AP/IB Chemistry NotesName:

Chapter 4.1–4.3: Solution Formation & Composition

  1. Water, The Common Solvent
  1. The water molecule is

- due to "V" shape (105 º angle)

- due to higher of the oxygen atom

  1. Hydration - process by which an ionic solid is as the and are broken apart due to the attraction to the polar water molecule. Water molecules then surround the anions and cations of the solute to hold them in solution. This process can be shown in a chemical equation :

Particle Diagrams for Solutions:

Molecular substances (i.e. sugar, use a simple particle to represent the complex molecule)

Ionic substances (i.e. salt, consider sodium chloride and represent the ions)

  1. Solubility

- definition : The that will dissolve in a given amount of solvent at a certain set of conditions.(e.g. the solubility of PbI2 is 0.012 mol/L at 25 ºC)

  1. The solubility of a solute will vary according to :

- the attraction between

- the attraction between

- the attraction between

  1. Water will generally dissolve (unless the ions of the ionic compound have a greater attraction for each other than for water) and polar molecular compounds.

  1. The Nature of Aqueous Solutions : Strong and Weak Electrolytes
  1. Solute - substance which is in a solution
  2. Solvent - substance in a solution. In a solution of two liquids, the liquid that has the greater volume is considered to be the solvent.
  3. Electrolytes - a substance that will form a solution that will

- usually (charged ions conduct current)

- some molecular compounds are electrolytes due to the formation of ions in aqueous solution:

example:

  1. Strong electrolytes** - form and form a very solution

- substances generally dissociate to a great extent (dilute solutions probably dissociate close to 100%)

- includes: - soluble that form ions

e.g. NaCl(s)  Na+(aq) + Cl-(aq)

- strong - form hydrogen ions in solution(HCl, HBr, HI, HNO3, H2SO4, HClO4) (HClO3 is sometimes included in this list)

e.g.

e.g. (note only first H ionizes to a great

extent under normal circumstances)

- strong (e.g. group 1A hydroxides and heavy Group II hydroxides (Ca, Sr and Ba)

e.g.

Particle diagram for major species in solution:

  1. Weak Electrolytes** - form in solution resulting in poor conducting solutions

- substances dissociate to a

- include

e.g. acetic acid : HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)

- Note the double arrow indicating the reaction does not proceed all the way to the right to form the products. The is actually far to the left indicating that only a few of the acetic acid molecules dissociate.

Particle diagram for major species in solution:

  1. Nonelectrolytes** - substances that do dissolve in water, but do not form ions and therefore .

Particle diagram for major species in solution:

**Note : Do not confuse solubility (how much will dissolve) with strength of electrolytes (how much of the solute will dissociate).

  1. The Composition of Solutions
  1. Qualitative

- dilute - little solute compared to the amount of solvent

- concentrated - much solute as compared to the amount of solvent

Represent dilute & concentrated solutions using particle diagrams:

  1. Quantitative

-Molarity (M) - Molarity is defined as the , as seen in the following equation: M = mol/L

-note : molarity of ions may be different than the molarity of the entire solute

  • e.g. A 1.0 M solution of Mg(NO3)2 has a 2.0 M concentration of NO3- ions

-standard solution - a solution whose is accurately known

  1. Preparing Solutions:

-Determining how much of a particular solid compound to dissolve to make a desired solution.

  1. Use molarity as a . Use volume and molarity to find .
  1. Remember to convert the volume to first!
  2. Example – You want to make 250 mL a 0.1 M solution of AgNO3

You will need to dissolve moles AgNO3 in 250 mL of water to make your solution.

  1. Use the of the compound to calculate the mass that should be dissolved.
  1. Example (con't) – Find the number of grams of AgNO3 that is equal to 0.025 moles.

=

  1. Diluting a solution to make a solution of a different molarity
  2. The idea is very similar to preparing a solution with a solid compound.
  3. We start again by determining the number of that will be in the desired solution.
  1. Example – You want to make 300 mL of 0.5 M HCl. You have 12 M HCl to work with.

=

  1. Next we determine the of concentrated solution that will be needed.
  1. Example (con't) – Using the same setup, we solve for the volume of 12 M HCl that will be needed to obtain 0.15 moles HCl.

=

  1. Once you have found the volume of concentrated solution needed, the concentrated solution to the volume of the desired solution.
  1. Example (con't) – To make our solution we should dilute .
  2. Remember that the molarity * volume of the desired solution will equal the molarity * volume of the concentrated solution. Since you do not change the number of moles present during the dilution process the number of mol present before and after the dilution will be equal. That is to say:

AP/IB Chemistry Notes: 4.4 – 4.7Name:

Chapter 4 – Classifying Reactions: Precipitation Reactions

  1. Types of Chemical Reactions: There are many, many types of chemical reactions. We are not going to go over every classification. We will discuss some classes that will help you identify and predict the products of many common reactions.
  2. Precipitation reactions: (An example of a "" or

"" reaction).

  1. Precipitation Reactions form a when two solutions are combined. An example is the combining aqueous potassium chromate with aqueous lead nitrate to form the precipitate lead chromate (still used in school bus paint!!)
  2. When ionic compounds dissolve in water, they dissociate into cations and anions (review particle models). Certain cations and anions are attracted to each other more than the water and form an ionic solid, which precipitates from the solution. Other ions remain dissolved.
  1. Predicting products of precipitation reactions:
  1. To identify the precipitate or predict the formation of a precipitate the solubilities of compounds can be used. These trends/rules should already be memorized!

- Most nitrate and acetate salts are soluble.

- Most salts containing the alkali metal ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium ion (NH4+) are soluble.

- Most chloride, bromide and iodide salts are soluble. Notable exceptions are salts containing the ions Ag+, Pb2+ and Hg22+.

- Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, HgSO4 and CaSO4.

- Most hydroxide salts are only slightly soluble. Important soluble hydroxides are NaOH and KOH. Ca(OH)2, Sr(OH)2, and Ba(OH)2 are somewhat soluble*.

- Most sulfide, carbonate, chromate, and phosphate salts are only slightly soluble**.

* Note Group 2 trends : As you go down the group sulfate solubility decreases and hydroxide solubility increases.

** Slightly soluble compounds will form precipitates using "normal" concentrations.

  1. Describing Reactions in Solution
  1. For reactions involving ionic compounds, we can write the reaction as a (or formula equation). This shows the normal (complete) formulas of all compounds:.

K2CrO4(aq) + Pb(NO3)2(aq)  PbCrO4(s) + 2 KNO3(aq)

  1. We can rewrite the same reaction as a - Shows a picture of what actually occurs in solution

- strong electrolytes represented as ions in solution

- weak and non- electrolytes still written in molecular (non-ionized) aqeuous state.

2K+(aq) + CrO42-(aq) + Pb2+(aq) + 2NO3-(aq)  PbCrO4(s) + 2 K+(aq) + 2 NO3-(aq)

  1. A includes only the solution components involved in the reaction (spectator ions, which do not undergo change, are omitted):

  1. Stoichiometry of Precipitation Reactions – based on Chapter 3 stoichiometry concepts, but using molarity (concentration) relationships.
  2. Example (based on Exc. #51): 100.0 mL of 0.20 M silver nitrate is mixed with 100.0 mL of 0.15 M calcium chloride.
  3. Predict the solid product formed when the solutions are mixed. Write a balanced complete and net ionic equation.
  4. Determine the mass of solid product that will be formed in the reaction.
  5. Calculate the concentrations of each ion remaining in solution after precipitation is complete.
  1. AP Connection: Question #4
  2. Precipitation reactions are a common type of reaction found in Question 4 on the AP test. In Question 4, you are presented with information about 3 different reactions and asked to predict the products, writing a balanced Net Ionic Equation (4 points). A follow up question is then asked about the reaction (1 point). Three of these questions are presented. You must answer all three.
  3. The first key is to identify the reaction as a precipitation (or double replacement) reaction. Hint: Look for 2 aqueous ionic solutions being mixed. (be careful because this could also be a key for a redox reaction).
  4. Sample:

A solution of sodium hydroxide is added to a solution of lead(II) nitrate.

i) Write the balanced equation: [implied to mean balanced net ionic equation]

ii) If 1.0 L volumes of 1.0 M solutions of sodium hydroxide and lead(II) nitrate are mixed together, how many moles of product(s) will be produced? Assume the reaction goes to completion.

AP/IB Chemistry Notes: 4.8Name:

Chapter 4 – Classifying Reactions: Acid-Base Reactions

  1. Acid-Base Reactions
  1. Definitions :
  1. Arrhenius:

Acid - forms ions in solution (e.g HCl)

Base - forms ions in solution (e.g. NaOH)

  1. Brønsted-Lowry :

Acid - proton (H+) (e.g. HCl)

Base - proton (e.g. NH3 ; NH3 + H+  NH4+)

  1. General reaction : Acid + base(metallic hydroxide)  salt + water (neutralization reaction)
  1. e.g. HCl and NaOH

- molecular equation : HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

- complete ionic eqn. : H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)

- net ionic equation :

  1. Stoichiometry of Acid-Base Reactions
  1. You first want to examine the acid-base reaction (similar to predicting a precipitation reaction). Here are some general steps (they can and should vary depending on the problem):
  2. List the major species present in solution before the reaction occurs. Decide what reaction will occur (look for formation of water or gases)
  3. Write a balanced net ionic equation. (leave space for a BCA table)
  4. Calculate the moles of reactants. For solutions, use the volumes of the original solutions and their molarities (before mixing). Input into a BCA table.
  5. Determine the limiting reactant if appropriate.
  6. Analyze the problem and find the moles of reactant or product asked for.
  7. Convert to grams or volume of solution if asked for

*All problems are different. Don’t “force” a problem into a particular solution method.

  1. Acid-base titrations (volumetric analysis) – determine an unknown quantity through titration.
  1. Titration involves adding a precisely measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte).
  2. The titrant reacts with the analyte in a known manner, such as an acid-base reaction.
  3. An indicator marks the equivalence point (or stoichiometric point) where just the right amount of titrant has been added to completely react with the analyte. The endpoint is where the indicator actually changes color, which hopefully occurs near the equivalence point.
  1. There are numerous variations on the acid-base reaction. Be sure to read through the many examples in Section 4.8. We will consider some examples now.

AP/IB Chemistry Notes: 4.9 – 4.10Name:

Chapter 4 – Classifying Reactions: Oxidation–Reduction Reactions

  1. Oxidation-Reduction Reactions - reactions in which are transferred.

- causes a change in the oxidation state of an element

  1. Oxidation states - numbers are assigned to elements

- used to keep track of (not the same as charge)

Rules for assigning oxidation states : (These need to be memorized.)

- The oxidation state of an uncombined element is zero (includes diatomic elements H2,N2, O2, F2, Cl2, Br2 and I2).

- The oxidation state of a monatomic ion is the same as its charge (e.g. the sulfide ion , S2-, has an oxidation state of -2).

- Oxygen has an oxidation state of -2 in covalent compounds (except in peroxides (O22-) where each oxygen is assigned an oxidation state of -1).

- In covalent compounds hydrogen is assigned an oxidation state of +1. (Hydrogen has a -1 charge in hydrides such as lithium hydride (LiH) or sodium hydride (NaH).

- In compounds, fluorine always has an oxidation state of -1.

- The sum of the oxidation states of the elements in a neutral compound must equal zero.

- The sum of the oxidation states of the elements in a polyatomic ion must equal the charge on the polyatomic ion.

- Oxidation states may on rare occasions be nonintegers. For example in iron (III) oxide (Fe3O4), the iron has an oxidation state of 8/3 (eight-thirds).

  1. Characteristics of Redox Reactions :
  1. Oxidation

- of electrons

- an in oxidation state (or number)

- the substance oxidized is the agent (gives electrons to another substance)

  1. Reduction

- of electrons

- a in oxidation state (or number)

- the substance reduced is the agent (takes electrons away from another substance)

Familiar examples of Red-Ox Reactions (identify the electron transfers in each case):

  1. Single Replacement – A more active element replaces a less active element in a compound. Metals replace metals, non-metals replace non-metals. Use an activity series to help you predict.
  2. Example – A solution of silver nitrate reacts with copper solid:
  1. Synthesis – two or more compounds to form a new compound.
  1. Example – Lead(II) sulfide burns in the air forming lead(II) oxide and sulfur dioxide:
  1. Decomposition – One compound into smaller compounds or its component elements.
  1. Example - Heating Calcium hydroxide causes it to decompose into Calcium oxide and water:
  1. Balancing Redox Reactions
  1. By the half-reaction method :

In acidic solution

  1. Write separate oxidation and reduction reactions for the reaction.
  2. For each half reaction :

- balance all the elements except and

- balance atoms using H2O

- balance atoms using H+

- balance the charge using

  1. If necessary, balance electrons lost and gained in each half reaction by multiplying one or both half reactions by an integer.
  2. the half-reactions and out like species.
  3. Check to make sure both charges and elements (mass) are .

In Basic solution :

  1. Balance as in an acidic solution (see above).
  2. Add a number of OH- ions equal to the H+ ions present to both sides of each half reaction to form H2O.
  3. Eliminate the number of H2O molecules that appear on both sides of the equation.
  4. Check to make sure charges and elements are balanced.
  1. Stoichiometry of Red-Ox reactions: Volumetric analysis (titrations) can also be performed with redox reactions. Use methods similar to solving acid-base problems:
  2. Identify the important components of the solution/system and decide what will react.
  3. Write and balance the reaction equation (net ionic if appropriate).
  4. Convert given quantities to moles. Use a BCA analysis to determine unknown quantities. (Don’t forget to pay attention to limiting reactants if necessary).
  5. Convert answer to desired quantity (grams, solution volume, etc.)

AP/IB Chem1Ch. 4 Solutions