1. Chemical Reactions and Stoichiometry
Writing and balancing Chemical Equation
Classifying Chemical Reactions:
Precipitation Reactions
Acid-base Reactions
Decomposition Reactions (nonredox)
Oxidation-Reduction Reactions
Reaction Stoichiometry
Reaction Yields
Quantitative Chemical Analysis

2. Chemical Equation #1 4Fe(s) + 3 O2(g)  2Fe2O3(s)
Description of reaction:
Iron reacts with oxygen gas and forms solid iron(III) oxide:
Identity: reactants = iron (Fe) and oxygen gas (O2); product = iron(III) oxide
Chemical equation: Fe(s) + O2(g)  Fe2O3(s)
Balanced equation:
4Fe(s) + 3 O2(g)  2Fe2O3(s)

3. Chemical Equation #2 Description of reaction:
Phosphorus reacts with oxygen gas to form solid tetraphosphorus decoxide.
Equation: P(s) + O2(g)  P4O10(s)
Balanced eqn.: 4P(s) + 5 O2(g)  P4O10(s)

4. Chemical Equation #3 Description of reaction:
Propane gas (C3H8) is burned in air (excess of oxygen) to form carbon dioxide gas and water vapor;
Identity: reactants = C3H8(g) and O2(g);
products = CO2(g) and H2O(g);
Equation: C3H8(g) + O2(g)  CO2(g) + H2O(g);
Balanced equation:
C3H8(g) + 5 O2(g)  3CO2(g) + 4H2O(g)

5. Chemical Equation 4 Description of reaction:
Ammonia gas (NH3) reacts with oxygen gas to form nitrogen monoxide gas and water vapor;
Equation: NH3(g) + O2(g)  NO(g) + H2O(g);
Balancing the equation:
2NH3(g) + 5/2 O2(g)  2NO(g) + 3H2O(g);
Multiply throughout by 2 to get rid of the fraction:
4NH3(g) + 5 O2(g)  4NO(g) + 6H2O(g);

6. Balancing Chemical Equations
Rules for balancing equations:
Use smallest integer coefficients in front of each reactants and products as necessary; coefficient “1” need not be indicated;
The formula of the substances in the equation MUST NOT be changed.
Helpful steps in balancing equations:
Begin with the compound that contains the most atoms or types of atoms.
Balance elements that appear only once on each side of the arrow.
Next balance elements that appear more than once on either side.
Balance free elements last.
Finally, check that smallest whole number coefficients are used.

7. Reactions in Aqueous Solution
1. Double-Displacement Reactions
Precipitation reactions
Neutralization (or Acid-Base) reactions
2. Oxidation-Reduction (Redox) Reactions
Combination reactions
Decomposition reactions
Combustion reactions
Single-Replacement reactions
Reactions involving strong oxidizing reagents

8. Precipitation Reactions
Reactions that produce insoluble products (or precipitates) when two aqueous solutions are mixed.
Examples:
(1) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
(2) Pb(NO3)2(aq) + K2CrO4  PbCrO4(s) + 2KNO3(aq)
(3) BaCl2(aq) + H2SO4(aq)  BaSO4(s) + 2HCl(aq)
(4) 3AgNO3 (aq) + Na3PO4 (aq)  Ag3PO4(s) + 3NaNO3(aq)

9. Precipitation Reactions

10. Precipitates may have other colors

11. Predicting Formation of Precipitation
Can we predict if a precipitate will form when two solutions are mixed (reacted)?
How do we know if one or more of the products of a reaction will form solid (not water soluble)?
We need to know the water solubility properties of ionic compounds
Use “Solubility Rules”

12. Solubility Rules for Predicting Solid Products
Soluble salts:
All compounds of alkali metals and NH4+
All compounds containing nitrate, NO3-, and acetate. C2H3O2-, except silver acetate, which is sparingly soluble;
Most chlorides, bromides, and iodides, except AgX, Hg2X2, PbX2, and HgI2; where X = Cl-, Br-, or I-.
Most sulfates, except CaSO4, SrSO4, BaSO4, PbSO4 and Hg2SO4.
Insoluble or slightly soluble salts:
Most hydroxides (OH-), sulfides (S2-), carbonates (CO32-), chromates (CrO42-), and phosphate (PO43-), except those associated with the Group 1A metals or NH4+.

13. Predicting Precipitation Reactions
Learn to predict possible products of “ion-exchange reactions”.
Complete and balance the following equations and predict which products are not water soluble.
a) CaCl2(aq) + Na2CO3(aq)  ?
b) NH4NO3(aq) + MgSO4(aq)  ?
c) Pb(NO3)2(aq) + NaI(aq)  ?
d) AgNO3(aq) + K2CrO4(aq)  ?

14. Equations for Precipitation Reactions
Molecular equation:
Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)
Total Ionic equation:
Pb NO K+ + CrO42-  PbCrO4(s) + 2K+ + 2NO3-
(K+ and NO3- are spectator ions)
Net ionic equation:
Pb2+(aq) + CrO42-(aq)  PbCrO4(s)

15. Acid-Base (Neutralization) Reactions
Acid – a compound that produces hydrogen ions (H+) when dissolved in aqueous solution;
Base – a compound that produces hydroxide ions (OH-) in aqueous solutions.
Some examples of acids and strong bases:
Acids: HCl, HClO4, HNO3, H2SO4, H3PO4, and HC2H3O2;
Bases: NaOH, KOH, Ba(OH)2, and NH3.

16. Acid-Base Reactions Some example of acid-base reactions:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
H2SO4(aq) + KOH(aq)  H2O(l) + K2SO4(aq)
HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq)
2HClO4(aq) + Ba(OH)2(aq)  2 H2O(l) + Ba(ClO4)2(aq)

17. Equations for Acid-Base Reactions
An example of strong acid and strong base reaction:
Molecular equation:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Total ionic equation:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + H2O(l)
Net ionic equation:
H+(aq) + OH-(aq)  H2O(l)

18. Equations for Acid-Base Reactions
An example of week acid and strong base reaction:
Molecular equation:
HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l)
Total ionic equation:
HC2H3O2(aq) + Na+(aq) + OH-(aq)  Na+(aq) + C2H3O2-(aq) + H2O(l)
Net ionic equation:
HC2H3O2(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)

19. Equations for Acid-Base Reactions
An example of strong acid and weak base reaction:
Molecular equation:
HCl(aq) + NH3(aq)  NH4Cl(aq)
Total ionic equation:
H+(aq) + Cl-(aq) + NH3(aq)  NH4+(aq) + Cl-(aq)
Net ionic equation:
H+(aq) + NH3(aq)  NH4+(aq)

20. Reactions That Produce Gas
Reactions producing CO2 gas:
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g);
NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g);
Reaction that produces SO2 gas:
Na2SO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + SO2(g);
Reaction that produces H2S gas:
Na2S(aq) + 2HCl(aq)  2NaCl(aq) + H2S(g);

21. Oxidation-Reduction Reactions
Oxidation  loss of electrons and increase in oxidation number;
Reduction  gain of electrons and decrease in oxidation number;
Oxidation-reduction (or Redox) reaction  one that involves transfer of electrons from one reactant to the other;
Oxidizing agent  the reactant that gains electrons and got reduced;
Reducing agent  the reactant that loses electrons and got oxidized.

22. Types of Redox Reactions
Reactions between metals and nonmetals;
Combustion reactions (reactions with molecular oxygen);
Single replacement reactions;
Decomposition reactions that form free elements;
Reactions in aqueous solution involving oxidizing and reducing agents;
Disproportionation reactions.

23. Types of Redox Reactions
Reactions between metals and nonmetals:
4Al(s) O2(g)  2Al2O3(s);
3Mg(s) + N2(g)  Mg3N2(s);
Combustion reactions:
CH4(g) O2(g)  CO2(g) + 2H2O(g);
2C8H18(l) O2(g)  16CO2(g) H2O(g);
C2H5OH(l) O2(g)  2CO2(g) + 3H2O(g);

24. Types of Redox Reactions
Single-Replacement Reactions:
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g);
Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s);
Cl2(aq) + 2KBr(aq)  2KCl(aq) + Br2(aq);
Decomposition Reactions:
2HgO(s)  2Hg(l) + O2(g);
(NH4)2Cr2O7(s)  Cr2O3(s) + N2(g) + 4H2O(g);

25. Types of Redox Reactions
Reactions in aqueous solutions that involve strong oxidizing reagents:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + H2O(l);
Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq)  2Cr3+(aq) + 7H2O(l) + 3 O2(g);
2Cr(OH)4-(aq) + 3H2O2(aq) + 2 OH-(aq)  2CrO42-(aq) + 8H2O(l);
Disproportionation reaction:
Cl2(g) + 2NaOH(aq)  NaOCl(aq) + NaCl(aq) + H2O(l);
3Br2(aq) + 6NaOH(aq)  NaBrO3(aq) + NaBr(aq) + H2O(l);

26. Guidelines for Determining Oxidation Numbers of Elements
1. Atoms in the free elemental form are assigned oxidation number zero (0);
2. The sum of oxidation number in neutral molecules or formula units is 0;
the sum of oxidation number (o.n.) of atoms in a polyatomic ion is equal to the net charge of the ion (in magnitude and sign);
3. In their compounds, each Group IA metal is assigned an o.n. of +1; each Group IIA metal an o.n. of +2; boron and aluminum each an o.n. of +3, and fluorine an o.n. of –1;
4. Hydrogen is assigned an o.n. of +1 in compounds or polyatomic ions with nonmetals, and an o.n. of –1 in metal hydrides;
5. In compounds and polyatomic ions, oxygen is assigned an o.n. of -2, except in peroxides, in which its o.n. is –1;
6. In binary compounds with metals, chlorine, bromine, and iodine each has an o.n. of -1; sulfur, selenium, and tellurium each has an o.n. of -2.

27. Oxidation-Reduction Reactions
In the following equations, identify all reactions that are redox reactions:
1. 2KMnO4(aq) + 16HCl(aq)  2MnCl2(aq) + 2KCl(aq) + 5Cl2(aq) + 8H2O(l);
2. 2KClO3(s)  2KCl(s) O2(g);
3. CaCO3(s)  CaO(s) + CO2(g);
4. Mg(OH)2(s)  MgO(s) + H2O(g);
5. Mg(s) + ZnSO4(aq)  MgSO4(aq) + Zn(s);
6. Cr2O72-(aq) + 3C2H5OH(l) + 2H+(aq)  2Cr3+(aq) + 3CH3COOH(aq) + 4H2O(l)

28. Balancing Redox Reactions by Half-Equation Method
Example-1: Balance the following oxidation-reduction reaction in acidic solution:
MnO4-(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq)
Solution-1:
Note: the above equation is both not balanced and not complete. It only shows the components (reactants) that undergoes changes is oxidation numbers;
Redox reactions in acidic solution means that you need to add H+ ion in the equation, which produces water as one of the products.

29. Balancing Redox Reactions by Half-Equation Method
Solution-1 (continued)
Balancing the equation – first step, break up the equation into two half equations: oxidation and reduction half-equations:
MnO4-(aq)  Mn2+(aq)
In this case, all the four oxygen in MnO4- will become water in acidic solution. So, add enough H+ ion to form H2O with the four oxide ions in in MnO4-. The half-equation with all atoms balanced will look like this:
MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l)

30. Balancing Redox Reactions by Half-Equation Method
Solution-1 (continued)
Next step, check the total charges on both side of the half-equation; note that they are not equal. Add enough electrons on the side that has more positive charges (or less negative charges), so that the total charges on both sides become equal (in magnitude and sign).
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l);
Now we have a balanced reduction half-equation.
(How do you know it is “reduction” half-equation?)

31. Balancing Redox Reactions by Half-Equation Method
Solution-1 (continued)
Next, write the other half-equation and balance it:
Fe2+(aq)  Fe3+(aq) + e-; (this is oxidation half-equation)
To obtain the overall equation, we add the two balanced half-equations, but make sure the number of electrons on both half-equations are equal, so that they cancel out. The overall equation should not contain any electrons. In this case, we multiply the above oxidation half-equation by 5 and obtain:
5Fe2+(aq)  5Fe3+(aq) + 5e-;
Now, adding the two half-equations yields the following balanced net ionic equation:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l);

32. Balancing Redox Reactions by Half-Equation Method
Example-2: Balance the following oxidation-reduction reaction in acidic solution:
Cr2O72-(aq) + H2O2(aq)  Cr3+(aq) + O2(g) + H2O(l);
Solution-2: Write the two half-equations and balance them.
Reduction half-equation:
Cr2O72-(aq) H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l);
Oxidation half-equation:
H2O2(aq)  O2(g) + 2H+(aq) + 2e-;

33. Balancing Redox Reactions by Half-Equation Method
Solution-2 (continued)
Multiply the oxidation half-equation by 3 to make the number of electrons equal with that of the reduction-half equation.
3H2O2(aq)  3 O2(g) + 6H+(aq) + 6e-;
Then add the two half-equations, canceling all the electrons on both side of the equation, all the H+ ions on the right-hand side, and the same number of H+ ions on the left-hand side. The final (balanced) net ionic equation will be as follows:
Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq)  2Cr3+(aq) + 3O2(g) + 7H2O(l);

34. Balancing Redox Reactions by Half-Equation Method
Example-3: Balance the following oxidation-reduction reaction in basic solution:
Cr(OH)4-(aq) + H2O2(aq)  CrO42-(aq) + H2O(l);
Solution-3:
Write the two half-equations and balance them, adding electrons on either side as needed to balance the charges.
Note that this reaction is in basic solution, which means the overall equation should contains OH- ion instead of H+ ions.

35. Balancing Redox Reactions by Half-Equation Method
Solution-3 (continued):
Oxidation half-equation:
Cr(OH)4-(aq) + 4OH-(aq)  CrO42-(aq) + 4H2O(l) + 3e-;
Reduction half-equation:
H2O2(aq) + 2e-  2OH-;
Multiply the oxidation half-equation by 2 and the reduction half-equation by 3 to make the number of electrons in both half-equations equal.
2Cr(OH)4-(aq) + 8OH-(aq)  2CrO42-(aq) + 8H2O(l) + 6e-;
3H2O2(aq) + 6e-  6OH-;

36. Balancing Redox Reactions by Half-Equation Method
Solution-3 (continued):
Now add the two half-equations, canceling all the electrons on both sides of the equation, all the 6OH- ions on the right-hand side, and the same number of OH- ions on the left-hand side.
The overall balanced net ionic equation will appear as follows:
2Cr(OH)4-(aq) + 3H2O2(aq) + 2OH-(aq)  2CrO42-(aq) + 8H2O(l);

37. Stoichiometry
Stoichiometry = the quantitative relationships between one reactant to another, or between a reactant and products in a chemical reaction.
Interpreting balanced equations:
Example: C3H8(g) + 5 O2(g)  3CO2(g) + 4H2O(g);
The equation implies that:
1 C3H8 molecule reacts with 5 O2 molecules to produce 3 CO2 molecules and 4 H2O molecules; OR
1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O.

38. Stoichiometric Calculations
Mole-to-mole relationship:
Example: In the following reaction, if 6.0 moles of octane, C8H18, is completely combusted in excess of oxygen gas, how many moles of CO2 and H2O, respectively, will be formed? How many moles of O2 does it consumed?
Reaction: 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)
Calculations:
Mole CO2 formed = 6.0 mol C8H18 x (16 mol CO2/2mol C8H18) = 48 moles
Mole H2O formed = 6.0 mol C8H18 x (18 mol H2O/2mol C8H18) = 54 moles
Mole O2 consumed = 6.0 mol C8H18 x (25 mol O2/2mol C8H18) = 75 moles

39. Stoichiometric Calculations
Mass-to-mole-to-mole-to-mass relationship:
Example-1: In the following reaction, if 690 g of octane, C8H18, is completely combusted in excess of oxygen gas, how many grams of CO2 are formed?
Reaction: 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)
Calculation-1:
Moles C8H18 reacted = 690 g C8H18 x (1 mol/114.2 g) = 6.0 moles
Moles CO2 formed = 6.0 mol C8H18 x (16 mol CO2/2 mol C8H18)
= 48 moles CO2
Mass of CO2 formed = 48 mol CO2 x (44.01 g/mol) = 2.1 x 103 g

40. Stoichiometric Calculations
Mass-to-mole-to-mole-to-mass relationship:
Example-2: In the following reaction, if 690 g of octane, C8H18, is completely combusted in excess of oxygen gas, how many grams of H2O are formed?
Reaction: 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)
Calculation-2:
Moles C8H18 reacted = 690 g C8H18 x (1 mol/114.2 g) = 6.0 moles
Moles H2O formed = 6.0 mol C8H18 x (18 mol H2O/2 mol C8H18)
= 54 moles CO2
Mass of H2O formed = 54 mol H2O x (18.02 g/mol) = 970 g

41. Stoichiometric Calculations
Mass-to-mole-to-mole-to-mass relationship:
Example-3: In the following reaction, if 690 g of octane, C8H18, is completely combusted in excess of oxygen gas, how many grams of oxygen gas are consumed?
Reaction: 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g)
Calculation-3:
Moles C8H18 reacted = 690 g C8H18 x (1 mol/114.2 g) = 6.0 moles
Moles O2 consumed = 6.0 mol C8H18 x (25 mol O2/2 mol C8H18)
= 75 moles O2
Mass of H2O formed = 75 mol O2 x (32.00 g/mol) = 2.4 x 103 g

42. Stoichiometry Involving Limiting Reactant
one that got completely consumed in a chemical reaction before the other reactants.
Product yields depend on the amount of limiting reactant

43. Limiting Reactant

44. Limiting Reactants
Synthesis of H2O

45. A Reaction Stoichiometry
For example, in the synthesis reaction to produce water according to the following equation,
2H2(g) + O2(g)  2H2O(l),
2 moles of H2 are required to react completely with 1 mole of O2, in which 2 mole of H2O are formed.
If a reaction is carried out with 1 mole of H2 and 1 mole of O2, H2 will be the limiting reactant and O2 will be present in excess.
Only 1 mole of H2O will be produced.

46. Synthesis of NH3

47. Stoichiometry Calculations
Ammonia is produced by the reaction of N2 with H2 according to the following equation:
N2(g) + 3H2(g)  2NH3(g)
(a) If 4.0 moles of N2 and 9.0 moles of H2 are reacted, which reactant will be completely consumed?
(b) How many moles of NH3 are formed?
(c) How many moles of the excess reactant remains after the reaction?

48. Limiting Reactants and Reaction Yields
Ammonia is produced in the following reaction:
N2(g) + 3H2(g)  2NH3(g)
(a) If 118 g of nitrogen gas is reacted with 31.5 g of hydrogen gas, which reactant will be completely consumed at the end of the reaction? (b) How many grams ammonia will be produced when the limiting reactant is completely reacted and the yield is 100%? (c) How many grams of the excess reactant will remain (unreacted)?
(Answer: (a) N2; (b) 6.0 g; (c) g of NH3)

49. Theoretical, Actual and Percent Yields
Chem 1A Chapter 3 Lecture Outlines
Theoretical, Actual and Percent Yields
Theoretical yield:
yield of product calculated based on the stoichiometry of balanced equation and amount of limiting reactant (assuming the reaction goes to completion and the limiting reactant is completely consumed).
Actual Yield:
Yield of product actually obtained from experiment
Percent Yield = (Actual yield/Theoretical yield) x 100%

50. Limiting Reactant & Yields
In an ammonia production, the reactor is charged with N2 and H2 gases at flow rates of 805 g and 195 g per minute, respectively, at 227oC, and the reaction is as follows:
N2(g) + 3H2(g)  2 NH3(g)
(a) What is the rate (in g/min) that ammonia is produced if the yield is 100%? (b) If the reaction produces 915 g of NH3 per minute, calculate the percentage yield of the reaction.
(Answer: (a) g/min; (b) Yield = 93.5%)

51. Stoichiometry in Aqueous Solution
Example-1: Precipitation reaction
How many grams of BaSO4 will be formed when mL of M BaCl2 is reacted with mL of M Na2SO4?
Reaction: BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
Solution-1:
Find the limiting reactant by calculating the mole of each reactant, where the moles can be calculated by multiplying the volume (in liters) of the solution with the molarity.
Mole of BaCl2 = L x mol/L = mole;
Mole of Na2SO4 = L x mol/L = mole;

52. Stoichiometry in Aqueous Solution
Solution-1 (continued)
BaCl2 is the limiting reactant;
Therefore, mole of BaSO4 expected = mole;
Mass of BaSO4 formed = mol x g/mol
= g
If 2.05 g of barium sulfate was actually obtained, what is the percent yield?
Percent yield = (2.05 g/2.33 g) x 100% = 88.0%

53. Acid-Base Titration

54. Acid-Base Titration Example-1: Solution-1:
In a titration experiment, mL of M NaOH solution was required to neutralize acetic acid in a mL sample of vinegar. Calculate the molarity of acetic acid in the vinegar. If the density of vinegar is 1.0 g/mL, calculate the mass percent of acetic acid in the vinegar.
Reaction: HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq)
Solution-1:
Calculate mole of NaOH using its volume and molarity;
According to the equation, mole of acetic acid = mole of NaOH

55. Acid-Base Titration Solution-1 (continued)
Mole of NaOH used = L x mol/L
= mol;
Mole of HC2H3O2 reacted = mole;
Molarity of HC2H3O2 in vinegar = mol/ L
= M
Mole of HC2H3O2 in mL = L x mol/L
= mol
Mass of HC2H3O2 in mL = mol x g/mol
= g
Mass of mL vinegar = mL x 1.0 g/mL
= 1.0 x 102 g
Percent of acetic acid in vinegar = (4.981 g/ 1.0 x 102 g) x 100%
= 5.0% (by mass)

56. Acid-Base Reactions Example-2: Solution-2:
How many milliliters of M NaOH solution will be required to neutralize mL of M H2SO4?
Reaction: H2SO4(aq) + 2NaOH(aq)  2H2O(l) + Na2SO4(aq)
Solution-2:
Mole of H2SO4 present = L x mol/L = mol;
Mole of NaOH needed = mol x 2 = mol;
Volume of M NaOH needed to neutralize the acid
= ( mol)/( mol/L) = L
= mL

57. Acid-Base Reactions Example-3: Solution-3:
A 4.00-mL sample of sulfuric acid is diluted to mL mL of the dilute acid is then titrated with M NaOH solution. If mL of the base is required to neutralize the acid, calculate the molarity of the original sulfuric acid solution.
Reaction: H2SO4(aq) + 2NaOH(aq)  2H2O(l) + Na2SO4(aq)
Solution-3:
Mole of NaOH used = L x ( mol/L) = mole;
Mole of H2SO4 titrated = ½ ( mol) = mole
Molarity of dilute acid = ( mol)/( L) = M
Molarity of undiluted acid = M x mL/4.00 mL
= M

58. Uses of Reactions in Aqueous Solutions
1. Dissolving Insoluble Compounds
Fe2O3(s) + 6HNO3(aq)  2Fe(NO3)3(aq) + 3H2O(l);
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
2. Syntheses of Inorganic Compounds
AgNO3(aq) + NaBr(aq)  AgBr(s) + NaNO3(aq);
Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)
3. Extraction of Metals form Solution
Mg2+(aq) + Ca(OH)2(aq)  Mg(OH)2(s) + Ca2+(aq);

59. Blank Page

60. Water as a Universal Solvent
Water molecule is very polar;
Water interacts strongly with ionic and polar molecules;
Strong interactions enable water to dissolve many solutes – ionic and nonionic;
Solubility of ionic compounds depends on the relative strength of ion-dipole interactions between ions and water molecules and ionic bonds within the compounds;
Many ionic compounds dissolve in water because of strong ion-dipole interactions;
Polar nonionic compounds dissolve in water due to strong dipole-dipole interactions or hydrogen bonding;

61. Electrolytes and Nonelectrolytes
Electrolytes – solutions capable of conducting electric current - contain ions that move freely;
Nonelectrolytes – solutions not capable of conducting electric current - contains neutral molecules only;
Strong electrolytes – ionic compounds, strong acids and strong bases; they dissociate completely when dissolved in water, producing a lot of free ions;
Weak electrolytes – weak acids or weak bases; they only dissociate (ionize) partially when dissolved in water – solutions contain mostly neutral molecules and very little free ions.

62. Dissolution of an Ionic Compound

64. Strong and Weak Electrolytes
Examples of strong electrolytes – they ionize completely:
NaCl(aq)  Na+(aq) + Cl-(aq)
H2SO4(aq)  H+(aq) + HSO4-(aq)
Ca(NO3)2(aq)  Ca2+(aq) + 2NO3-(aq);
Examples of weak electrolytes – they do not ionize completely:
HC2H3O2(aq) ⇄ H+(aq) + C2H3O2-(aq)
NH4OH(aq) ⇄ NH4+(aq) + OH-(aq);
Mg(OH)2(s) ⇄ Mg2+(aq) OH-(aq)

65. Nonelectrolytes
Substances that do not ionize in aqueous solution are nonelectrolytes;
Most organic compounds are nonelectrolytes;
Solutions containing such substances cannot conduct electricity, because they do not have freely moving ions.
Examples of nonelectrolytes:
C6H12O6, C12H22O11, CH3OH, C2H5OH, C3H7OH, HOC2H4OH, etc.

66. Solution Concentrations
The concentration of a solution may be expressed in:
Percent by mass, percent by volume, or molarity;
Percent (by mass) = (Mass of solute/Mass of solution) x 100%
Percent (by volume) = (Vol. of solute/Vol. of solution) x 100%
Molarity = (Mol of solute/Liter of solution)
Mol of solute = Liters of solution x Molarity

67. Percent by Mass Example:
A sugar solution contains 25.0 g of sugar dissolved in g of water. What is the mass percent of sugar in solution?
Percent sugar = {25.0 g/(25.0 g g)} x 100%
= 20.0% (by mass)

68. Calculation of Mass from Percent
Example:
Seawater contains 3.5% (by mass) of NaCl. How many grams of sodium chloride can be obtained from 5.00 gallons of seawater? (1 gall. = L; assume density of seawater = 1.00 g/mL)
Mass of seawater =
5.00 gall x (3785 mL/gall.) x (1.00 g/mL) = g;
Mass of NaCl = g sw x (3.5/100) = 662 g

69. Percent by Volume Example:
A solution is prepared by mixing 150. mL of methanol, 100. mL of acetone, and 250. mL of water. What is the volume percent of methanol and acetone in solution?
Percent methanol = (150. mL/500. mL) x 100%
= 30.0% (by volume)
Percent acetone = (100. mL/500. mL) x 100%
= 20.0% (by volume)

70. Molar Concentration Example:
4.0 g of sodium hydroxide, NaOH, is dissolved in enough water to make a 100.-mL of solution. Calculate the molarity of NaOH.
Mole of NaOH = 4.0 g NaOH x (1 mole/40.0 g)
= 0.10 mole
Molarity of NaOH = 0.10 mol/0.100 L = 1.0 M

71. Calculation of Solute Mass in Solution
Example:
How many grams of NaOH are present in 35.0 mL of 6.0 M NaOH solution?
Mole of NaOH = (6.0 mol/L) x 35.0 mL x (1 L/1000 mL)
= 0.21 mol
Mass of NaOH = 0.21 mol x (40.0 g/mol)
= 8.4 g NaOH

72. Preparing Solutions from Pure Solids
From the volume (in liters) and molarity of solution, calculate the mole and mass of solute needed;
Weigh the mass of pure solute accurately;
Transfer solute into a volumetric flask of appropriate size;
Add deionized water to the volumetric flask, well below the narrow neck, and shake well to dissolve the solute.
When completely dissolved, add more distilled water to fill the flask to the mark and mix the solution well.

73. Preparing Solution from Solid
Example:
Explain how you would prepare 1.00 L of M NaCl solution.
Calculate mass of NaCl needed:
Mole of NaCl = 1.00 L x (0.500 mol/L) = mol
Mass of NaCl = mol x (58.44 g/mol) = 29.2 g
Preparing the solution:
Weigh 29.2 g of NaCl accurately and transfer into 1-liter volumetric flask. Fill the flask half way with distilled water, shake well until all solid has dissolved. Fill the flask to the 1-liter mark with more distilled water and mix the solution well by inverting the flask back and forth several times.

74. Preparing Solution from Stock
Calculate volume of stock solution needed using the formula: MiVi = MfVf (i = initial; f = final)
Measure accurately the volume of stock solution and carefully transfer to a volumetric flask of appropriate size;
Dilute stock solution with distilled water to the required volume (or to the “mark” on volumetric flask)
Mix solution well.
(Note: if diluting concentrated acid, place some distilled water in the flask, add the concentrated acid, and then add more distilled water to the required volume.)

75. Preparing Solution from Stock
Example:
Explain how you would prepare 1.0 L of 3.0 M H2SO4 solution from concentrated H2SO4, which is 18 M.
Calculate volume of concentrated H2SO4 needed:
Vol. of conc. H2SO4 = (1.0 L x 3.0 M/18 M) = 0.17 L = 170 mL
Preparing the solution:
Place some distilled water in the 1-liter volumetric flask (that would fill the flask to about a quarter full). Measure accurately 170 mL of conc. H2SO4 and transfer carefully to the volumetric flask that already contains some distilled water. Then fill the flask to the 1-liter mark with more distilled water and mix the solution well by inverting the flask back and forth several times.