1. Saturday Study Session 1 1st Class Reactions

2. 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.

3. Electrolytes

4. 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)
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5. 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)
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6. HCl --- Strong electrolyte 100% dissociation

7. 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.
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8. Precipitant

10. White Precipitant
AgCl, Mg(OH)2,
BaSO4

11. 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  Ca (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 (OH)-  Na2O + H2O ionic

12. Aluminum metal is added to solution of copper II chloride
2Al + 3CuCl2  2 AlCl3 + 3Cu molecular
2Al + 3Cu2+  2Al Cu ionic
Fluorine gas is bubbled into solution of sodium bromide
  F NaBr  2NaF + Br2 molecular
F Br-  2F- + Br2
Solutions of Hydrofluoric acid is added to ammonium hydroxide
HF + NH4OH  NH4F + H2O molecular
HF + NH4OH  NH F- + H2O ionic
Butane is burned in Air
2C4H O2  8CO H2O molecular and ionic

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

14. Stoichiometry Exercise 1
A 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

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

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

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

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

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(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).
mol Na3PO4 present to start and mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is g/mol, 1.1 g of Pb3(PO4)2 will form.
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20. All group I metals are soluble All nitrates are soluble 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb2(PO4)2(s) 6Na+ + 2PO Pb+2 + 6NO3- → 6Na+ + 6NO3- + Pb3(PO4)2 2PO Pb+2 → Pb3(PO4)2

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

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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 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 mol of Pb(NO3)2 present to start, then mol of nitrate ions are present. The total volume in solution is 30.0 mL. Therefore the concentration of nitrate ions = mol / L = 0.27 M.
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23. Step 1 – find moles of NO3- present
0.200 M Pb(NO3)2 = 2 NO3- x M = M
0.400 mol/L x L = mol of NO3-
Step 2 – find new volume after mixing solutions
20.0 ml ml = 30.0 ml or L
Step 3 - find molarity of NO3-
mol / .003 L = 0.27 M

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(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 mol / .03 L = M
The concentration of phosphate ions left in solution after the reaction is complete is M.
Phosphate ions directly participate in the chemical reaction to make the precipitate. There were mol of Na3PO4 present to start, therefore there was mol of phosphate ions present to start mol of phosphate ions were used up in the chemical reaction, therefore mol of phosphate ions is leftover ( – mol).
The total volume in solution is 30.0 mL. Therefore the concentration of phosphate ions = mol / L = M.
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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+.
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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 mol of ions (0.400 L × 0.10 M × 2). b) contains mol of ions (0.300 L × 0.10 M × 3). c) contains 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.
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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.
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28. 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.

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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
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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 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 + Na2SO 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.
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31. Titration – Laboratory Procedure
To determine the concentration
of a solution
Redox reaction or acid base
Neutralization reaction

32. 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
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33. Reaction of Sodium and Chlorine
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34. Find the oxidation states for each of the elements in each of the following compounds:
K2Cr2O7
CO32-
MnO2
PCl5
SF4

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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
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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
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37. 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 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?

38. . A g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in sufficient Water to produce 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) MnO4-(aq) C2O42-(aq)  2 Mn2+(aq) CO2(g) H2O(l)
The volume of the M KMnO4(aq) required to reach the equivalence point of the titration was 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 mL of prepared solution.
 iv. Calculate the mass percent of the BeC2O4 in the impure g sample