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Ch. 4 Types of Chemical Reactions and Solution Stoichiometry.

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Presentation on theme: "Ch. 4 Types of Chemical Reactions and Solution Stoichiometry."— Presentation transcript:

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2 Ch. 4 Types of Chemical Reactions and Solution Stoichiometry

3 Solute A solute is the dissolved substance in a solution. A solvent is the dissolving medium in a solution. Solvent Salt in salt water Sugar in soda drinks Carbon dioxide in soda drinks Water in salt waterWater in soda

4 Saturation of Solutions  A solution that contains the maximum amount of solute that may be dissolved under existing conditions is  saturated.  A solution that contains less solute than a saturated solution under existing conditions is  unsaturated.  A solution that contains more dissolved solute than a saturated solution under the same conditions is  supersaturated.

5 An electrolyte is:  A substance whose aqueous solution conducts an electric current. A nonelectrolyte is:  A substance whose aqueous solution does not conduct an electric current. Try to classify the following substances as electrolytes or nonelectrolytes… Definition of Electrolytes and Nonelectrolytes

6 1.Pure water 2.Tap water 3.Sugar solution 4.Sodium chloride solution 5.Hydrochloric acid solution 6.Ethyl alcohol solution 7.Pure, solid sodium chloride Electrolytes?

7 ELECTROLYTES:NONELECTROLYTES: Tap water (weak) NaCl solution HCl solution Pure water Sugar solution Ethanol solution Pure, solid NaCl But why do some compounds conduct electricity in solution while others do not…? Answers …

8 Ionic CompoundsDissociate NaCl(s)  AgNO 3 (s)  MgCl 2 (s)  Na 2 SO 4 (s)  AlCl 3 (s)  Na + (aq) + Cl - (aq) Ag + (aq) + NO 3 - (aq) Mg 2+ (aq) + 2 Cl - (aq) 2 Na + (aq) + SO 4 2- (aq) Al 3+ (aq) + 3 Cl - (aq)

9 The reason for this is the polar nature of the water molecule… Positive ions associate with the negative end of the water dipole (oxygen). Negative ions associate with the positive end of the water dipole (hydrogen). Ions tend to stay in solution where they can conduct a current rather than re-forming a solid.

10 Covalent acids form ions in solution, with the help of the water molecules. For instance, hydrogen chloride molecules, which are polar, give up their hydrogens to water, forming chloride ions (Cl - ) and hydronium ions (H 3 O + ). Some covalent compounds IONIZE in solution

11 Other examples of strong acids include:  Sulfuric acid, H 2 SO 4  Nitric acid, HNO 3  Hydriodic acid, HI  Perchloric acid, HClO 4 Strong acids such as HCl are completely 100% ionized in solution.

12 Many of these weaker acids are “organic” acids that contain a “carboxyl” group. The carboxyl group does not easily give up its hydrogen. Weak acids such as lactic acid usually ionize less than 5% of the time.

13 Other organic acids and their sources include: o Citric acid – citrus fruit o Malic acid – apples o Butyric acid – rancid butter o Amino acids – protein o Nucleic acids – DNA and RNA o Ascorbic acid – Vitamin C This is an enormous group of compounds; these are only a few examples. Because of the carboxyl group, organic acids are sometimes called “carboxylic acids”.

14 Sugar (sucrose – C 12 H 22 O 11 ), and ethanol (ethyl alcohol – C 2 H 5 OH ) do not ionize - That is why they are nonelectrolytes! However, most covalent compounds do not ionize at all in solution.

15 Molarity The concentration of a solution measured in moles of solute per liter of solution. M = mol L

16 Preparation of Molar Solutions Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?  Step #1: Ask “How Much?” (What volume to prepare?) 1.500 L  Step #2: Ask “How Strong?” (What molarity?) 0.500 mol 1 L  Step #3: Ask “What does it weigh?” (Molar mass is?) 58.44 g 1 mol = 43.8 g

17 Serial Dilution It is not practical to keep solutions of many different concentrations on hand, so chemists prepare more dilute solutions from a more concentrated “stock” solution. Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? M stock V stock = M dilute V dilute (11.6 M)(x Liters) = (3.0 M)(0.250 Liters) x Liters = (3.0 M)(0.250 Liters) 11.6 M = 0.065 L

18 A. Single Replacement Reactions Replacement of:  Metals by another metal  Hydrogen in an acid by a metal  Hydrogen in water by a metal  Halogens by more active halogens A + BX  AX + B BX + Y  BY + X  Ex. Mg(s) + HCl(aq)  MgCl 2 (aq) + H 2 (g)  Ex. 2 Li(s) + 2 H 2 O(l)  2 LiOH(aq) + H 2 (g)

19 The Activity Series of the Metals Lithium Potassium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron Nickel Lead Hydrogen Bismuth Copper Mercury Silver Platinum Gold Metals can replace other metals provided that they are above the metal that they are trying to replace. Metals above hydrogen can replace hydrogen in acids. Metals from sodium upward can replace hydrogen in water

20 The Activity Series of the Halogens Fluorine Chlorine Bromine Iodine Halogens can replace other halogens in compounds, provided that they are above the halogen that they are trying to replace. 2NaCl(s) + F 2 (g)  2NaF(s) + Cl 2 (g) MgCl 2 (s) + Br 2 (g)  ??? No Reaction ???

21 Practice problems - Answers are unbalanced!  1. Mg + FeCl 3  Fe + MgCl 2 2. Sodium is added to water. Na + H 2 O  H 2 + NaOH 3. Lithium is added to hydrochloric acid Li + HCl  H 2 + LiCl 5. Chlorine gas is bubbled into a solution of potassium iodide Cl 2 + KI  I 2 + KCl 4. Zinc is added to a solution of sodium chloride Zn + NaCl  N.R. 6. Chlorine gas is bubbled into a solution of potassium fluoride Cl 2 + KF  N. R.

22 Double Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY  AY + BX One of the compounds formed is usually a precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

23 Double replacement forming a precipitate… Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2KNO 3 (aq) Pb 2+ (aq) + 2 NO 3 - (aq) + 2 K + (aq) +2 I - (aq)  PbI 2 (s) + 2K + (aq) + 2 NO 3 - (aq) Pb 2+ (aq) + 2 I - (aq)  PbI 2 (s) Double replacement (ionic) equation Complete ionic equation shows compounds as aqueous ions Net ionic equation eliminates the spectator ions

24 Solubility Rules – Mostly Soluble IonSolubilityExceptions NO 3 - SolubleNone ClO 4 - SolubleNone Alkali metals SolubleNone NH 4 + SolubleNone Cl -,Br -, I - Soluble Pb 2+, Ag +, Hg 2 2+ SO 4 2- Soluble Ca 2+, Ba 2+, Sr 2+, Pb 2+, Ag +, Hg 2+ C2H3O2-C2H3O2-C2H3O2-C2H3O2-Soluble Ag +

25 Solubility Rules – Mostly Insoluble IonSolubilityExceptions CO 3 2- Insoluble Group IA and NH 4 + PO 4 3- Insoluble Group IA and NH 4 + OH - Insoluble Group IA and Ca 2+, Ba 2+, Sr 2+ S 2- Insoluble Groups IA, IIA, and NH 4 + CrO 4 -2 Insoluble Group IA & NH 4 +, Ca 2+, Sr 2+ SO 3 -2 Insoluble Group IA and NH 4 +

26 D.R. Practice problems 1. KBr(aq) + AgNO 3 (aq)  AgBr(s) + KNO 3 (aq) 2. Silver nitrate + potassium chromate  2AgNO 3 (aq) + K 2 CrO 4 (aq)  AgCrO 4 (s) + 2KNO 3 (aq) 3. Ammonium chloride + cobalt (II) sulfate  2NH 4 Cl(aq) + CoSO 4 (aq)  (NH 4 ) 2 SO 4 (aq) + CoCl 2 (aq) N.R. 4. Lithium hydroxide + sodium chromate 2LiOH(aq) + Na 2 CrO 4 (aq)  2NaOH(aq) + Li 2 CrO 4 (s) 5. Zinc acetate + cesium hydroxide  Zn(C 2 H 3 O 2 ) 2 (aq) + 2CsOH(aq)  Zn(OH) 2 (s) + 2CsC 2 H 3 O 2 (aq) 6. What is the net ionic equation for the rxn above? Zn +2 (aq) + OH - (aq)  Zn(OH) 2 (s)

27 Unstable Compounds!!! (own note paper) Ammonium hydroxide – –NH 4 OH Carbonic Acid – –H 2 CO 3 Sulfurous acid – –H 2 SO 3 Sulfide salts (ex. Na 2 S ) from acid (H + ) All break down to form other products! –NH 4 OH  NH 3 (g) + H 2 O(l) –H 2 CO 3  CO 2 (g) + H 2 O(l) –H 2 SO 3  SO 2 (g) + H 2 O(l) –S -2  H 2 S(g)

28 Unstable Examples: 1. Sodium sulfite + hydrochloric acid Na 2 SO 3 (aq) + 2HCl(aq)  H 2 SO 3 (aq) + 2NaCl(aq) H 2 SO 3 (aq)  H 2 O(l) + SO 2 (g) Na 2 SO 3 (aq) + 2HCl(aq)  H 2 O(l) + SO 2 (g) + 2NaCl(aq) 2. Ammonium sulfate + sodium hydroxide (NH 4 ) 2 SO 4 (aq) + NaOH(aq)  2 NH 4 OH(aq) + Na 2 SO 4 (aq) 2 NH 4 OH(aq)  2 NH 3 (g) + H 2 O(l) (NH 4 ) 2 SO 4 (aq) + 2NaOH(aq)  2NH 3 (g) + 2H 2 O(l) + Na 2 SO 4 (aq) What is the net ionic equation for the reaction above? NH 4 + (aq) + OH - (aq)  NH 3 (g) + H 2 O(l)


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