Saturday Study Session 1 1st Class Reactions

Things to remember Solids, liquids, and gases can NOT be broken into ions SOLUTION – if it says a solution, then it CAN be broken into ions if it is soluble in water. Only ionic compounds can become separate ions in a solution. The 6 strong acids (HCl, HBr, HI, HNO3, HClO4, H2SO4) and the strong bases (group 1 + OH-and Ba, Sr, Ca + OH-) are always written as separate ions.

Electrolytes

Formula Equation (Molecular Equation) Gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solution. Reactants and products generally shown as compounds. Use solubility rules to determine which compounds are aqueous and which compounds are solids. AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) Copyright © Cengage Learning. All rights reserved

Complete Ionic Equation All substances that are strong electrolytes are represented as ions. Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) AgCl(s) + Na+(aq) + NO3-(aq) Copyright © Cengage Learning. All rights reserved

HCl --- Strong electrolyte 100% dissociation

Net Ionic Equation Includes only those solution components undergoing a change. Show only components that actually react. Ag+(aq) + Cl-(aq) AgCl(s) Spectator ions are not included (ions that do not participate directly in the reaction). Na+ and NO3- are spectator ions. Copyright © Cengage Learning. All rights reserved

Precipitant

White Precipitant AgCl, Mg(OH)2, BaSO4

Opening activity CaO + H2O  Ca(OH)2 molecular equation Solid calcium oxide is added to water   CaO + H2O  Ca(OH)2 molecular equation CaO + H2O  Ca2+ + 2 (OH)- net ionic Potassium chlorate(s) is heated 2KClO3 2KCl + 3O2 molecular and ionic Calcium carbonate(s) is heated CaCO3 CaO + CO2 molecular and ionic Sodium hydroxide(aq) is heated 2NaOH  Na2O + H2O molecular   2Na+ + 2(OH)-  Na2O + H2O ionic

Aluminum metal is added to solution of copper II chloride 2Al + 3CuCl2  2 AlCl3 + 3Cu molecular 2Al + 3Cu2+  2Al3+ + 3 Cu ionic   Fluorine gas is bubbled into solution of sodium bromide   F2 + 2NaBr  2NaF + Br2 molecular F2 + 2Br-  2F- + Br2 Solutions of Hydrofluoric acid is added to ammonium hydroxide HF + NH4OH  NH4F + H2O molecular HF + NH4OH  NH4 + + F- + H2O ionic Butane is burned in Air 2C4H10 + 13O2  8CO2 + 10H2O molecular and ionic

Stoichiometry Exercise 1 A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution?

Stoichiometry Exercise 1 A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? 1.57 M

Problem #1 Step 1 - find molecular mass of K3PO4 212.274 g/mol Step 2 - convert from g of K3PO4 to mol 500.0 g / 212.274 g/mol = 2.355 mol Step 3 - divide mol by liters to get molarity 2.355 mol / 1.50 L = 1.57 M

Exercise 2 What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution?

Exercise 2 What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution? 60.0 mL

M1V1 = M2V2 (2 M)(X) = (0.800M)(150 mL) X = (.800M)(150mL) (2.00M) X = 60.0 mL

Copyright © Cengage Learning. All rights reserved (Part I) Exercise 3 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What precipitate will form? lead(II) phosphate, Pb3(PO4)2 What mass of precipitate will form? 1.1 g Pb3(PO4)2 The balanced molecular equation is: 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s). 0.0030 mol Na3PO4 present to start and 0.0040 mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore 0.0013 mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is 811.54 g/mol, 1.1 g of Pb3(PO4)2 will form. Copyright © Cengage Learning. All rights reserved

All group I metals are soluble All nitrates are soluble 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb2(PO4)2(s) 6Na+ + 2PO4-3 + 3Pb+2 + 6NO3- → 6Na+ + 6NO3- + Pb3(PO4)2 2PO4-3 + 3Pb+2 → Pb3(PO4)2

R 2PO4-3 + 3Pb+2 → 1Pb3(PO4)2 I .0030 .0040 0 C -2x -3x +1x E .000033 excess 0 0.0013 0.0040 -3x = 0 therefore x = 0.001333333 0 + 1x = 0.0013333 mol Pb3(PO4)2 Molecular mass of Pb3(PO4)2 equals 811.55 g/mol 0.0013 mol x 811.55 g/mol = 1.1 g of Pb3(PO4)2

Copyright © Cengage Learning. All rights reserved Exercise 4 (Part II) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of nitrate ions left in solution after the reaction is complete? Nitrate 0.27 M The concentration of nitrate ions left in solution after the reaction is complete is 0.27 M. Nitrate ions are spectator ions and do not participate directly in the chemical reaction. Since there were 0.0040 mol of Pb(NO3)2 present to start, then 0.0080 mol of nitrate ions are present. The total volume in solution is 30.0 mL. Therefore the concentration of nitrate ions = 0.0080 mol / 0.0300 L = 0.27 M. Copyright © Cengage Learning. All rights reserved

Step 1 – find moles of NO3- present 0.200 M Pb(NO3)2 = 2 NO3- x 0.200 M = 0.400 M 0.400 mol/L x .0020 L = .00080 mol of NO3- Step 2 – find new volume after mixing solutions 20.0 ml + 10.0 ml = 30.0 ml or 0.003 L Step 3 - find molarity of NO3- 0.0008 mol / .003 L = 0.27 M

Copyright © Cengage Learning. All rights reserved (Part III) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of phosphate ions left in solution after the reaction is complete? 0.011 M Excess was .000033 mol / .03 L = .011 M The concentration of phosphate ions left in solution after the reaction is complete is 0.011 M. Phosphate ions directly participate in the chemical reaction to make the precipitate. There were 0.0030 mol of Na3PO4 present to start, therefore there was 0.0030 mol of phosphate ions present to start. 0.0027 mol of phosphate ions were used up in the chemical reaction, therefore 0.00033 mol of phosphate ions is leftover (0.0030 – 0.0027 mol). The total volume in solution is 30.0 mL. Therefore the concentration of phosphate ions = 0.00033 mol / 0.0300 L = 0.011 M. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Which of the following ions form compounds with Pb2+ that are generally soluble in water? a) S2– b) Cl– c) NO3– d) SO42– e) Na+ a), b), and d) all form precipitates with Pb2+. A compound cannot form between only Pb2+ and Na+. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Which of the following solutions contains the greatest number of ions? 400.0 mL of 0.10 M NaCl. 300.0 mL of 0.10 M CaCl2. 200.0 mL of 0.10 M FeCl3. 800.0 mL of 0.10 M sucrose. a) contains 0.080 mol of ions (0.400 L × 0.10 M × 2). b) contains 0.090 mol of ions (0.300 L × 0.10 M × 3). c) contains 0.080 mol of ions (0.200 L × 0.10 M × 4). d) does not contain any ions because sucrose does not break up into ions. Therefore, letter b) is correct. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution? Add water to the solution. Pour some of the solution down the sink drain. c) Add more sodium chloride to the solution. d) Let the solution sit out in the open air for a couple of days. e) At least two of the above would decrease the concentration of the salt solution. For letter a), adding water to the solution will increase the total volume of solution and therefore decrease the concentration. For letter b), pouring some of the solution down the drain will not change the concentration of the salt solution remaining. For letter c), adding more sodium chloride to the solution will increase the number of moles of salt ions and therefore increase the concentration. For letter d), water will evaporate from the solution and decrease the total volume of solution and therefore increase the concentration. Therefore, since only letter a) would decrease the concentration, letter e) cannot be correct. Copyright © Cengage Learning. All rights reserved

After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of 38 percent. The correct value for the percentage of water in the hydrate is 51 percent. Which of the following is the most likely explanation for this difference? a. Strong initial heating caused some of the hydrate sample to spatter out of the crucible. b. The dehydrated sample absorbed moisture after heating. c. The amount of the hydrate sample used was too small. d. The crucible was not heated to constant mass before use. e. Excess heating caused the dehydrated sample to decompose.

Copyright © Cengage Learning. All rights reserved Acid–Base Reactions (Brønsted–Lowry) “Bro-Pro” Brønsted deals with Protons Acid—proton donor Base—proton acceptor Conjugate Acid – formed when base gains a H+ Conjugate Base – formed when acid loses a H+ HF + NH3  NH4+ + F- Acid Base Conjugate Conjugate Acid Base Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of 0.500 M sulfuric acid to reach the endpoint? 1.00 mol NaOH 2NaOH + 1H2SO4 → Na2SO4 + 2H2O It takes 2 mol of OH- to react with 1 mol 2H+ (.500M)(1.00L) = .500 mol H+ so .500 mol x 2 = The balanced equation is: H2SO4 + 2NaOH → 2H2O + Na2SO4. 0.500 moles of sulfuric acid is present to start. Due to the 1:2 ratio in the equation, 1.00 mol of NaOH would be required to exactly react with the sulfuric acid. 1.00 mol of sodium hydroxide would be required. Copyright © Cengage Learning. All rights reserved

Titration – Laboratory Procedure To determine the concentration of a solution Redox reaction or acid base Neutralization reaction

Redox Characteristics Transfer of electrons Transfer may occur to form ions Oxidation – increase in oxidation state (loss of electrons); reducing agent Reduction – decrease in oxidation state (gain of electrons); oxidizing agent Copyright © Cengage Learning. All rights reserved

Reaction of Sodium and Chlorine Copyright © Cengage Learning. All rights reserved

Find the oxidation states for each of the elements in each of the following compounds: K2Cr2O7 CO32- MnO2 PCl5 SF4

Copyright © Cengage Learning. All rights reserved Find the oxidation states for each of the elements in each of the following compounds: K2Cr2O7 CO32- MnO2 PCl5 SF4 K = +1; Cr = +6; O = –2 C = +4; O = –2 Mn = +4; O = –2 P = +5; Cl = –1 S = +4; F = –1 K2Cr2O7; K = +1; Cr = +6; O = -2 CO32-; C = +4; O = -2 MnO2; Mn = +4; O = -2 PCl5; P = +5; Cl = -1 SF4; S = +4; F = -1 Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Which of the following are oxidation-reduction reactions? Identify the substance oxidized and reduced. Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l) 2CuCl(aq) CuCl2(aq) + Cu(s) Zn – reducing agent; HCl – oxidizing agent c) CuCl acts as the reducing and oxidizing agent Copyright © Cengage Learning. All rights reserved

Free Response 2003B Answer the following questions that relate to chemical reactions. (a) Iron(III) oxide can be reduced with carbon monoxide according to the following equation. Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) A 16.2 L sample of CO(g) at 1.50 atm and 200.°C is combined with 15.39 g of Fe2O3(s). How many moles of CO(g) are available for the reaction? (ii) What is the limiting reactant for the reaction? Justify your answer with calculations. (iii) How many moles of Fe(s) are formed in the reaction?  

. A 0.345 g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in sufficient Water to produce 100.0 mL of solution. A 20.0 mL portion of this solution was titrated with KMnO4(aq). The balanced equation for the reaction that occurs is:   16H+(aq) + 2 MnO4-(aq) + 5 C2O42-(aq)  2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) The volume of the 0.0150 M KMnO4(aq) required to reach the equivalence point of the titration was 17.80 mL i. Identify the the substance oxidized and the substance reduced in the titration reaction.  ii. For the titration at the equivalence point, calculate the number of moles of each of the following that reacted:   a. MnO4- b. C2O42- iii. Calculate the total number of moles of C2O42- that were present in the 100.0 mL of prepared solution.  iv. Calculate the mass percent of the BeC2O4 in the impure 0.345 g sample