Experiment 7: Experiment 7: Acids, Bases Acids, Bases

and Saltsand SaltsGroup # 4 TCD3Group # 4 TCD3

Mateo, Nachura, San DiegoMateo, Nachura, San Diego

Definition of TermsDefinition of Terms• Electrolytes – substances that, when

dissolved in water, results in a solution which conducts electricity

• Nonelectrolytes – substances that, when dissolved in water, gives a solution that does not conduct electricity

Definition of TermsDefinition of Terms• Acids

– Arrhenius: substance that yields H+ when dissolved in water

– Bronsted-Lowry: Proton donor– Lewis: Electron-pair acceptor

Definition of TermsDefinition of Terms• Bases

– Arrhenius: A substance that yields OH- when dissolved in water

– Bronsted-Lowry: Proton acceptor– Lewis: Electron-pair donor

Definition of TermsDefinition of Terms• Salts – An ionic compound made up

of a cation (not H+) and an anion (not OH- or O2-)

Definition of TermsDefinition of Terms• Neutralization reaction – a reaction

between an acid and a base• Titration – The gradual addition of a

solution of accurately known concentration to another solution of unknown concentration until the chemical reaction between the two solutions is complete

ElectrolysisElectrolysis• 5 drops of each was put in a test

tube and tested with indicators – NaOH – NH4Cl– HCl– HC2H3O2

– NaCl– C2H5OH– C2H22O11

Reagents0.1 M

Solutions

Classification (acid, base or neutral)

pH Classification of acid/base (weak or strong)

Conductivity (good,

weak, none)

Classification (strong, weak non-electrolyte)

NaOH Base 14 Strong Good Strong

NH4Cl Acid 3 Strong Good Strong

HCl Acid 1 Strong Good Strong

HC2H3O2 Acid 2 Strong (SHOULD HAVE BEEN WEAK)

Weak Weak

NaCl Neutral 6 Weak Acid (or Neutral) Good Strong

H2O Neutral 7 Neutral None Non-electrolyte

Sucrose Neutral 5 Weak acid (or Neutral) None Non-electrolyte

Ethanol Neutral 6 Weak acid (or Neutral) None Non-electrolyte

HOAc + NaOH

Acid 5 Weak acid (SHOULD HAVE BEEN BASIC)

XXX XXX

HCl + NaOH

Base 13 Strong (SHOULD HAVE BEEN NEUTRAL)

XXX XXX

ResultsResults

ExplanationExplanationBlue litmus paper

Red litmus paper

Congo Red

Phenolphthalein

pH paper

Acid Red Red Blue Colorless

<7

Base Blue Blue Red Pink >7

What happened.What happened.• They were also tested using the

conductivity apparatus

Reagents0.1 M

Solutions

Classification (acid, base or neutral)

pH Classification of acid/base (weak or strong)

Conductivity (good,

weak, none)

Classification (strong, weak non-electrolyte)

NaOH Base 14 Strong Good Strong

NH4Cl Acid 3 Strong Good Strong

HCl Acid 1 Strong Good Strong

HC2H3O2 Acid 2 Strong (SHOULD HAVE BEEN WEAK)

Weak Weak

NaCl Neutral 6 Weak Acid (or Neutral) Good Strong

H2O Neutral 7 Neutral None Non-electrolyte

Sucrose Neutral 5 Weak acid (or Neutral) None Non-electrolyte

Ethanol Neutral 6 Weak acid (or Neutral) None Non-electrolyte

HOAc + NaOH

Acid 5 Weak acid (SHOULD HAVE BEEN BASIC)

XXX XXX

HCl + NaOH

Base 13 Strong (SHOULD HAVE BEEN NEUTRAL)

XXX XXX

ResultsResults

ExplanationExplanation• Electrolytes are substances which

ionize or dissociate in water. The ions are attracted to electrodes in the conductivity apparatus. The movement of ions sets up an electric current and electricity is produced.

ExplanationExplanation• In general:

– acids and bases are electrolytes– strong acids and strong bases: strong

electrolytes– weak acids and weak bases: weak

electrolytes– neutral, ionic substances can also

conduct electricity

Preparation of 1 M NaOHPreparation of 1 M NaOH• Formula

– M = n/L, where M is the molarity of the substance, n is the number of moles, and V is the volume in liters

– g = n x MW (molecular weight)

Preparation of 1 M NaOHPreparation of 1 M NaOH• Calculation

– 1 M NaOH = n / 0.1 L– n = 0.1 mol NaOH– g NaOH = 0.1 mol NaOH x

(40gNaOH/1mol NaOH)– g NaOH = 4 grams

Preparation of 0.1 M NaOHPreparation of 0.1 M NaOH• Formula

– M1V1=M2V2

– M is the molarity of the substance– V is the volume in liters

Preparation of 0.1 M NaOHPreparation of 0.1 M NaOH• Calculation

– M1V1=M2V2

– (0.1M)(100ml) = (1M)(V)– V = 10 ml

Titration of an Acid with a Titration of an Acid with a BaseBase

• Two drops of phenolphthalein was added to the solution of the unknown acid with a volume of 60 ml

• It was carefully titrated with the standardized solution (0.1 M NaOH) until the solution turned a light pink color

Titration of an Acid with a Titration of an Acid with a BaseBase

• This light pink color means that the equivalence point has been reached, meaning the acid has been completely neutralized by the base.

Titration of an Acid with a Titration of an Acid with a BaseBase

Volume of 0.10 M NaOH 11 ml

Volume of unknown acid solution 60 ml

Molarity of unknown acid solution 0.0183 M

Computation:

0.1 M NaOH = (n/.011 L)

n = 1.1 x 10-3 mol

Munknown = (1.1 x 10-3)/(0.060 ml)

M = 0.0183 molar

QuestionsQuestions1. From your results, what

relationship can you draw between the acidity/basicity of a substance and its electrolyte property?

• All acids and bases are electrolytes. The strength of its conductivity is determined by its strength in acidity/basicity. Strong acids and strong bases are strong electrolytes while weak acids and weak bases are weak electrolytes. Not all neutral substances are electrolytes. Neutral, ionic substances are electrolytes and can conduct electricity.

2. Why are electrolytes conductors of electricity?

• Electrolytes are conductors of electricity because they can ionize in water. The ions are attracted to the electrodes, and the movement of the ions sets up an electric current, generating electricity. The electric current set up is equivalent to the flow of electrons along a metal wire.

3. Calculate the pH of the following solutions:

a. 0.001 M HCl[H+] = 0.001 MpH = -log (0.001M)pH = 3

b. 0.005 M NaOH[OH-] = 0.005 MpOH = -log(0.005M)pOH = 2.30pH = 11.70

c. 0.10 M NH4OH solution (Kb = 1.8 x 10-5)

Kb = [NH4+][OH-] = 1.8 x 10-5[NH4OH]Kb = x2 = x2 = 1.8x10-5

0.10-x 0.10x2 = 7.2 x 10-6x = 2.7 x 10-3 MpOH = -log (2.7 x 10-3) = 2.57pH = 14-2.57pH = 11.43

d. 0.02 M HCOOH (Ka = 1.8 x 10-4)

Ka = [COOH-][H+] = 1.8 x 10-4[HCOOH]Ka = x2 = 1.8 x 10-5

0.02-x0 = x2 + (1.8 x 10-5)x – 0.02(1.8 x 10-5)x = -6.09 x 10-4 and/or 5.91 x 10-4x = 5.91 x 10-4 pH = -log(5.91x10-4)pH = 3.23

ConclusionConclusion• Indicators and pH paper are used to

classify acids, bases, and salts.• The conductivity apparatus is used to

classify electrolytes as weak or strong.

• Bases with the highest pHvalues are classified as strong and are good conductors of electricity.

• Acids with the lowest pH values are classified as strong and are good conductors of electricity.

• Weak acids and bases are weak conductors of electricity.

ConclusionConclusion

ConclusionConclusion• Neutral solutions are solutions which

have a pH of 7. Not all neutral solutions can conduct electricity.

ConclusionConclusion• By determining the endpoint of an

acid-base titration the concentration of an unknown solution can be known.