Chapter 4. Reactions in Aqueous Solution. Solutions. 4.1 General Properties of Aqueous Solutions

Size: px

Start display at page:

Download "Chapter 4. Reactions in Aqueous Solution. Solutions. 4.1 General Properties of Aqueous Solutions"

George Underwood
6 years ago
Views:

1 Chapter 4 in Solution 4.1 General Properties of Solutions Solutions Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes.

and solvates them. This process is called dissociation.

2 Solutions Dissociation When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them. This process is called dissociation. An electrolyte is a substances that dissociates into ions when dissolved in water.

Hydration Process Water molecules surround the

dissociates into ions when dissolved in water.

3 Hydration Process Water molecules surround the ions. Note the positions of the positive and negative charges. Solutions An electrolyte is a substance that dissociates into ions when dissolved in water. A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

Molecular compounds tend to be nonelectrolytes, except for acids and bases.

4 Electrolytes and Nonelectrolytes Soluble ionic compounds tend to be electrolytes. Molecular compounds tend to be nonelectrolytes, except for acids and bases. Electrolytes A strong electrolyte dissociates completely when dissolved in water. A weak electrolyte only dissociates partially when dissolved in water.

5 Electrolytes Strong acids Strong bases Strong Electrolytes Are

KCl(s) KCl(aq) K+(aq) + Cl (aq) A weak electrolyte is partially dissociated in water.

6 Strong Electrolytes Are Strong acids Strong bases Soluble ionic salts MSOffice1 A strong electrolyte is completely dissociated in water. KCl(s) KCl(aq) K+(aq) + Cl (aq) A weak electrolyte is partially dissociated in water. NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH (aq) Non-electrolytes do not form ions. C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq)

슬라이드 12 MSOffice1 I had to use the line option to recreate the arrows so that the weak electrolyte could have the proper equilibrium arows., 2006-07-29 4.

7 슬라이드 12 MSOffice1 I had to use the line option to recreate the arrows so that the weak electrolyte could have the proper equilibrium arows., Precipitation When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

8 Precipitation Solubility Rules How can a precipitation reaction be predicted?

9 Metathesis (Exchange) Metathesis comes from a Greek word that means to transpose. It appears as though the ions in the reactant compounds exchange, or transpose, ions: AgNO 3 (aq) + KCl(aq) AgCl(s) + KNO 3 (aq) Solution Chemistry It is helpful to pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution). If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction.

10 Types of Chemical Molecular equation (components written as compounds): AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) Complete Ionic Equation (electrolytes written as ions): all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. Ag + (aq) + NO 3 (aq) + Na + (aq) + Cl (aq) AgCl(s) + Na + (aq) + NO 3 (aq) Net ionic equation (ions participating in the reaction Ag + (aq) + Cl (aq) AgCl(s) (Na + and NO 3 are spectator ions.) Writing Net Ionic Equations 1. Write a balanced molecular equation. 2. Dissociate all strong electrolytes. 3. Cross out anything that remains unchanged from the left side to the right side of the equation. 4. Write the net ionic equation with the species that remain.

4.3 Acids, Bases, & Neutralization Acids The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H + when dissolved in water.

There are only seven strong acids: Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO 3 ) Sulfuric (H 2 SO 4 ) Chloric (HClO 3 ) Perchloric (HClO 4 ) Bases

11 4.3 Acids, Bases, & Neutralization Acids The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H + when dissolved in water. Both the Danish chemist J. N. Brønsted and the British chemist T. M. Lowry defined them as proton donors. There are only seven strong acids: Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO 3 ) Sulfuric (H 2 SO 4 ) Chloric (HClO 3 ) Perchloric (HClO 4 ) Bases Arrhenius defined bases as substances that increase the concentration of OH when dissolved in water. Brønsted and Lowry defined them as proton acceptors. The strong bases are the soluble metal salts of hydroxide ion: Alkali metals Calcium Strontium Barium

water: CH 3 COOH(aq) + NaOH(aq) CH 3 COONa(aq) + H 2 O(l) When a strong acid reacts with a strong base,

12 Acid-Base : the acid donates a proton (H + ) to the base. Neutralization Generally, when solutions of an acid and a base are combined, the products are a salt and water: CH 3 COOH(aq) + NaOH(aq) CH 3 COONa(aq) + H 2 O(l) When a strong acid reacts with a strong base, the net ionic equation is HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) H + (aq)+ Cl (aq)+ Na + (aq) + OH (aq) Na + (aq)+ Cl (aq)+ H 2 O(l) H + (aq) + OH (aq) H 2 O(l)

13 Neutralization with Gas-Forming Some metathesis reactions do not give the product expected. In this reaction, the expected product (H 2 CO 3 ) decomposes to give a gaseous product (CO 2 ): CaCO 3 (s) + HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) NaHCO 3 (aq) + HBr(aq) NaBr(aq)+ CO 2 (g) + H 2 O(l) Similarly, when a sulfite reacts with an acid, the products are a salt, sulfur dioxide, and water: SrSO 3 (s) + 2HI(aq) SrI 2 (aq) + SO 2 (g) + H 2 O(l) This reaction gives the predicted product, but you had better carry it out in the hood, or you will be very unpopular! But just as in the previous examples, a gas is formed as a product of this reaction: Na 2 S(aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + H 2 S(g)

Oxidation Numbers : To determine if an oxidation reduction reaction has occurred, we assign an oxidation number to each element in a neutral

14 4.4 Oxidation-Reduction 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. Oxidation Numbers : To determine if an oxidation reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity. 1. Elements in their elemental form have an oxidation number of The oxidation number of a monatomic ion is the same as its charge. 3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.

15 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. 4. The sum of the oxidation numbers in a neutral compound is The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. Sample Problem Assign oxidation numbers to all atoms in the following: CO 2, K 2 Cr 2 O 7, PCl 5 CO 2 oxygen is -2 2 = -4, making C +4. K 2 Cr 2 O 7 oxygen -2 7 = -14. K (group 1A) is +1 2 = +2 2Cr must equal +12 = each Cr = +6 PCl 5 Cl (group VIIA) = -1 5 = -5 P = +5

16 Oxidation of Metals by Acids & Bases Displacement In displacement reactions, ions oxidize an element. The ions, then, are reduced. Displacement In this reaction, silver ions oxidize copper metal: Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s)

17 Displacement The reverse reaction, however, does not occur: Cu 2+ (aq) + 2Ag(s) x Cu(s) + 2Ag + (aq) Activity Series

18 Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution: Molarity (M) = moles of solute volume of solution in liters Mixing a Solution To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute. The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

19 Sample Problem What is the molarity of 100 ml solution containing 15.6g of NaOH? 1 mol NaOH 15.6g NaOH mol NaOH 40.0g NaOH mol NaOH M = 3.90 M NaOH 1 L 100 ml 1000 ml Sample Problem How many moles of HCl are contained in 250 ml of a molar solution? 1 L mol HCl 250. ml mol HCl 1000 ml 1 L

20 Other Concentration Units Other Concentration Units ppm ppb ppt

7 10 M Cr 8 6+ Dilution One can also dilute a more concentrated solution by Using a pipet to

21 Sample Problem What is the molarity of a solution that is 4.0 ppb Cr 6+? 4.0 g 4.0 ppb L soln g 1 g Cr 1mol Cr 6 6+ L soln. 10 g 52.0 gcr M Cr 8 6+ Dilution One can also dilute a more concentrated solution by Using a pipet to deliver a volume of the solution to a new volumetric flask, and Adding solvent to the line on the neck of the new flask Pearson Education, Inc.

Dilution The molarity of the new solution can be determined from the equation M c V c = M d V d, Moles solute before dilution = moles solute after dilution where M c and M d are the molarity of the

22 Dilution The molarity of the new solution can be determined from the equation M c V c = M d V d, Moles solute before dilution = moles solute after dilution where M c and M d are the molarity of the concentrated and dilute solutions, respectively, and V c and V d are the volumes of the two solutions. How many ml of 15.0 M nitric acid solution should be dilute to make 250. ml of a 0.50 M solution? V initial Cfinal Vfinal Cinitial 0.50 M 250. ml 15.0M 8.3 ml 4.6 Using Molarities in Stoichiometric Calculations

23 Titration Titration is an analytical technique in which one can calculate the concentration of a solute in a solution. Equivalence point The point when all reactants are completely consumed. End point Close to the equivalence point marked by an indicator. Titration

24 Sample Problem What volume of M HCl solution is need to neutralize 25.0 ml of a M KOH solution? The balanced molecular equation is: HCl(aq) + KOH(aq) KCl(aq) + H 2 O(l) mol KOH 1mol OH 1 L 25.0 ml 1 L 1 mol KOH 1000 ml mol OH at neutralization, mol OH mol H 56.3 ml HCl + - mol H - + 1mol HCl 1 L 1000 ml + 1 mol H mol HCL 1 L Homeworks 필수숙제 : 18, 24, 36, 46, 68, 86, 115 추가연습 20, 22, 32, 40, 48, 50, 58, 62, 74

Chapter 4 Aqueous Reactions and Solution Stoichiometry

Chapter 4 Aqueous Reactions and Solution Stoichiometry Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Solutions Chapter 4 and Solution Stoichiometry AP Chemistry 2014-15 North Nova Education Centre