Solution Stoichiometry

Chapter 4

Parts of a solution

Solution- homogeneous mixture

  • Solute – what gets dissolved
  • Solvent – what does the dissolving; present in the greatest amount

Soluble – can be dissolved

Insoluble- cannot be dissolved

Miscible- liquids dissolve into one another

Immiscible – liquids do not dissolve in one another

Aqueous solutions

  • Dissolved in water
  • Water is a good solvent because the molecules are polar

The oxygen atoms have a partial negative charge

The hydrogen atoms have a partial positive charge

The angle is 104.5

Dissociation

  • the process of breaking ions of salts apart in water
  • Ions have charges and attract the opposite partial charges on water molecules-see diagram 4-1.
  • Represented by dissociation equation:

CaCl2(s) Ca+2(aq) + 2 Cl-1(aq)

Na3PO4(s) 3 Na+1(aq) + PO4-3(aq)

Solubility

  • How much a substance will dissolve in a given amount of water, usually grams of solute /100 ml of solvent
  • Water can dissolve ionic compounds
  • Water can also dissolve non-ionic compounds as long as they are polar molecules (LIKE DISSOLVES LIKE)
  • For example: water (polar molecule) dissolves ethanol,CH3CH2OH (polar molecule)

Conductivity

A physical property describing the ability to conduct an electric current

  • Electricity is moving charges
  • The ions that are dissolved can move
  • Solutions that can conduct electricity are called ELECTROLYTES
  1. Strong electrolytes : completely dissociate(fall apart into ions) so many ions present and thus conduct well
  • all soluble ionic compounds
  • strong acids and strong bases
  1. Weak electrolytes :partially dissociate (fall apart into ions) so few ions present and thus conducts electricity slightly
  • weak acids and weak bases
  1. Non-electrolytes- don’t dissociate so no ions and thus no electricity is conducted; examples: sugar, ethanol (CH3CH2OH)
  • covalent (molecular) compounds

Types of solutions

Acids- covalent compounds that form hydrogen ions (H+) when dissolved in water; also known as a proton donor

  • Strong acids fall apart completely yielding so many ions and are considered a strong electrolyte

MEMORIZE THESE: H2SO4 HNO3 HCl, HBr, HI, HClO3, and HClO4

  • Weak acids don’t dissociate completely yielding few ions and are considered weak electrolytes.

Bases- ionic or covalent compounds that form hydroxide ions (OH-) when dissolved in water; also known as a proton acceptor

  • Strong bases fall apart into many ions and thus a strong electrolyte

MEMORIZE THESE: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) 2, Sr(OH) 2 and Ba(OH) 2

  • Weak bases don’t dissociate completely yielding few ions and are considered weak electrolytes.

Measuring Solutions

Concentration- how much is dissolved

Molarity = Moles of solute

Liters of solution

  • Abbreviated M
  • 1M = 1 mol solute per 1 liter of solution
  • 1.0 M NaCl is really 1.0 mol Na+1 ions and 1.0 mol Cl-1 ions

Example 1

Calculate the molarity of a solution with 34.6 g of NaCl in 125 ml of solution.

Example 2

How many grams of HCl would be required to make 50.0 ml of a 2.7M solution?

Example 3

What would the concentration be if you used 27g of CaCl2 to make 500. ml of solution? What is the concentration of each ion?

Example 4

Calculate the concentration of a solution made by dissolving 45.6 g of Fe2 (SO4) 3 to 475 ml. What is the concentration of each ion?

Making Solutions

  • important to know the number of moles of solute
  • standard solution – concentration of solution is accurately known
  • To make a solution: add solute to volumetric flask and add enough water to needed volume

Example 5

Describe how to make 100.0 ml of a 1.0 M K2Cr2O4 solution.

Example 6

Describe how to make 250. ml of a 2.0 M copper (II) sulfate dihydrate solution.

Dilutions

  • Adding more solvent to a known solution
  • The moles of the solute stay the same
  • Moles = M x V so M1 V1 = M2 V2

Initial = Final

moles = moles

  • Stock solution is a solution of known concentration used to make a more dilute solution

Example 7

What volume of a 1.7M solution is needed to make 250. ml of a .50 M solution?

Example 8

18.5 ml of 2.3 M HCl is addedto 250 ml of water. What is the concentration of the solution?

Example 9

18.5 ml of 2.3 M HCl is diluted to 250 ml of water. What is the concentration of the solution?

Example 10

You have a 4.0 M stock solution. Describe how to make a 1.0L of a .75 M solution.

Types of Reactions

Metathesis Reactions: double replacement reactions where positive ions swap places with each other

1.Precipitation reactions

  • When aqueous solutions of ionic compounds are mixed together , forming an insoluble solid called a precipitate,
  • Must determine the precipitate using the solubility rules: see handout or wiki for help on memorizing these
  • If you’re not part of the solution, you are part of the precipitate.

Example:

AgNO3 (aq) + KCl (aq) 

The molecular equation lists the reactants and products in their molecular

Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl(s) + K+ (aq) + NO3- (aq)

•The completeionic equationshows all strong electrolytes (strong acids, strong bases, and soluble ionic salts) dissociated into their ions.

•This more accurately reflects the species that are found in the reaction mixture.

Ag+(aq) + Cl-(aq) AgCl(s)

•To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.

•The only things left in the equation are those things that change (i.e., react) during the course of the reaction.

•Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions.

Example 11: Write the molecular, complete ionic and net ionic forms

(NH4)2SO4 (aq) + Ba(NO3)2(aq)

Example 12: Write the molecular, complete ionic and net ionic forms

Iron (III) sulfate and potassium sulfide react

Stoichiometry of Precipitation Reactions

1.identify the species and write a molecular or net ionic equation

2.calculate the moles of reactants

3.determine which is limiting and calculate the moles of product

4.convert to desired unit

Example 15

What mass of solid is formed when 100.00 ml of .100 M barium chloride is mixed with 100.00 ml of .100 M sodium hydroxide?

Example 16

What volume of .204 M HCl is needed to precipitate the silver from 50.0 ml of .0500 M silver nitrate solution?

2.Acid – Base Reactions(neutralization reactions)

  • An acid is proton donor (H+)
  • A base is proton acceptor (OH-)
  • In an acid-base reaction, the acid donates a proton (H+) to the base.
  • Generally, when solutions of an acid and a base are combined, the products are a salt and water

HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)

What is the net ionic equation for the reaction of a strong acid, HCI(aq) and a strong base, NaOH(aq)?

H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq)  Na+ (aq) + Cl- (aq) + H2O (l)

H+ + OH-  H2O (net ionic)

Example 17 Write a net ionic reaction for the neutralization reaction seen below

Mg(OH)2, and HCl react

Titrations

  • The analytical technique in which one can calculate the concentration of a solute in a solution.
  • A solution of known concentration called the titrant is added to a solution of unknown concentration called the analyte until the equivalence point is reached.
  • An indicator is added to the analyte.
  • End point is when the color of the indicator changes which signifies equivalence point (the acid has neutralized the base(equal mole amounts)

Stoichiometry of Titrations

1. Identify the acid & base & write a balanced chemical equation

2. Convert the moles of titrant to analyte.

3. Use the molarity equation to calculate the molarity or volume of

the analyte.

Example 18

A 50.00 ml sample of aqueous Ca(OH)2 requires 34.66 ml of .0980 M nitric acid for neutralization. What is the [Ca(OH)2] ?

Example 19

75.0 ml of .25 M HCl is mixed with 225 ml of .055 M Ba(OH)2. What is the concentration of the excess H+ or OH-?

3.Oxidation- Reduction Reactions (REDOX)

  • An oxidation-reduction reaction involves the transfer of electrons
  • An oxidation occurs when an atom or ion loses electrons.
  • A reduction occurs when an atom or ion gains electrons.
  • One cannot occur without the other
  • To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity( a way of keeping track of electrons)
  • Not necessarily true of what is in nature but it works
  • Need the rules for assigning them- MEMORIZE
  • Elements in their elemental form have an oxidation number of 0.
  • The oxidation number of a monatomic ion is the same as its charge.
  • Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.
  • Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.
  • Fluorine always has an oxidation number of −1.
  • The other halogens have an oxidation number of

−1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.

7.The sum of the oxidation numbers in a neutral

compound is 0.

8.The sum of the oxidation numbers in a

polyatomic ion is the charge on the ion.

Example 19

Assign oxidation numbers to each element in the following:

a) CO3-2b) NO3-1 c) SF6 d) FeCl3 e) H3S

Oxidation – Reduction

  • Mnemonic: OIL RIG or LEO the lion says GER
  • Oxidation is the loss of electrons
  • Reduction is the gain of electrons

2 Na + Cl2 2 NaCl

Sodium is being ______

Chlorine is being ______

  • Oxidation means an increase in oxidation state- lose electrons
  • Reduction means a decrease in oxidation state- gain electrons

Example 21

Identify the substance being oxidized and substance being reduced in the following reactions:

a)Fe2O3 + 3CO  2 Fe + 3 CO2

b)SO3-1 + H+1 + MnO4-1  SO4-1 + H2O + Mn+2

Half Reactions

  • All redox reactions can be thought of as happening in 2 halves
  • One produced electrons- oxidation half
  • The other requires electrons – reduction half

Example 22

Write the half reactions for the following:

  1. Na + Cl2  Na+1 + Cl-1
  1. SO3-1 + H+1 + MnO4-1  SO4-1 + H2O + Mn+2

Balancing Redox Reactions

In Acidic Solution

  • In aqueous solutions the key is the number of electrons produced must be the same as those required
  • For reactions in acidic solution, follow an 8 step procedure
  1. Write separate half reactions
  2. For each half reaction balancer all reactants except H and O
  3. Balance O using H2O
  4. Balance H using H+1
  5. Balance charge using e-1
  6. Multiply each equation to make electrons equal
  7. Add equations and cancel identical species
  8. Check that the charges and elements are balanced

Example 23

Balance the following redox reactions in acidic solution using the half-reaction method

a) Cr(OH)3 + OCl-1 + OH-1  CrO4-2 + Cl-1 + H2O

b) MnO4-1 + Fe+2  Mn+2 + Fe+3

Basic Solution

  • Do everything you would with an acid but add one more step
  • Add enough OH-1 to both sides to neutralize the H+1

Example 23

Balance the following redox reaction in basic solution using the half-reaction method.

a) CrI3 + Cl2  CrO4-1 + IO4-1 + Cl-1

REDCOX Titrations

  • Same as other titrations
  • The permanganate ion is used often because it is its own indicator. MnO4-1 is purple, Mn+2 is colorless. When reaction solution remains clear, MnO4-1 is gone
  • Chromate ion is also useful, but color change, orange yellow to green is harder to detect.

Example 24

A 20.0 ml sample of Na2SO3 was titrated with 36.30 ml of .05130 M K2Cr2O7 in presence of H2SO4. Calculate the molarity of Na2SO3.

8 H+ + Cr2O7-2 + 3 SO3-2 2 Cr+3 + 3 SO4-2 + 4 H2O

Example 25

The Fe content of Fe ore can be determined by titration of standard KMnO4 solution. The Fe ore is dissolved in excess HCl and the Fe is reduced to Fe+2 ions. The solution is then titrated with KMnO4, producing Fe+3 ions and Mn+2 ions in acidic solution. If it requires 41.95 ml of .205 M KMnO4 to titrate a solution made with 6.128 g of Fe ore, what percent of the ore was Fe?

MnO4-1 + 5 Fe+2 + 8 H+1 Mn+2 + 5 Fe+3 + 4 H2O