Types of Chemical Reactions & Solution Stoichiometry Chapter 4

Water is the dissolving medium, or solvent. 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? C6H14 H20 I2 C6H14 H20 I2 NaNO3 H20

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.

pure H2O, sugar solution, glycerol Electrolytes Strong - conduct current efficiently and are soluble salts, strong acids, and strong bases. NaCl, KNO3, HNO3, NaOH Weak - conduct only a small current and are weak acids and weak bases. HC2H3O2, aq. NH3, tap H2O Non - no current flows and are molecular substances pure H2O, 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 BaCl2 dissolves, the Ba2+ and Cl- ions are randomly dispersed in the water. BaCl2 is a strong electrolyte.

Acids Strong acids - dissociate completely (~100 %) to produce H+ in solution HCl, H2SO4, HNO3, HBr, HI, & HClO4 Weak acids - dissociate to a slight extent (~ 1 %) to give H+ in solution HC2H3O2, HCOOH, HNO2, & H2SO3

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 NH3(aq) + HOH(l) <---> NH4+(aq) + OH-(aq)

An aqueous solution of sodium hydroxide which is a strong bases dissociating almost 100 %.

Acetic acid(CH3COOH) 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. CaCl2 HCl Fe(NO3)3 KBr (NH4)2Cr2O7

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(NO3)2 are present in 25. 00 mL of a 0 How many moles of Co(NO3)2 are present in 25.00 mL of a 0.75 M Co(NO3)2 solution?

Molarity Calculations How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO3)2 solution? (25.00mL)(1L/1000mL)(0.75mol Co(NO3)2/1L) (2 mol NO3-/1 mol Co(NO3)2) = 3.8 x 10-2 mol NO3-

mol Calculate the moles of each of the ions in 40.00 ml of the following solution. .20 M Na2CO3 P106 Q1,-,7, 8,9a-c

Calculate the moles of potassium ions in 50 Calculate the moles of potassium ions in 50.00 ml of the following solution. 2 M K3P2

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 flask add a small quantity of distilled HOH dissolve the solute by gently swirling the flask add 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.

Steps involved in making a standard solution.

Steps to dilute a stock 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)

Make a solution demo

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 H2SO4 solution? use dilution formula

What volume of 3 M sulfuric acid must be used to prepare. 5 L of a What volume of 3 M sulfuric acid must be used to prepare .5 L of a .25 M H2SO4 solution?

Types of Solution Reactions Precipitation reactions AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Acid-base reactions NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) Oxidation-reduction reactions Fe2O3(s) + 2Al(s)  2Fe(l) + Al2O3(s)

1.      Which of the following substances would you expect to be insoluble in water? Barium hydroxide Hydrochloric acid Magnesium sulfate Ammonium nitrate Silver chloride Lithium carbonate Calcium carbonate Barium sulfate Ammonium acetate Lead I Chloride Sodium hydroxide Ammonium nitrate Silver nitrate

Solubility Using the solubility rules, predict what will happen when the following pairs of solutions are mixed. a) KOH(aq) & Mg(NO3)2(aq) b) Na2SO4(aq) & Pb(NO3)2(aq) c) KNO3(aq) & BaCl2(aq) Mg(OH)2(s) forms PbSO4(s) forms No precipitate forms

Describing Reactions in Solution 1. Molecular equation (reactants and products as compounds) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) 2. Complete ionic equation (all strong electrolytes shown as ions) Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)  AgCl(s) + Na+(aq) + NO3(aq)

Describing Reactions in Solution (continued) 3. Net ionic equation (show only components that actually react) Ag+(aq) + Cl(aq)  AgCl(s) Na+ and NO3 are spectator ions.

Write the balanced complete ionic and net ionic equations: CuSO4(aq) + BaCl2(aq) → P108 q 28

Sodium Sulfate and Lead II Nitrate

Old limiting problem to compare If 68.5 g of CO(g) is reacted with 8.60 g of H2(g), what is the theoretical yield of methanol that can be produced? __H2(g) + __CO(g) ---> __CH3OH(l)

Precipitation Calculations When aqueous solutions of Na2SO4 & Pb(NO3)2 are mixed. Calculate the mass of the percipitate formed when 1.25 L of 0.0500 M Pb(NO3)2 & 2.00 L of 0.0250 M Na2SO4 are mixed. 1.

Precipitation Calculations Continued (1.25L)(0.0500mol Pb(NO3)2/1L) = 0.0625 molPb(NO3)2 (2.00L)(0.0250mol Na2SO4/1L) = 0.0500 molNa2SO4 4.(0.0625 mol Pb(NO3)2 (1mol Na2SO4/1mol Pb(NO3)2 ) = 0.0625 mol SO42- Na2SO4 is the limiting reactant. 5. (0.0500mol Na2SO4)(1mol PbSO4/1mol Na2SO4) (303.3g/1mol PbSO4) = 15.2 g PbSO4

PRACTICE What mass of precipitate should result when 0.550 L of 0.500 M aluminum nitrate solution is mixed with 0.240 L of 1.50 M sodium hydroxide solution?

answer 9.3g

What is the net ionic equation? What mass of precipitate should result when 0.350 L of 0.200 M aluminum nitrate solution is mixed with 0.540 L of .50 M sodium hydroxide solution? What is the net ionic equation? P108 29-32

What volume of 0.415 M silver nitrate will be required to precipitate as silver bromide all the bromide ion in 35.0 mL of 0.128 M calcium bromide?

answer

How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 25.0 mL of 0.350 M aluminum sulfate?

(93.8 mL barium nitrate) answer

Acid-Base Calculations What volume of a 0.100M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? Cl-, Na+, 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 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 49,a,b 53,54 practice for unit Textbook 181-183 15a 17a 23a 29a-c 35ab 39 try 45b Additional questions 36,a,b 41 40 18,b 12 24

Redox Reactions Redox reactions are reactions in which electrons are transferred. Decomposition and synthesis reactions may be redox. Single replacement reactions are always redox. Double replacement reactions are never redox. Combustion reactions are always redox.

reminder ,Do redox in redox

OIL RIG Oxidation Is Loss. Reduction Is Gain.

Redox Oxidizing agent is the electron acceptor--usually a nonmetal. Reducing agent is the electron donor--usually a metal. CH4(g) + 2O2(g) ----> CO2(g) + 2HOH(g) Carbon is oxidized. Oxygen is reduced. CH4 is the reducing agent. O2 is the oxidizing agent.

Rules for Assigning Oxidation States 1. Oxidation state of an atom in an element = 0 2. Oxidation state of monatomic element = charge 3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1) 4. H = +1 in covalent compounds 5. Fluorine = 1 in compounds 6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

Determining Oxidation States SF6 [NO3]- +6 -6 = 0 +5 -6 = -1 (-1 for each F) (-2 for each O)

Assign the oxidation states to each element in the following. CO2 NO3- H2SO4 Fe2O3 Fe3O4

Oxidizing agent Reducing agent Substance oxidized Substance reduced in the following reactions Fe (s) + O2(g) ® Fe2O3(s) Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g) SO3- + H+ + MnO4- ® SO4- + H2O + Mn+2

All redox reactions can be thought of as happening in two halves. One produces electrons - Oxidation half. The other requires electrons - Reduction half. Write the half reactions for the following. Na + Cl2 ® Na+ + Cl- SO3- + H+ + MnO4- ® SO4- + H2O + Mn+2

Reviewbook 110 50-52

Balancing by Half-Reaction Method 1. Write separate reduction, oxidation reactions. 2. For each half-reaction: - Balance elements (except H, O) - Balance O using H2O - Balance H using H+ - Balance charge using electrons

Balancing by Half-Reaction Method (continued) 3. If necessary, multiply by integer to equalize electron count. 4. Add half-reactions. 5. Check that both elements and charges are balanced.

Balancing By Half-Reaction Acidic Solution H+(aq) + Cr2O72-(aq) + C2H5OH(l) ---> Cr3+(aq) + CO2(g) + HOH(l) Red Cr2O72-(aq) ---> Cr3+(aq) Ox C2H5OH(l) ---> CO2(g) Red 2(6e- + 14H+(aq) + Cr2O72-(aq) ---> 2Cr3+(aq) + 7HOH(l)) Ox C2H5OH(l) + 3HOH(l) ---> 2CO2(g) + 12H+(aq) + 12e- 16H+(aq) + 2Cr2O72-(aq) + C2H5OH(l) ---> 4Cr3+(aq) + 11HOH(l) + 2CO2(g) 12+ = 12+

Half-Reaction Method - Balancing in Base 1. Balance as in acid. 2. Add OH that equals H+ ions (both sides!) 3. Form water by combining H+, OH. 4. Check elements and charges for balance.

Balancing By Half-Reaction Basic Solution Ag(s) + CN-(aq) + O2(g) ---> Ag(CN)2-(aq)(Basic) Ox CN-(aq) + Ag(s) ---> Ag(CN)2-(aq) Red O2(g) ---> Ox 4(2CN-(aq) + Ag(s) ---> Ag(CN)2-(aq) + e-) Red O2(g) + 4H+(aq) + 4e- ---> 2HOH(l) 8CN-(aq) + 4Ag(s) + O2(g) + 4H+(aq) ---> 4Ag(CN)2-(aq) + 2HOH(l)

Balancing By Half-Reaction Basic Solution(Continued) 8CN-(aq) + 4Ag(s) + O2(g) + 4H+(aq) + 4OH-(aq) ---> 4Ag(CN)2-(aq) + 2HOH(l) + 4OH-(aq) 8CN-(aq) + 4Ag(s) + O2(g) + 4HOH(l) ---> 4Ag(CN)2-(aq) + 2HOH(l) + 4OH-(aq) 8CN-(aq) + 4Ag(s) + O2(g) + 2HOH(l) ---> 4Ag(CN)2-(aq) + 4OH-(aq) 8- = 8-