chapter 4 solution chemistry and the hydrosphere

Chapter 4

Solution Chemistry and

the Hydrosphere

Earth: The Water Planet• About 70% of the Earth is covered with

water, with 97% residing in oceans

• Earth’s early atmosphere may have been formed from the gases released by volcanic activity.

• As the Earth cooled the vapor in the atmosphere condensed and rain fell. This filled the depressions in the planet’s crust. We think…

Temperature and Salinity (halinity) in the Oceans

Salinity (halinity)

Temperature and Salinity (halinity) vary with ocean depth

• Solutions are homogeneous mixtures of two or more substances.

• Solvent – the substance present in a solution in the greatest proportion (in number of moles).

• Solute - the substance dissolved in the solvent.

Aqueous Solution (aq) – a solution where water is the solvent.

Solution Terms

Solute + Solvent = Solution

Solutions can be either Homogeneous or Heterogeneous (like oceans which are turbid).

Solutions

• Solutions are homogeneous mixtures of two or more substances.

• The solvent is the substance in greatest quantity.

• Solutes are the smaller quantity ingredients (usually) dissolved in the mixture.

Aqueous SolutionsWater is the dissolving medium

104.5o

Some Properties of Water

• Water is “bent” or V-shaped.

• Water is a molecular compound.

• Water is a polar molecule.

• Hydration occurs when ionic compounds dissolve in water.

Hydration of a solute Ion (Mg)

Concentration – ratio of the quantity of solute to either the mass or volume of the solution or solvent in which the solute is dissolved.

Consequently, there are many different concentration terms depending on:

1. What is used to identify the quantity of solute (e.g. moles, mass, volume, etc.)

2. Whether the denominator of the ratio is the solvent or solution.

3. Whether mass or volume is used as the unit in the denominator.

How much of a solute dissolves?

Concentration of Solutions

• Molarity (M) = moles of solute per volume of solution in liters:

M

M

molaritymoles of soluteliters of solution

HClmoles of HCl

liters of solution3

62

• Molality (m): moles solute/kg solvent• ppm: parts per million; mg solute/kg

solution• ppb: mg solute/kg solution• ppt: ng solute/kg solution• % by weight:

(grams of solute/total g solution) x 100%

• Mole fraction: mole solute/total moles in

solution

Other Common Units of Concentration

Average Concentrations of the 11 Major Constituents of Seawater.

Ions g/kg mmol/kg mmol/L

Na+ 10.781 468.96 480.57

K+ 0.399 10.21 10.46

Mg2+ 1.284 52.83 51.14

Ca2+ 0.4119 10.28 10.53

Sr2+ 0.00794 0.0906 0.0928

Cl- 19.353 545.88 559.40

SO42- 2.712 28.23 28.93

HCO3- 0.126 2.06 2.11

Br- 0.0673 0.844 0.865

B(OH)3 0.0257 0.416 0.426

F- 0.00130 0.068 0.070

Total 35.169 1119.87 1147.59

ProblemWhat is the molarity of an aqueous solution prepared by adding 36.5 g of barium chloride (208.233g/mol) to enough water to make 750.0 mL of solution?

Determining the Number of Moles of Solute

• # moles Solute = (Molarity)(# Liters of Solution)

• n = M x volume (in Liters)

Molarity = # of moles of solute# of liter of solution

Problem

How many grams of aluminum nitrate (212.996 g/mol) are required to make 500.0 mL of a 0.0525 M aqueous solution?

What is the molarity of nitrate ions in a 0.0525 M solution of aluminum nitrate?

Problem

Dilution of Concentrated Solutions

• In dilution, a volume of stock solution is In dilution, a volume of stock solution is obtained and more solvent is addedobtained and more solvent is added The number of moles of solute is constant in The number of moles of solute is constant in

a dilution.a dilution. # moles solute(stock) = Molarity# moles solute(stock) = Molarityss x volume x volumess

• The dilute solution uses the moles of The dilute solution uses the moles of solute with additional amounts of solvent.solute with additional amounts of solvent. # moles solute(dilute) = Molarity# moles solute(dilute) = Molaritydd x volume x volumedd

MMss x V x Vss = = MMdd x V x Vdd

Dilutions

Practice Hydrochloric acid is obtained in 12.0 Hydrochloric acid is obtained in 12.0 MM stock stock

solution. What volume of stock solution is solution. What volume of stock solution is required to make 500.0 mL of a 0.145 required to make 500.0 mL of a 0.145 MM dilute dilute solution?solution?

Electrolytes

Strong - conduct current efficiently

Examples: Aqueous solutions of NaCl, HNO3, HCl

Electrolytes

Weak - conduct only a small current (vinegar, tap water)

Non Electrolyte

A solution in which no ionization occurs. There is no conduction of electrical current.

Examples: Aqueous solutions of sugar, ethelyne glycol

Acid-Base Reactions• Bronsted-Lowry acids are proton (H+)

donors.

• Bronsted-Lowry bases are proton acceptors.

• Free hydrogen ions don’t exist in water because they strongly associate with a water molecule to create a hydronium ion (H3O+) (a hydrated proton).

Acid-Base Reactions

• A neutralization reaction takes place when an acid reacts with a base and produces a solution of a salt and water.

• A salt is made up of the cation characteristic of the base and the anion characteristic of the acid.

• Example: HCl + NaOH ---> NaCl + H2O

Strong Acids and Bases• A strong acid or strong base is completely

ionized in aqueous solution.

• HCl, HBr, HI, HNO3, HClO4 and H2SO4 are all strong acids. All other acids are assumed to be weak acids.

• A weak acid or weak base only partially ionizes in aqueous solution.

• Amphiprotic substances can behave as either a proton acceptor or a proton donor. Water is an example.

Types of Equations

• Molecular Equations have reactants and products written as undissociated (not ionized) molecules.

HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(l)

• Overall Ionic Equations show all the species, both ionic and molecular present in aqueous solution for the reaction.

H+ + Cl- + Na+ + OH- --> Na+ + Cl- + H2O

Continued• Strong acids and strong bases are written

as the corresponding ions in an overall ionic equation.

• Net Ion Equations describe the actual chemical reaction occurring.

H+ + OH- ----> H2O• The Na+ and Cl- ions are spectator ions

in this reaction, because they are unchanged by the reaction (as is the solvent).

ProblemWrite a balanced molecular equation and a net ionic equation for the following reactions:

a. Solid magnesium hydroxide reacts with a solution of sulfuric acid.

ProblemWrite a balanced molecular equation and a net ionic equation for the following reactions:

a. Ammonia gas reacts with hydrogen chloride gas.

Precipitation reactions:1. A solid product (often colloidal) is formed from a

reaction in solution.2. The General Solubility Rules can be used to

predict whether precipitates will form when mixing solutions of ionic compounds.

3. Rules are summarized on the following two slides.

For example:

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Notice that solid silver chloride is formed (a precipitate).

GENERAL IONIC SOLUBILITY RULES (see Table 4.5, p. 149)

Soluble Compounds: Exceptions:1. All salts of (Na+), (K+) and (NH4

+).

2. All (Cl‾ ), (Br‾ )and (I‾ ) [halide salts] Halide salts of Ag+, Hg2

2+, Pb2+

3. All (F‾ ) salts Fluoride salts of Mg2+, Ca2+, Sr2+, Ba2+, Pb2+

4. All (NO3 ‾), (ClO3 ‾), (ClO4 ‾ ), Acetates of Ag+ and Hg22+

(C2H3O2 ‾) only moderately soluble

5.All sulfate salts (SO42 ‾) Sr2+, Ba2+, Pb2+,

(Ca2+, Ag+ are moderately soluble)

GENERAL IONIC SOLUBILITY RULES

Poorly Soluble Salts: Exceptions:

6. All (CO32), (PO4

3 ), Na +, K +, NH4 +

(CrO42 ), & (C2O4

2 )

7. All (S2 ) Group 1 & 2 cations and NH4

+

8. All (OH ) & (O2 ) Group 1 & NH4 +,

(Ca 2+, Sr2+ and Ba 2+ are moderately soluble)

Determining Whether a Precipitate will Form

Does a precipitate form when sodium chloride is mixed with silver nitrate? If so what is the precipitate?

CaCl2(aq) ---> Ca+2(aq) + 2Cl-(aq)

Na2CO3 ---> 2Na+(aq) + CO3-2(aq)

Net Ionic Equations• Soluble ionic compounds are called

strong electrolytes and completely ionize in aqueous solution.

• Write the balanced net ionic equation when sodium sulfate reacts with barium acetate.

Types of Solutions

• A saturated solution contains the maximum concentration of solute that can dissolve in it (for a given T, V and P).

• A supersaturated solution contains more than the quantity of a solute that is predicted to be soluble in a given volume of solution at a given temperature.

A Saturated Solution Example

Supersaturated Solution

Sodium acetate precipitates from a supersaturated solution.

ProblemWhat mass of barium sulfate (233.390g/mo) is produced when 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron (III) sulfate?

Rules for Assigning Oxidation States

1. The oxidation number of elements in a neutral molecule sum to zero or sum to charge of the ion in an ion.

2. Oxidation state of an atom in an element = 0

3. Oxidation state of monatomic ion = charge

4. Fluorine = 1 in all compounds

5. Hydrogen = +1, Oxygen = 2 in most compounds (except in peroxides where oxygen = 1)

6. Unless combined with O, or F the halogens are -1.

Practice Problems

SOSO22

CrOCrO442-2-

NHNH33

ClOClO33--

SFSF66

ClCl22

Oxygen is -2Oxygen is -2 and Sulfur is +4 and Sulfur is +4

Oxygen is -2Oxygen is -2 and Chromium is +6 and Chromium is +6

Oxidation-Reduction Reactions

Fe2O3(s) + Al(s) Fe(l) + Al2O3(s)

M X

X-M+

e-Oxidized Reduced

Loses e- Gains e-

Oxidation State Reduction State

Increases Increases

Reducing Agent Oxidizing Agent

Fe2O3(s) + Al(s) Fe(l) + Al2O3(s)

Oxidation-Reduction Half Reactions

• When copper wire is immersed in a solution of silver nitrate it is oxidized.

• Cu ---> Cu2+ + 2 e-

• Ag+ + e- ---> Ag

• Silver ion is reduced.

Balancing by Half-Reaction Method in Acid Solution

1. Write separate reduction, oxidation half-reactions.

2. For each half-reaction:a. Balance elements (except H, O)

b. Balance O using H2O

c. Balance H using H+

d. Balance charge using electrons

Balancing by Half-Reaction Method (continued)

3. If necessary, multiply by an integer to equalize electron count.

4. Add half-reactions.

5. Check that elements and charges are balanced.

PracticeBalance the following equation in acid solution:

BrBr--(aq) + MnO(aq) + MnO44--(aq) Br(aq) Br22(l) + Mn(l) + Mn2+2+(aq)(aq)

Half-Reaction Method - Balancing in Basic Solution

1. Balance as in acid.

2. Add OH to both sides of the reaction that equals the number of H+ ions.

3. Form water by combining H+, OH.

4. Check elements and charges for balance.

PracticeBalance the following oxidation-reduction reaction in basic solution.

Ag(s) + CN- + O2(g) ----> Ag(CN)2-(aq)

4.102. A method for determining the quantity of dissolved oxygen in natural waters requires a series of redox reactions. Balance the following chemical equations in that series under the conditions indicated:a. Mn2+(aq)+ O2(g) → MnO2(s) (basic solution)b. MnO2(s) + I–(aq) → Mn2+(aq) + I2(s) (acidic solution)c. I2(s)+ S2O3

2–(aq) → I–(aq) + S4O62–(aq) (neutral solution)

Magnetite, Fe3O4, could also be written as FeO۰Fe2O4 so is referred to as a ferrous-ferric oxide (another example of a mixed oxidation state).

Magnetite is the most magnetic of naturally occurring minerals (lodestones).

Crystals of magnetite have been found in some bacteria, the brains of bees, termites, birds and humans.

Titration Terms• A titration is a volumetric analytical method

used to determine the concentration of an unknown solution by reacting it with a standard solution.

• A standard solution is a solution of known concentration.

• The equivalence point in a titration is reached when enough standard solution has been added to completely react with the unknown solution.

• The end point in a titration is reached when an indicator changes color or a specific pH or potential is reached.

Titration ExampleH2SO4 + 2 NaOH ---> Na2SO4 + 2 H2O

H2SO4

NaOH

End Point

Stoichiometry Calculations

H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O

What is the concentration of sulfuric acid if 15.00ml of it reacts with 18.45 mL of a 0.0973 M NaOH solution?

PracticeIf 30.34 mL of a 0.135 M solution of hydrochloric acid were required to neutralize 25.00 mL of a sodium hydroxide solution. What is the molarity of the sodium hydroxide solution?

ChemTour: Molarity

Click to launch animation

PC | Mac

Students learn to calculate the molar concentration of a solution.

ChemTour: Dilutions


PC | Mac

Students are introduced to the concept of dilution and making standard solutions from a stock solution. Students arrange a series of solutions in order of increasing concentration based upon the intensity of their color. Using the ordered series of solutions with concentrations given, students predict the concentration of unknown solutions.

ChemTour: Saturated Solutions


PC | Mac

This ChemTour explains the dynamics of saturated solutions.

Dilutin of HCl

An HCl solution of pH 3.0 is diluted by a factor of 10 with water. What is the new pH of the solution?

A) < 4.0 B) 4.0 C) > 4.0

Dilutin of HCl

Consider the following arguments for each answer and vote again:

A. Although the concentration of HCl in solution will decrease by a factor of 10, more HCl dissociates so the concentration of H3O+ decreases less.

B. Since HCl is a strong acid, dilution by a factor of 10 will result in a decrease in the H3O+ concentration from 0.001 M to 0.0001 M, giving a pH of 4.0.

C. Dilution of a strong acid produces a weak acid, whose pH is higher than 4.0.

Titraction of HCl and NaHO

To the left is a plot that shows the pH of an HCl solution as a function of the added volume of 0.011 M NaOH. Which of the following plots would correspond to the same titration but using 0.022 M NaOH?

A) B) C)

Titraction of HCl and NaHO


A. The shape of both titration curves is the same, but the pH for titration with the stronger base should be higher at every point on the curve.

B. The titration curve stays the same up to the equivalence point, but the pH will be higher when there is excess strong base.

C. The curve will be shifted to the left because only half the volume of 0.022 M NaOH will be required to reach the equivalence point.

Dilution of Acidic Acid

An acetic acid (HAc) solution of pH 3.0 is diluted by a factor of 10 with water. What is the new pH of the solution?

A) < 4.0 B) 4.0 C) > 4.0

Dilution of Acidic Acid


A. Dilution of an acetic acid solution will drive the HAc/Ac- equilibrium toward further ionization of HAc. Therefore, the final pH will be between 3 and 4.

B. Dilution of an acidic solution with pH 3.0 by a factor of 10 will result in a decrease in the H3O+ concentration from 0.001 M to 0.0001 M, giving a pH of 4.0.

C. Dilution of a weak acid solution with pH 3.0 will give a less acidic solution than the dilution of a strong acid solution with the same pH.

pH of HCl,H2SO4,and NaOH Solutions

Which of the following three solutions would have the highest pH?

A) 10-3 M NaOH B) 10-6 M H2SO4 C) 10-12 M HCl

pH of HCl,H2SO4, and NaOH Solutions


A. Of the three solutions, only the NaOH solution is basic, so its pH must be the highest.

B. Although H2SO4 is a strong acid, it dissociates in water to form the base SO4

2-, making its pH higher than that of the other two solutions.

C. Since HCl dissociates completely in water, the concentration of H3O+ is 10‑12 M for this solution. Therefore, the pH of the solution is 12.

chapter 4 solution chemistry and the hydrosphere

Documents