Good mornin’Lot’s of things to pick up today:

Quiz ScantronQuiz version that you tookLab SheetFinal Study Guide Dry Erase BoardMarker

10 minutes to look over missed quiz questions & answer the Catalyst

pHTest Tube [H+] (mol/L) pH

1 0.01 <42 0.001 <43 0.0001 44 0.00001 55 0.000001 6

• How does the molarity of H+ change?• How does the pH change?• Why are the pH’s for test tubes 1 and 2 indicated as “<4”?• What is an Arrhenius acid? What is an Arrhenius base?• What is a Bronsted-Loewry acid? What is a Bronsted-Loewry

base?

Example Brønsted Acids and Bases:

Here, H2O acts as a Brønsted acid by donating a proton to NH3 which acts as a

Brønsted base.Conjugate base and conjugate acid?

NH3 + HOH NH4+ + OH-

pHpH is a scale used to reflect the concentration of H+

ions!Acids and bases are aqueous solutionsWater can act as an acid and a base:

HOH <--> H+ + OH- Why is water neutral?

Acids dissociate and increase the number of protons (H+) in solution

Bases dissociate and increase the number of hydroxide (OH-) in solution either directly (Arrhenius) or indirectly (Bronsted-Loewry)

Neutralization ReactionRemember that this is just an acid base reaction

What would a product of a neutralization reaction be?

Just like the stoichiometry you’ve been doing, in lab we can use this fact to calculate the molarity of solution whose concentration is unknown

TitrationDefinition◦ Analytical method

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

standard solution

unknown solutionCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Buret

stopcock

Erlenmeyer flask

Titration VocabularyTitrant◦ The substance added to the analyte in a

titration (a standard solution)Analyte◦ The substance being analyzed

Equivalence point◦ The point in a titration at which the

quantity of titrant is exactly sufficient for stoichiometric reaction with the analyte.

Acid-Base Titration

Titrant

Analyte

If the concentration of the titrant is known, then the unknown concentration of the analyte can be determined.

Buret Reading

Why do chemists use titrations??

Quantitative analysis — used to determine the amounts or concentrations of substances present in a sample by using a combination of chemical reactions and stoichiometric calculations

Acidic, basic, or neutral??

The “perfect pink” for a titration with phenolphthalein

Indicator - changes color to indicate pH change

Volume base added

Example… phenolphthalein is colorless in acid and pink in basic solution

pH

Endpoint =7

pink

point at which exactly enough reactant has been added for the solution to be neutralized and no more

Equivalence point (endpoint)Point at which equal

amounts of H3O+ and OH- are present in solution.

Determined by…indicator color changedramatic change in pH

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Titration

moles H3O+ = moles OH-

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

White Board Review!SolubilityMolarityDilutionsSolutions StoichiometryAcid and Base

Problem #1What mass of (NH4)2SO4 is required to make 1.25

L of a 0.250 M solution of NH4+?

Answer: 20.6 g (mm=132.14g/mol)

Problem #2If 25 g of KCl is added to 50 g of water at 40°C, the solution would be:

1)Unsaturated2)Saturated3)Supersaturated

Problem #3Calculate the molarity of a solution prepared by

dissolving 4.1 g of solid KBr in enough water to make 1.10 L of solution

Answer: 0.031 M (mm=119.02 g/mol)

Problem #4One way to determine the amount of chloride ion in a

a water sample is to titrate the sample of standard AgNO3 solution to produce solid AgCl.

Ag+(aq) + Cl-(aq) AgCl(s)If a 25.0 mL water sample requires 27.2 mL of 0.104 M

AgNO3 in such a titration, what is the concentration of Cl- in the sample?

Answer: 0.113 M

Problem #5What volume of of a a 5.00M Ca(NO3)2 solution is

needed to prepare 465 mL of a 0.250 M Ca(NO3)2 solution?

Answer: 23.3 mL

Problem #6Calculate the mass of sodium iodide that must

be added to 425.0 mL of a 0.100 M lead (II) nitrate solution to precipitate all of the lead (II) ions as lead (II) iodide.

Answer: 12.7 g NaI