CHAPTER 4 AP CHEMISTRY

PRECIPITATION PROBLEMS Water Highly polar Ionic and polar compounds are attracted to the positive and/or negative ends of the water molecule. This is called HYDRATION (a hydration reaction is something different) When a precipitate is formed there is a zero net charge Solubility rules (handout) What happens when a ionic compound dissolves in water? They break up into their ion form Cation (positive ion) and anion (negative ion) What is a precipitate? A solid is formed from two aqueous solutions Using the solubility rules predict what will happen CuCl2 + (NH4)2SO4 Ba(NO3)2 + Na2CO3

NET IONIC EQUATIONS Net ionic equations include those species that participate in the reaction What are spectator ions? Ions that do not participate in the reaction You must show that atoms and charges are balanced Write a net ionic equation for the following: KOH and FeCl2 MgCl2 and AgNO3 (NH4)2SO4 and LiBr

ACID/BASE An acid is a species that Has hydrogen ions in a water solution Has a sour taste Reacts with bases Changes an indicators color A base is a species that Has hydroxide ions in a water solution Has a bitter taste Reacts with acids Strong acids Ionizes 100% in water Strong electrolyte HCl, HClO4, HClO3, HNO3, H2SO4, HBr, HI Weak acids Ionizes slightly into ions HB(aq) <=> H+ (aq) + B-(aq) HF, any carbon containing acid

CONTINUED Weak acid with water CH3COOH + H2O <=> H3O+ + CH3COO- A strong base in water Dissociates 100% KOH --> K+ + OH- Hydroxides with group one, Ba2+, Sr2+, and Ca2+ (slightly) Weak bases in water Dissociates slightly B + H2O <=> BH+ + OH- F-, H3PO4, NH3, CH3NH2

ELECTROLYTES Solute - substance Solvent - water, what you put solute in Electrolytes- conduct electricity Strong - strong acids and bases, soluble salts Weak - weak acids and bases, insoluble or slightly soluble salts Nonelectroytes - molecules, nonpolar covalent compounds ARRHENIUS Ions are responsible for conducting electricity Down played convictions to get PhD then crusaded to get theory accepted Received Nobel prize 1903

EXAMPLES Weak acid - weak base Can be acidic, basic or neutral depending on which is stronger, acid or base Strong acid - strong base Forms a salt and is a strong electrolyte Neutralization reaction, pH is neutral H+ (aq) + OH-(aq)  H2O(l) Weak acid - strong base Forms a base CH3COOH + OH-  H2O + CH3COO- Strong acid - weak base HCl + NH3  NH4Cl H+ (aq) + Cl- (aq) + NH3(aq) --> NH4+(aq) +Cl- (aq) H+ (aq) + NH3(aq) --> NH4+(aq)

NET IONIC REACTIONS Carbonic acid with sodium hydroxide H2CO3 + NaOH H2CO3 + Na+ + OH- -->H2O(l) + Na+ + CO32- H2CO3(aq) + OH-(aq) --->H2O(l) + CO32-(aq) Dimethylamine with sulfuric acid H3C-NH-CH3 Hydroiodic acid with barium hydroxide

OXIDATION-REDUCTION REACTIONS An oxidation-reduction reaction involves transfer of electrons Addition of O2 or H2 often occur LEO goes GER Oxidized or oxidation Species that lose electrons oxidation number goes up Cl- --> Cl2 Reduced or reduction Species that gain electrons or oxidation number reduced Cu+ + Al --> Cu + Al3+ Oxidizing agent What causes oxidation Reducing agent What causes reduction

OXIDATION NUMBER RULES Elements found in nature is zero Cu, O2 Monoatomic ions same as charge Cl-, Mg2+ Fluorine always 1- Oxygen 2- Except with fluorine 2+ Peroxide 1-, H2O2 Superoxide ½ -, KO2 Hydrogen 1+ With group one elements then a hyride 1- Sum of oxidation numbers of a neutral compound is zero The most electronegative element number is the same as the ion charge Sum of oxidation numbers of a polyatomic ion is same as its charge

REDOX What is the oxidation number of carbon in CaC2O4? Of Chromium in Cr2O72-? The concept of oxidation numbers explain oxidation and reduction Oxidation: increase in oxidation number (lose electrons) Reduction: decreases in oxidation number (gain electrons) Balancing redox equations The half-equation method Split the equation into two half-reactions One for reduction and the other for oxidation Balance one of the half-reactions; balance atom other than O or H first, next use water to balance oxygen, then add hydrogen ions to balance the hydrogen, add electrons to balance charge Balance other half-reaction Combine the two in such away as to eliminate all the electrons

PROBLEMS Consider the unbalanced equation Cr3+(aq) + Cl- (aq) ---> Cr(s) + Cl2(g) Ions can only be free in an aqueous solution The reaction between MnO4- and Fe2+ (acidic). The reaction can be represented by the following (unbalanced) equation MnO4- + Fe2+ ---> Mn2+ + Fe3+ Reaction between MnO4- and I-(basic). The reaction can be represented by the following unbalanced equation MnO4- + I- --> I2 + MnO2

MOLARITY Number of moles of solute per liter of solution M = number of mol/L Symbol [ ] means molar concentration Calculate the molarity of a solution when 16.3 g of ammonium sulfate is added to a 500. mL volumetric flask and distilled water is added until the 500. mL mark is reached. 16.3g/ 1 mol / 1000 mL /________ / 132.1g / 1 L / 500. mL .247 M How many moles of 18.0 M sulfuric acid are there in 35.0 mL of the solution? 18.0 mol / 35.0 mL / 1L_______ L / / 1000 mL .630 moles H2SO4

EXAMPLES How many moles of chloride ions are in 1.75 L of 1.0 x 103- M zinc chloride? Formalin is an aqueous solution of formaldehyde (HCHO). At high concentrations it is used as a preservative for biological specimen's. How many grams of formaldehyde are contained in 2.5 L of 12.3 M formalin? Moles of solute after dilution = moles of solute before dilution M1V1 = M2V2 When aqueous solutions of sodium hydroxide and chromium (III) nitrate are mixed a green precipitate is formed. 45.00 mL of 0.200 M NaOH and 22.50 mL of 0.150 M Cr(NO3)3. Calculate the mass of precipitate. A solution of permanganate reacts with iron (II) in solution to give manganese (II) and iron (III) ions. How many mL of 0.213 M KMnO4 solution will react completely with 50.00 mL of 0.128 M Fe(NO3)2?

TITRATION A volumetric analysis involving titration, where a measured volume of a known concentration of an acid or base is required to find the exact concentration (the volume must be known) of an unknown base or acid Equivalent point Where exact stoichometry amounts of each reactant are present. (same number of H+ and OH- ) NO EXCESSES End point Where the indicator changes color. Immediately stop adding titrate.

TITRATION QUESTION A 0.951 g sample of a mixture of CaCl2 and NH4NO3 was analyzed by dissolving the sample in water and then completely precipitating the calcium ion as CaC2O4. The CaC2O4 was dissolved in sulfuric acid (MUST KNOW THIS ACID), and the resulting oxalic acid was titrated with a KMnO4 solution. The titration required 38.36 mL of 0.0283 M KMnO4 to reach equivalence point. Manganese (II) is the reduction product of MnO4- and CO2 is the oxidation product of H2C2O4. 1) Redox equation 2) How many moles of H2C2O4 were titrated? 3) How many moles of CaCl2 were in the original sample? 4) What was the percent mass of CaCl2 in the original sample? find mass from #3 and divide by .951 and X 100

THE END

MnO4-  Mn2+ 8H+ + MnO4-  Mn2+ + 4H2O 8H+ + 5e- + MnO4-  Mn2+ + 4H2O H2C2O4  CO2 H2C2O4  2CO2 + 2H+ + 2e- 16H+ + 10e- + 2MnO4-  2Mn2+ + 8H2O 5H2C2O4  10CO2 + 10H+ + 10e- 6H+ + 2MnO4- + 5H2C2O4  2Mn2+ + 8H2O + 10CO2

38.36 mL of 0.0283 M KMnO4 .03836L / 0.0283 moles KMnO4 / L 1.09 X 10-3 mol KMnO4 1.09 X 10-3 mol KMnO4 / 5 moles H2C2O4 / 2 moles KMnO4 2.71 X 10-3 moles H2C2O4 END

look at the flow of ions from one compound to another there is one C2O42- in one H2C2O4 that came from the compound CaC2O4, which is a one to one ratio the Ca came from CaCl2 which only gives you 1 Ca so the number of moles is the same as part 2