1. CHAPTER 8 REACTIONS IN AQUEOUS SOLUTIONS

2. PREDICTING PRODUCTS WITH AQUEOUS REACTANTS Driving Force for reactants to form certain products Most common:formation of a solid formation of water formation of gas transfer of electrons

3. FORMATION OF A PRECIPITATE Vocabulary: Precipitation – formation of a solid in chemical reaction Precipitate - solid formed from chemical reaction Precipitation reaction – reaction in which a solid is formed and separates from solution Ex: K 2 (CrO 4 )(aq) + Ba(No 3 ) 2 (aq) → K(NO 3 )(aq) + Ba (CrO 4 )(s) Notice that 2 aqueous reactants produced a solid and an aqueous product

4. GENERAL RULES FOR SOLUBILITY Most nitrates are soluble Most salts of Na, K, and NH 4 are soluble Most chlorides are soluble exceptions: AgCl, PbCl 2, and Hg 2 Cl 2 Most sulfates are soluble exceptions: BaSO 4,PbSO 4,CaSO 4 ------------------------------------------------------------------------------------------------------------------------------------------------------------ Most OH compounds are only slightly soluble exceptions: KOH, Ba(OH) 2, Ca(OH) 2 are moderately soluble Most sulfides, carbonates and phosphates are only slightly soluble

5. HOW DO YOU KNOW WHAT PRODUCTS WILL FORM 1)You need to know the nature of the reactants Let’s take Ba(NO 3 ) 2 (aq) -the (aq) means that Ba(No 3 ) 2 has been dissolved in water -the formula indicates Ba +2 and NO 3 -1 ions **(the ions separate (dissociate) when dissolved in water) -the formula also indicates 2 NO 3 ions form with every 1 Ba ion -these ions are good electrolytes-conductors of electricity Let’s take K 2 CrO 4 -the (aq) means that K 2 CrO 4 also dissolves in water -the formula also indicates dissociation - these ions are also good electrolytes What we have now: Ba(NO 3 ) 2 (aq) + K 2 CrO 4 → Products (?)

6. PREDICTING THE PRODUCTS So what are the possibilities? We know that it’s a double displacement reaction We also know that K when placed in water will totally dissolve so a product containing K would not form a solid therefore the products must include the Ba compound forming the precipitate Knowing the solubility of the ions is important in determining which product will form the precipitate -soluble solid = solid readily dissolves in water -insoluble/slightly soluble = only a small amount dissolves in water

7. PRACTICE Problem: When an aqueous solution of silver nitrate is added to an aqueous solution of potassium chloride, a white solid forms. Identify this precipitate an write the balance equation. So what you have is: AgNO 3 (aq) + KCl (aq) → white solid This is a double displacement reaction so we know the products: AgCl and KNO 3 The question now becomes which one of these is the precipitate To figure out look at the solubility rules Rule 1: Most Nitrates are soluble Rule 2: Most salts containing K are soluble Rule 3: Most chlorides are soluble with exceptions==AgCl This means that AgCl MUST be the precipitate AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq)

8. PRACTICE Using the solubility rules predict the products and indicate the states of the products a) KNO 3 (aq) and BaCl 2 (aq) b) Na 2 SO 4 (aq) and Pb(NO 3 ) 2 (aq) c)KOH(aq) and Fe(NO 3 ) 3 (aq) a)According to the solubility rules nitrates and chlorides are soluble in water SO the ions will stay dissolved in water AND no chemical reaction takes place b)According to the solubility rules nitrates are soluble BUT sulfates are only slightly soluble SO Pb(So 4 ) will form the precipitate c)According to the solubility rules nitrates are soluble BUT hydroxides are only slightly soluble SO Fe(OH) will form the precipitate

9. PRACTICE 2 Predict whether a solid will form when the following pairs of solution are mixed. If so, identify the solid and write a balanced equation for the reaction. 1.Ba(NO 3 ) 2 (aq) and NaCl(aq) 2.Na 2 S(aq) and Cu(NO 3 ) 2 (aq) 3.NH 4 Cl(aq) and Pb(NO 3 ) 2 (aq )

10. PRACTICE 2 ANSWERS 1.No solid forms 2.Na 2 S(aq) + Cu((NO 3 ) 2 (aq) → CuS(s) + 2Na(NO 3 )(aq) 3.2NH 4 Cl(aq) + Pb(NO 3 ) 2 (aq) → PbCl 2 (s) + 2NH 4 (NO 3 )(aq)

11. DESCRIBING REACTIONS IN AQUEOUS SOLUTION 3 Types 1)Molecular Equation – shows overall reactions may or may not show actual states of R and P Ex: K 2 CrO 4 (aq) + Ba(NO 3 ) 2 →BaCrO 4 (s) + 2KNO 3 (aq) 2) Complete Ionic Equation - shows all R and P that are strong electrolytes Ex: 2K + (aq) + CrO 4 (aq) - + Ba + (aq) +2NO 3 (aq) - →BaCRO 4 (s) + 2K + (aq) + 2NO 3 - (aq) 3) Net Ionic Equation – shows all components that undergo a change Spectator ions (ions that do not participate directly the reaction) are not included Ex: Ba +2 (aq) + CrO 4 -2 (aq) →BaCrO 4 (s)

12. CLASSWORK/HOMEWORK Page 253 1-6 Page 271-272 1-16

13. Acids:substance that produces H+ ions when dissolved in water (proton donor) taste sour good electrolytes litmus paper turns red phenolphthalein is colorless reacts with metals to form H↑ and metal corrosive Ex: HCl, H 2 SO 4,HNO 3 Bases: substance that produces OH- ions when dissolved in water (proton acceptor) tastes bitter strong bases-good electrolytes weak bases- poor electrolytes litmus paper turns blue phenolphthalein is pink emulsifies fats and oils Ex: NaOH, KOH, Ca(OH) 2 ACIDS AND BASES

14. WHAT OCCURS WHEN ACIDS AND BASES REACT Salts and water is produced Consider a Neutralization Reaction because of the stability of water Individual properties of the acid and bases are lost to the salt Ex: HCl(aq) + Na(OH)(aq) →H 2 O(l) +NaCl(aq) Because HCl, NaOH and NaCl can completely dissolve in water then the Complete ionic equation is: H + (aq) +Cl - (aq) +Na + (aq) + OH - (aq) →H 2 0(l) + Na + (aq) + Cl - (aq) **** Spectator Ions include: Cl - and Na + Net Ionic equation is:H + (aq) + OH - (aq) → H 2 O(l)

15. PRACTICE Nitric Acid id a strong acid. Write the molecular, complete ionic and net ionic equations for the reaction of aqueous acid and aqueous potassium hydroxide. Molecular equation: HNO 3 (aq) + KOH(aq) →H 2 O(l) + KNO 3 (aq) Complete Ionic equation: H + (aq) + NO3 - (aq) + K + (aq) + OH - (aq) →H 2 O(l) + K + (aq) +NO 3 - (aq) Net Ionic Equation: H + (aq) + OH - (aq) → H 2 O(l) NOTE: The products will always include H 2 O and a SALT that may be soluble or insoluble.

16. REACTIONS OF METALS AND NONMETALS Considered Oxidation-Reduction Reactions (Redox)= any chemical change in which the elements undergo a change in oxidation number. This involves the transfer of electrons when forming compounds Ex: Na + Cl →NaCl 1)Na metal has a charge of O; Cl nonmetal has a charge of O 2)However, when they combine they have charges Na +1 and Cl -1 3)Because the Na loses an electron to the chlorine and gets a (+) charge it is said to be “oxidized”. Na 0 →Na + +1e - 4)Because the Cl gains an electron from the Na and gains a (-) charge it is said to be “reduced”. Cl 0 + 1e - →Cl -

17. RULES FOR ASSIGNING OXIDATION NUMBERS 1)All uncombined elements have an oxidation number of zero. 2)A monoatomic ion has an oxidation number equal to its charge. 3)Fluorine has an oxidation number of -1 in all compounds. 4)Oxygen has an oxidation number of -2 in all compounds. Exception: Peroxide H 2 O 2 Oxygen Oxidation number is -1. 5)Hydrogen has an oxidation number of +1. Exception: when in combination with metals it is -1 (NaH) 6)The sum of the oxidation number of all atoms in a neutral compound is equal to zero.

18. PRACTICE For each reaction, show how electrons are gained or lost. Label the reduced and oxidized element. 1)2 Na(s) + Br 2 (l) →2NaBr(s) 2)2Ca(s) + O 2 (g) →2CaO(s) Solution: 1)Oxidized: Na → Na + + 1e - Reduced: Br + 1e - →Br - 2)Oxidized: Ca→Ca +2 + 2e - Reduced: O + 2e - →O -2

19. CLASSWORK/HOMEWORK Cw: Lab: Precipitation Reactions—Complete 1 & 2 CW/HwRead Pages 254-262 and Complete section review : Page 262 1-6