1. Determining Limiting Reagents Guided Practice Problem
Part of the SO2 that is introduced into the atmosphere ends up being converted to sulfuric acid, H2SO4. The net reaction is:
2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq)
How much H2SO4 can be formed from 5.0 mol of SO2, 1.0 mol O2, and an unlimited quantity of H2O?

2. Determining Limiting Reagents Guided Practice Problem
Consider the following reaction:
2Na3PO4(aq) + 3Ba(NO3)2(aq)  Ba3(PO4)2(s) + 6NaNO3(aq)
Suppose that a solution containing 3.50 g of Na3PO4 is mixed with a solution containing 6.40 g of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed? What is the % yield, if experimentally, only 4.70 g were obtained from the reaction?

3. Chapter 4 Aqueous Reactions and Solution Stoichiometry
CHEMISTRY The Central Science 9th Edition
Chapter 4 Aqueous Reactions and Solution Stoichiometry

4. Molarity: Moles of solute per liter of solution.
Solution Composition
Solutions are homogenous mixtures of two or more substances:
Solute: present in smallest amount and is the substance dissolved in the solvent.
Solvent: present in the greater quantities and is used to dissolve the solute.
Example: NaCl dissolved in Water (water = Solvent and NaCl = Solute)
Change concentration by using different amounts of solute and solvent.
Molarity: Moles of solute per liter of solution.

5. Concentrations of Solutions
Formula for Molarity
The most widely used way of quantifying concentration of solutions in chemistry. Molarity is generally represented by the symbol M and defined as the number of moles of solute dissolved in a liter of solution.

6. Class Guided Practice Problem
Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate, Na2SO4, in enough water to form 125 mL of solution.
Class Practice Problem
Calculate the molarity of a solution made by dissolving 5.00 g of NaCl in sufficient water to form L of solution.

7. Class Guided Practice Problem
Determining Mass using Molarity
Key concept: If we know: molarity and liters of solution, we can calculate moles (and mass) of solute.
Class Guided Practice Problem
How many grams of C6H12O6 are required to make 100 mL of M C6H12O6?
Class Practice Problem
How many grams of NaCl are required to make a 1 L of M NaCl?

8. Concentration of Diluted Solutions
Solutions are routinely prepared in stock solutions form.
Example: 12 M HCl
Solution of lower concentrations are prepared by adding more solvent (e.g., water), a process called dilution.
We recognize that the number of moles are the same in dilute and concentrated solutions.
Hence, moles solute before dilution = moles solute after dilution
So:
MdiluteVdilute = MconcentratedVconcentrated or
Mfinal Vfinal = MinitialVinitial

9. Class Guided Practice Problem Class Practice Problem
How much 3.0 M H2SO4 would be required to make 500 mL of 0.10 M H2SO4?
How many milliliters of 5.0 M K2Cr2O7 solution must be diluted in order to prepare 250 mL of 0.10 M solution?
Class Practice Problem

10. General Properties of Aqueous Solutions Electrolytic Properties
Aqueous solutions, solutions in water, have the potential to conduct electricity.
The ability of the solution to conduct depends on the number of ions in solution.
There are three types of solution:
Strong electrolytes,
Weak electrolytes, and
Nonelectrolytes.

11. General Properties of Aqueous Solutions
Electrolytic Properties

12. General Properties of Aqueous Solutions
Molecular Compounds in Water
Molecular compounds in water (e.g., CH3OH): no ions are formed.
If there are no ions in solution, there is nothing to transport electric charge.

13. General Properties of Aqueous Solutions
Ionic Compounds in Water
Ions dissociate in water (NaCl).
In solution, each ion is surrounded by water molecules.
Transport of ions through solution causes flow of current.
Other substances that are not ionic compound dissociate in water to form ions.
For example: (HCl)

14. General Properties of Aqueous Solutions
Strong and Weak Electrolytes
Strong electrolytes: completely dissociate in solution.
For example:
Weak electrolytes: produce a small concentration of ions when they dissolve.
These ions exist in equilibrium with the unionized substance.

15. General Properties of Aqueous Solutions
Some General Terms
Acids - substances that able to ionize in solution to form hydrogen ion (H+) and increase the concentration of H+ in the solution.
For example, HCl dissociate in water to form H+ and Cl- ions.
Bases - are substances that can react with or accept H+ ions.
For example, OH- will accept H+ from HCl forming H2O.
Salts - are ionic compounds that can be formed by replacing one or more of the hydrogen ions of an acid by a different positive ion.
For example, NaCl instead of HCl.

16. General Properties of Aqueous Solutions
Identifying Strong and Weak Electrolytes
Most salts are strong electrolytes (NaCl, CaCO3.
Most acids are weak electrolytes. However, HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4 are strong acids.
The common strong bases are the hydroxides, Ca(OH)2, of the alkali metals and the heavy alkaline earth metals.
Most other substances are nonelectrolytes.

17. Precipitation Reactions Exchange (Metathesis) Reactions
When two solutions are mixed and a solid is formed, the solid is called a precipitate.
Metathesis reactions involve swapping ions in solution:
AX + BY  AY + BX.
HCl + NaOH  NaCl + H2O
Metathesis reactions will lead to a change in solution if one of three things occurs:
an insoluble solid is formed (precipitate),
formation of either a soluble weak or nonelectrolytes,
an insoluble gas is formed.

18. Class Practice Problem
Write a balanced equation for the reaction between phosphoric acid, H3PO4, and potassium hydroxide, KOH.
H3PO4 + 3KOH  3H2O + K3PO4
AX BY  AY + BX

19. Precipitation Reactions
Writing Reaction Equations
Ionic equation: used to highlight reaction between ions.
Molecular equation: all species listed as molecules:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Complete ionic equation: lists all ions:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)
Net ionic equation: lists only unique ions:
H+(aq) + OH-(aq)  H2O(l)

20. Class Guided Practice Problem
Write the net ionic equation for the reactions that occur when solutions of KOH and Co(NO3)2 are mixed.
2OH- + Co2+  Co(OH)2

21. Acid-Base Reactions
Acids with one acidic proton are called monoprotic (e.g., HCl).
Acids with two acidic protons are called diprotic (e.g., H2SO4).
Acids with many acidic protons are called polyprotic.

22. Identifying Strong and Weak Electrolytes
Acid-Base Reactions
Identifying Strong and Weak Electrolytes
Water soluble and ionic = strong electrolyte (probably).
Water soluble and not ionic, but is a strong acid (or base) = strong electrolyte.
Water soluble and not ionic, and is a weak acid or base = weak electrolyte.
Otherwise, the compound is probably a nonelectrolyte.

23. Strong and Weak Electrolyte Summary
Acid-Base Reactions
Strong and Weak Electrolyte Summary

24. Neutralization Reactions and Salts
Acid-Base Reactions
Neutralization Reactions and Salts
Neutralization occurs when a solution of an acid and a base are mixed:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Notice we form a salt (NaCl) and water.
Salt = ionic compound whose cation comes from a base and anion from an acid.
Neutralization between acid and metal hydroxide produces water and a salt.

25. Acid-Base Reactions with Gas Formation
Sulfide and carbonate ions can react with H+ in a similar way to OH-.
2HCl(aq) + Na2S(aq)  H2S(g) + 2NaCl(aq)
2H+(aq) + S2-(aq)  H2S(g)
HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2O(l) + CO2(g)

26. Oxidation-Reduction Reactions
Oxidation and Reduction
When a metal undergoes corrosion it loses electrons to form cations:
Ca(s) +2H+(aq)  Ca2+(aq) + H2(g)
Oxidized: atom, molecule, or ion becomes more positively charged.
Oxidation is the loss of electrons.
Reduced: atom, molecule, or ion becomes less positively charged.
Reduction is the gain of electrons.

27. Oxidation-Reduction Reactions
Oxidation and Reduction

28. Oxidation-Reduction Reactions
Oxidation Numbers
Oxidation number for an ion: the charge on the ion.
Oxidation number for an atom: the hypothetical charge that atom would have if it was an ion.
Oxidation numbers are assigned by a series of rules:
If the atom is in its elemental form, the oxidation number is zero. E.g., Cl2, H2, P4.
For a monoatomic ion, the charge on the ion is the oxidation state.

29. Oxidation-Reduction Reactions
Oxidation Numbers
Nonmetal usually have negative oxidation numbers:
Oxidation number of O is usually –2. The peroxide ion, O22-, has oxygen with an oxidation number of –1.
Oxidation number of H is +1 when bonded to nonmetals and –1 when bonded to metals.
The oxidation number of F is –1.
The sum of the oxidation numbers for the atom is the charge on the molecule (zero for a neutral molecule).

30. Oxidation-Reduction Reactions
Oxidation of Metals by Acids and Salts
Metals are oxidized by acids to form salts:
Mg(s) +2HCl(aq)  MgCl2(aq) + H2(g)
During the reaction, 2H+(aq) is reduced to H2(g).
Metals can also be oxidized by other salts:
Fe(s) +Ni2+(aq)  Fe2+(aq) + Ni(s)
Notice that the Fe is oxidized to Fe2+ and the Ni2+ is reduced to Ni.

31. Oxidation-Reduction Reactions
Activity Series
Some metals are easily oxidized whereas others are not.
Activity series: a list of metals arranged in decreasing ease of oxidation.
The higher the metal on the activity series, the more active that metal.
Any metal can be oxidized by the ions of elements below it.

33. Solution Stoichiometry and Chemical Analysis
There are two different types of units:
laboratory units (macroscopic units: measure in lab);
chemical units (microscopic units: relate to moles).
Always convert the laboratory units into chemical units first.
Grams are converted to moles using molar mass.
Volume or molarity are converted into moles using M = mol/L.
Use the stoichiometric coefficients to move between reactants and product.

34. Solution Stoichiometry and Chemical Analysis

35. Solution Stoichiometry and Chemical Analysis
Titrations

36. Solution Stoichiometry and Chemical Analysis
Titrations
Suppose we know the molarity of a NaOH solution and we want to find the molarity of an HCl solution.
We know:
molarity of NaOH, volume of HCl.
What do we want?
Molarity of HCl.
What do we do?
Take a known volume of the HCl solution, measure the mL of NaOH required to react completely with the HCl.

37. Solution Stoichiometry and Chemical Analysis
Titrations
What do we get?
Volume of NaOH. We know molarity of the NaOH, we can calculate moles of NaOH.
Next step?
We also know HCl + NaOH  NaCl + H2O. Therefore, we know moles of HCl.
Can we finish?
Knowing mol(HCl) and volume of HCl (20.0 mL above), we can calculate the molarity.