Data tables 1 and 2, as shown on the last page of this lab, into your data notebook.

CALCULATIONS

Part A. Titration of Base of Unknown Concentration

To calculate the concentration of the unknown base we must begin with a balanced equation. The reaction between hydrochloric acid and sodium hydroxide is:

HCl + NaOH → NaCl + H2O

Stoichiometrically we see that one mole of the acid reacts with one mole of the base. Because of this one-to-one relationship we can use the following formula to calculate the unknown concentration:

Macid Vacid = Mbase Vbase

Rearrange the equation to solve for the unknown concentration of the base:

M base=M acid×V acidV base

For each trial in Part A, determine the molarity of the NaOH solution, [NaOH]. Show your calculations in a table similar to the one shown below. Calculate the average for your trials.

Table 3. Calculating the concentration of the sodium hydroxide solution.

Trial Calculations [NaOH]

M base=M acid×V acidV base

1

2

3

4

Average -----

Collect the data from the rest of the class. Calculate the class average:

Table 4. Class data for the concentration of the sodium hydroxide solution.

Group [NaOH]

1

2

3

etc.

Average

Part B. Titration of Vinegar

The reaction between sodium hydroxide and vinegar – acetic acid, HC2H3O2 – is represented by:

HC2H3O2 + NaOH NaC2H3O2 + H2O

Again there is a 1:1 relationship between the acid and the base. As before we can determine the concentration of the unknown solution – in this case the acetic acid – if we know the volume and molarity of the base and the volume of the acid used:

M acid=Mbase×V baseV acid

Using the molarity of the base you calculated in Part A of the lab, determine the molarity of the acetic acid. Show your calculations in Table 5, which you should copy into your notebook.

Collect the data from the rest of the class. Calculate the class average:

Table 5. Calculating the concentration of the vinegar solution.

Trial Calculations [HC2H3O2]

M acid=M base×V baseV acid

1

2

3

4

Average -----

Table 6. Class data for the concentration of the vinegar solution.

Group [NaOH]

1

2

3

etc.

Average

CONCLUSIONS AND QUESTIONS

1. How did the results for each of your trials for the titration of the sodium hydroxide compare? Were the results similar or did they vary a great deal?

2. What are some of the major sources of error with this experiment?

3. The volume of water added during this experiment – to rinse droplets of acid from the buret or as water dded to the acid in the flask – does not affect the calculations and thus does not need to be accounted for. Why not?

Data Tables.

Table 1. Titration of NaOH with Unknown Concentration

Trial 1 Trial 2 Trial 3 Trial 4

HCl NaOH HCl NaOH HCl NaOH HCl NaOH

initial volume

final volume

volume used

Table 2. Titration of Vinegar

Trial 1 Trial 2 Trial 3 Trial 4

Vinegar NaOH Vinegar NaOH Vinegar NaOH Vinegar NaOH

initial volume - - - -

final volume - - - -

volume used 10.0 10.0 10.0 10.0