Presentation on theme: "Types of Chemical Reactions & Solution Stoichiometry Chapter 4."— Presentation transcript:
Types of Chemical Reactions & Solution Stoichiometry Chapter 4
Aqueous Solutions Water is the dissolving medium, or solvent.
Some Properties of Water Water is able to dissolve so many substances because: -Water is “bent” or V-shaped. -The O-H bonds are covalent. -Water is a polar molecule. -Hydration occurs when salts dissolve in water.
Water is a polar molecule because it is a bent molecule. The hydrogen end is + while the oxygen end is -, Delta ( ) is a partial charge--less than 1.
Polar water molecules interact with the positive and negative ions of a salt, assisting in the dissolving process. This process is called hydration.
A Solute -dissolves in water (or other “solvent”) -changes phase (if different from the solvent) -is present in lesser amount (if the same phase as the solvent)
A Solvent -retains its phase (if different from the solute) -is present in greater amount (if the same phase as the solute)
Solubility The general rule for solubility is: “Like dissolves like.” Polar water molecules can dissolve other polar molecules such as alcohol and, also, ionic substances such as NaCl. Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.
Miscibility Miscible -- two substances that will mix together in any proportion to make a solution. Alcohol and water are miscible because they are both polar and form hydrogen bonds. Immiscible -- two substances that will not dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and vinegar is polar.
Solubility How does the rule “Like dissolves like.” apply to cleaning paint brushes used for latex paint as opposed to those used with oil-based paint? C 6 H 14 H 2 0 I 2 C 6 H 14 H20I2H20I2H20I2H20I2 NaNO 3 H 2 0
Electrolytes & Nonelectrolytes An electrolyte is a material that dissolves in water to give a solution that conducts an electric current. A nonelectrolyte is a substance which, when dissolved in water, gives a nonconducting solution.
Electrolytes Strong - conduct current efficiently and are soluble salts, strong acids, and strong bases. NaCl, KNO 3, HNO 3, NaOH NaCl, KNO 3, HNO 3, NaOH Weak - conduct only a small current and are weak acids and weak bases. HC 2 H 3 O 2, aq. NH 3, tap H 2 O Non - no current flows and are molecular substances pure H 2 O, sugar solution, glycerol
Electrical conductivity of aqueous solutions. a) strong electrolyte b) weak electrolyte c) nonelectrolyte in solution. Svante Arrhenius first identified these electrical properties.
When BaCl 2 dissolves, the Ba 2+ and Cl - ions are randomly dispersed in the water. BaCl 2 is a strong electrolyte.
Acids Strong acids -dissociate completely (~100 %) to produce H + in solution HCl, H 2 SO 4, HNO 3, HBr, HI, & HClO 4 Weak acids - dissociate to a slight extent (~ 1 %) to give H + in solution HC 2 H 3 O 2, HCOOH, HNO 2, & H 2 SO 3
HCl is completely ionized and is a strong electrolyte.
Bases Strong bases - react completely with water to give OH ions. sodium hydroxide NaOH (s) ---> Na + (aq) + OH - (aq) Weak bases - react only slightly with water to give OH ions. ammonia NH 3(aq) + HOH (l) NH 4 + (aq) + OH - (aq)
An aqueous solution of sodium hydroxide which is a strong bases dissociating almost 100 %.
Acetic acid(CH 3 COOH) exists in water mostly as undissociated molecules. Only a small percent of the molecules are ionized.
Write the equation of the dissolving of the following compounds. CaCl 2 HCl Fe(NO3) 3 KBr (NH 4 ) 2 Cr 2 O 7
Molarity Molarity (M) = moles of solute per volume of solution in liters:
Molarity Calculations Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. (
Molarity Calculations Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution. (1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl) (1000mL/1L) = 1.60M HCl
How many moles of Co(NO 3 ) 2 are present in 25.00 mL of a 0.75 M Co(NO 3 ) 2 solution?
Molarity Calculations How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO 3 ) 2 solution? (25.00mL)(1L/1000mL)(0.75mol Co(NO 3 ) 2 /1L) (2 mol NO 3 - /1 mol Co(NO 3 ) 2 ) = 3.8 x 10 -2 mol NO 3 -
mol Calculate the moles of each of the ions in 40.00 ml of the following solution..20 M Na 2 CO 3 P106 Q1,-,7, 8,9a-c
Calculate the moles of potassium ions in 50.00 ml of the following solution. 2 M K 3 P 2 2 M K 3 P 2
Standard Solution A standard solution is a solution whose concentration is accurately known. Standard solutions are made using a volumetric flask as follows: mass the solute accurately and add it to the volumetric flaskmass the solute accurately and add it to the volumetric flask add a small quantity of distilled HOHadd a small quantity of distilled HOH dissolve the solute by gently swirling the flaskdissolve the solute by gently swirling the flask add more distilled HOH until the level of the solution reaches the mark on the neckadd more distilled HOH until the level of the solution reaches the mark on the neck invert the capped volumetric 25X to thoroughly mix the solution.invert the capped volumetric 25X to thoroughly mix the solution.
Common Terms of Solution Concentration Stock - routinely used solutions prepared in concentrated form. Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl) Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl)
Dilution of Stock Solutions When diluting stock solutions, the moles of solute after dilution must equal the moles of solute before dilution. Stock solutions are diluted using either a measuring or a delivery pipet and a volumetric flask.
Dilution Calculations What volume of 6 M sulfuric acid must be used to prepare 1 L of a 3.0 M H 2 SO 4 solution? use dilution formula use dilution formula
What volume of 3 M sulfuric acid must be used to prepare.5 L of a.25 M H 2 SO 4 solution?
Types of Solution Reactions -Precipitation reactions AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) -Acid-base reactions NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) -Oxidation-reduction reactions Fe 2 O 3 (s) + 2Al(s) 2Fe(l) + Al 2 O 3 (s)
1. Which of the following substances would you expect to be insoluble in water? Barium hydroxideHydrochloric acid Magnesium sulfateAmmonium nitrate Silver chlorideLithium carbonate Calcium carbonateBarium sulfate Ammonium acetateLead I Chloride Sodium hydroxideAmmonium nitrate Silver nitrate
Solubility Using the solubility rules, predict what will happen when the following pairs of solutions are mixed. a) KOH (aq) & Mg(NO 3 ) 2(aq) b) Na 2 SO 4(aq) & Pb(NO 3 ) 2(aq) c) KNO 3(aq) & BaCl 2(aq) Mg(OH) 2(s) forms PbSO 4(s) forms No precipitate forms
Describing Reactions in Solution 1.Molecular equation (reactants and products as compounds) AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) 2.Complete ionic equation (all strong electrolytes shown as ions) Ag + (aq) + NO 3 (aq) + Na + (aq) + Cl (aq) AgCl(s) + Na + (aq) + NO 3 (aq)
Describing Reactions in Solution (continued) 3.Net ionic equation (show only components that actually react) Ag + (aq) + Cl (aq) AgCl(s) Na + and NO 3 are spectator ions.
Write the balanced complete ionic and net ionic equations : CuSO 4 (aq) + BaCl 2 (aq) → P108 q 28
Old limiting problem to compare If 68.5 g of CO (g) is reacted with 8.60 g of H 2(g), what is the theoretical yield of methanol that can be produced? __H 2(g) + __CO (g) ---> __CH 3 OH (l)
Precipitation Calculations When aqueous solutions of Na 2 SO 4 & Pb(NO 3 ) 2 are mixed. Calculate the mass of the percipitate formed when 1.25 L of 0.0500 M Pb(NO 3 ) 2 & 2.00 L of 0.0250 M Na 2 SO 4 are mixed. 1.
Acid-Base Calculations What volume of a 0.100M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? 1.Cl -, Na +, 2.2. H + (aq) + OH - (aq) ----> HOH (l) (25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.100mol) = 87.5 mL HCl solution
What volume of a 0.500M HCl solution is needed to neutralize 32.0 mL of 0.250 M LiOH?
Key Titration Terms Titrant - solution of known concentration used in titration Analyte - substance being analyzed Equivalence point - enough titrant added to react exactly with the analyte Endpoint - the indicator changes color so you can tell the equivalence point has been reached.
Ammonium sulfate is manufactured by reacting sulfuric acid with potassium hydroxide. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L potassium hydroxide solution if 50.0 mL of sulfuric acid is used?
Titration Calculations Titration 49,a,b 53,54 practice for unit Textbook 181-183 practice for unit Textbook 181-18315a17a23a29a-c35ab 39 try 45b Additional questions 36,a,b 41 40 18,b 12 24
reminder,Do redox in redox reminder,Do redox in redox