Chapter 4: Types of Chemical Reactions and Solution Stoichiometry Mr. Redmond

Aqueous Solutions dissolved in water Properties: Electrical Conductivity Strong Electrolytes- conducts well Weak Electrolytes- conducts little Nonelectrolytes- does not conduct

Strong Electrolyte Substances that completely ionize when dissolved in water Types: Soluble Salts Strong Acids Strong Bases

Soluble Salts NaCl dissolved in water Practically all of the sample becomes Na+ and Cl- ions

Acids (Sour) Arrhenius states: an acid is a substance that produces H+ ions (protons) when it is dissolved in water HCl, HNO3, and H2SO4 all virtually ionize when placed in water, therefore are considered strong acids.

Bases (Bitter) When dissolved in water it produces OH-, hydroxide. Strong bases: NaOH, KOH

Weak Electrolytes When placed in water they produce few ions. Weak Acid Weak Base

Weak Acid Acetic Acid, dissociates only to a slight extent in an aqueous solution. Only about 1% of its molecules ionize. HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

Weak Base Ammonia solution is basic because it produces some OH- ions NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Nonelectrolytes Substances that dissolve but do not dissolve into ions that can conduct electricity Ethanol, C2H5OH and sucrose, C12H22O11

Molarity Molarity= moles of solute per volume of solution in liters M = Molarity = (moles of solute)/(liters of solution) Solute- a substance dissolved in a liquid Solution- a homogeneous mixture

Sample Ex. 4.2 You dissolve 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution. Find the molarity.

Sample Ex. 4.2 cont. First we calculate the number of moles of HCl...... Change volume from mL to L....... Now divide and find Molarity.......

Sample Ex. 4.4 Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2 Formula: ZnCl2(s) Zn2+(aq) + 2Cl-

Sample Ex. 4.4 cont. You have 1.0 x 10-3 M ZnCl2 After it dissociates you have 1.0 x 10-3 M Zn2+ and 2.0 x 10-3 M Cl- Use the known information to calculate moles of Cl-

Sample Ex. 4.5 Typical blood serum is about 0.14 M NaCl. What volume of blood serum contains 1.0 mg NaCl?

Sample Ex. 4.5 cont. How many moles of NaCl are in 1 mg NaCl? What volume of 0.14 M NaCl solution contains the number of moles of NaCl in 1 mg NaCl?

Sample Ex. 4.6 To analyze the alcohol content of a certain wine, a chemist needs 1.0 L of 0.200 M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be weighed out to make this solution?

Sample Ex. 4.6 cont. Determine the moles of potassium dichromate required. Convert moles into grams

Dilutions Most times when you purchase chemicals for a lab they come in a concentrated form call stock solutions. Water is added to these to give the desired molarity of the particular solution

Dilution Important to understand that: moles of solute after dilution = moles of solute before dilution For example: if we need 500 mL of 1.00 M acetic acid from a 17.4 M stock solution of acetic acid, we would need to dilute it.

Dilution First, determine the number of moles of acetic acid in the final solution...... We then use the volume of 17.4 M acetic acid that contains the calculated amount of moles We then use this information to solve for the volume

Sample Ex. 4.7 What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution?

Sample Ex. 4.7 Determine moles of H2SO4 in 1.5 L of a 0.10 M H2SO4 Find the volume of 16 M H2SO4 that contains 0.15 mol H2SO4

Dilution The equation: M1V1 = M2V2 , may help you remember that the moles before dilution is equal to the moles after dilution Repeat the last example using this equation

Dilution M1V1 = M2V2 M1=16M, M2= 0.10M, V2=1.5 L Solve for V1

Types of Reactions Precipitation reactions Acid-base reaction Oxidation-reduction reaction

Precipitation Reactions Sometimes when two solutions are mixed an insoluble substance is formed, this is called a precipitation reaction The solid substance that separates from the solution is called the precipitate

Precipitation Reactions Say we perform a precipitation reaction with an aqueous solution of potassium chromate K2CrO4(aq), which is yellow, and mix it with a colorless aqueous solution containing barium nitrate Ba2(NO3)2(aq) When these are mixed a yellow precipitate forms

Precipitation Reactions What is the yellow solid forming?

Precipitation Reactions Remember in virtually every case, when a solid containing ions dissolves in water, the ions separate From this we know that we have K+, CrO42-, Ba2+, and NO3-

Precipitation Reactions The reaction is: K2CrO4(aq) + Ba(NO3)2(aq) product It may be easier to see what is happening if we write the equation a little different 2K+(aq)+CrO42-(aq)+Ba2+(aq)+2NO3-(aq) product What are the different ways these ions could combined?

Precipitation Reactions Here are our choices: K2CrO4, KNO3, BaCrO4, or Ba(NO3)2 We can quickly exclude K2CrO4 and Ba(NO3)2 since they are the reactants It takes a little background knowledge to figure the rest out

Precipitation Reactions K+ ion and the NO3- ion are both colorless, thus KNO3 would precipitate white. CrO42- ion is yellow as noted earlier thus the yellow solid must be BaCrO4 But what happened to K+ and NO3-?

Precipitation Reactions The other ions are left dissolved in the solution. If we removed our precipitate and then dissolved the water we would find KNO3(s) left in the beaker

Precipitation Reactions

Sample Ex. 4.8 Use the table just given to solve the problems. Predict what will happen when you mix the following pairs. KNO3(aq) and BaCl2(aq) Na2SO4(aq) and Pb(NO3)2(aq) KOH(aq) and Fe(NO3)3(aq)

Sample Ex. 4.8 Split up the ions and pair them. The table show that KCl and Ba(NO3)2 are soluble in water, thus no precipitate forms. NaNO3 is soluble but PbSO4 is not. K+ and NO3- salts are soluble but Fe(OH)3 is only slightly soluble, thus it will precipitate

Describing Reactions in Solutions formula equation complete ionic equation net ionic equation

Formula Equation The typical equation we normally write K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq)

Complete Ionic Equation In a complete ionic equation, all the substances that are strong electrolytes are represented as ions 2K+(aq)+CrO42-(aq)+Ba2+(aq)+2NO3-(aq) BaCrO4(s) + 2K+ + 2NO3-

Net Ionic Equation A net ionic equation only includes those solution components directly involved in the reaction. We do not write the spectator ions (K+ and NO3-) Ba2+(aq) + CrO42-(aq) BaCrO4(s)

Sample Ex. 4.9 For the following reaction, write the formula equation, the complete ionic equation, and the net ionic equation. Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate

Stoichiometry of Precipitation Reaction Just like in Chapter 3, we first convert all quantities into moles Second, we form the proper molar ratio using the coefficients of the balanced equation Use the limiting reactant when necessary

Stoichiometry of Precipitation Reaction Special attention needs to be given to what reaction actually occurs. Use the complete ionic formula and your background knowledge of the reactants to determine the reaction We must use the molarity and volume to determine the moles of reactant

Sample Ex. 4.10 Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.

Sample Ex. 4.10 Write the complete ionic equation Refer to the solubility table Determine what the spectator ions are and what compound is the precipitate Convert molarity and volume to moles Use the proper molar ratio then convert to grams

Acid-Base Reactions Arehenius’ concept of acids and bases Acids produce H+ ions when dissolved in water Bases produce OH- ions when dissolved in water Not all acids or bases have H+ or OH- so a more general definition is needed

Acid-Base Reactions Bronsted-Lowery Model Acids are proton donors Bases are proton acceptors How do we predict acid-base reactions?

Acid-Base Reactions Take the example of mixing solutions of HCl and NaOH. We know that HCl is a strong acid and NaOH is a strong base. Write the ions that are present and look at the solubility table, will NaCl precipitate? Because water is a nonelectrolyte, large quantities of H+ and OH- ions cannot coexist in the solution. What happens?

Acid-Base Reactions They will react to for H2O. H+(aq) + OH-(aq) H2O(l) This is the net ionic equation for the reaction that occurs when aqueous HCl and NaOH are mixed.

Acid-Base Reactions What if we now mix solutions of HC2H3O2 and KOH, acetic acid is a weak acid while KOH is a strong base. What ions are available? OH- is a proton acceptor and has such a great affinity for protons that it can strip them from the acetic acid

Acid-Base Reactions The net ionic equation for the reaction is OH-(aq) + HC2H3O2(aq) H2O(l) + C2H3O2-(aq) What do we learn from this? The hydroxide ion is such a strong base that for purposes of stoichiometric calculations it can be assumed to react completely with any weak acid that we will encounter in this course.

Acid-Base Reactions List the ions present and decide what reaction will occur Write the balance net ionic equation Calculate moles of reactant Determine limiting reactant Calculate required moles of reactant or product Convert to grams or volume as required

Sample Ex. 4.13 In a certain experiment, 28.0 mL of 0.250 M HNO3, and 53.0 mL of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH- ions in excess after the reaction goes to completion?

Sample Ex. 4.13 What ions are available to react? Check solubility table and write net ionic equation Compute the amount of H+ and OH- present What is the molar ratio between H+ and OH-?

Sample Ex. 4.13 What is the limiting reactant? What amount of OH- or H+ are in excess? What is the total volume of the solution after mixing the two solutions? Use the new information to calculate the concentration.

Acid-Base Titrations Volumetric analysis is a technique for determining the amount of a certain substance by doing a titration. A titration involves delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance to be analyzed (the analyte)

Acid-Base Titrations When you have added just enough titrant to react exactly with the analyte you have reached the equivalence point, also called the stoichiometric point This point is often marked by an indicator, a substance that is added at the beginning of the titration that changes color at or very near the equivalence point

Acid-Base Titrations The point where the indicator actually changes color is called the end point of the titration The goal is to choose an indicator such that the end point is exactly at the equivalence point

Acid-Base Titrations To have a successful titration you need to meet three requirements: The exact reaction between titrant and analyte must be known and rapid The equivalence point must be marked accurately The volume of titrant required to reach the equivalence point must be known exactly

Acid-Base Titrations When the analyte is base or an acid, the required titrant is a strong acid or base respectively. We call this an Acid-Base Titration. Phenolphthalein is a common indicator for acid-base titration which is clear in an acidic solution but pink in a basic solution

Acid-Base Titrations The end point here occurs one drop of base beyond the equivalence point

Sample Ex. 4.14 A student carries out an experiment to standardize (determine the exact concentration of) a sodium hydroxide solution. To do this, the student weighs out a 1.3009 g sample of potassium hydrogen phthalate (KHC8H4O4, or KHP) which has a molar mass of 204.22 g/mol. KHP has one acidic hydrogen. KHP is dissolved in distilled water and phenolphthalein is added as an indicator.

Sample Ex. 4.14 He then titrates the solution with sodium hydroxide to phenolphthalein’s end point. The difference between the final and initial buret readings indicates 41.20 mL of sodium hydroxide solution is required to react exactly with 1.3009 g KHP. Calculate the concentration of the sodium hydroxide solution.

Sample Ex. 4.14 What ions are initially present Now consider a solution with those ions, what will the OH- do to the HC8H4O4-? Write the net ionic equation and determine the molar ratio. Determine moles of KHP and use the molar ratio Calculate Molarity of NaOH

Sample Ex. 4.15 An environmental chemist analyzed the effluent from an industrial process know to produce the compounds carbon tetrachloride and benzoic acid (HC7H5O2), a weak acid that has one acidic hydrogen atom per molecule. A sample of this effluent weighing 0.3518 g was shaken with water, and the resulting aqueous solution required 10.59 mL of 0.1546 M NaOH for neutralization. Calculate the mass percent of HC7H5O2 in the original sample.

Sample Ex. 4.15 What ions are present? Write the net ionic equation What happens when you have a weak acid in a solution with OH-? What is the molar ratio? Find moles of OH- From the molar ratio calculate the mass of HC7H5O2 then calculate mass percent

Oxidation-Reduction Reaction Also called redox reactions, occurs when one or more electrons are transfered Examples, reaction to produce NaCl, photosynthesis, combustion reaction, etc.

Redox Reaction Oxidation state (oxidation number)- provides a way to keep track of electrons in redox rxns, particularly those containing covalent substances. Oxidation state of atoms in covalent compounds are obtained by arbitrarily assigning the electrons which are normally shared to an atom

Assigning Electrons If the two atoms are identical then the electrons are shared equally If two different atoms are involved, the electrons are assigned to the atom that has a stronger attraction to electrons, for example H2O Note that the oxidation state is an imaginary charge

Rules for Oxidation State

Sample Ex. 4.16 Assign oxidation states to all atoms in the following CO2 SF6 NO3-

Non-Integer Oxidation States Apply our rules to Fe3O4 We first assign O a -2 So the three iron atoms must give us +8 We find it to be +8/3 Although not common, we need to realize that it is possible to have non-integer oxidation states

Characteristics of Redox Rxns Redox rxns are characterized by the transfer of electrons Sometimes it is less obvious though, CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Note the oxidation state of O2 is 0 because it is in elemental form. Notice that Carbon goes from -4 in CH4 to +4 in CO2, this is accounted for by the loss of 8 e-

Characteristics of Redox Rxns Also, each O changes from oxidation state of 0 to -2 in H2O and CO2 We can show this by using the equations CH4 CO2 + 8e- 2O2 + 8e- CO2 + 2H2O

Characteristics of Redox Rxns Oxidation is an increase in oxidation state (the loss of an electron) Reduction is a decrease in oxidation state (the gain of an electron) A helpful reminder LEO GER

Characteristics of Redox Rxns Looking at the example 2Na(s) + Cl2(g) 2NaCl(s) We see that sodium is oxidized while chloride is reduced We say that Cl2 is the oxidizing agent (electron acceptor) and Na is the reducing agent (electron donor)

Sample Ex. 4.17 When powdered aluminum metal is mixed with pulverized iodine crystals and a drop of water is added to help the reaction get started, the resulting reaction produces a great deal of energy. The mixture bursts into flames, and a purple smoke of I2 vapor is produced from the excess iodine. The reaction is: 2Al(s) + 3I2(s)-->2AlI3(s)

Sample Ex. 4.17 Identify which atoms are oxidized and reduced and what are the oxidizing and reducing agents?

Sample Ex. 4.18 Metallurgy, the process of producing a metal from its ore, always involves oxidation-reduction reactions. In the metallurgy of galena (PbS), the principal lead containing ore, the first step is the conversions of the lead sulfide to its oxide: 2PbS(s) + 3O2-->2PbO(s) + 2SO2(g)

Sample Ex. 4.18 The oxide is then treated with carbon monoxide to produce the free metal: PbO(s) + CO(g)-->Pb(s) + CO2(g) For each reaction identify the atoms that are oxidized and reduced, and specify the oxidizing and reducing agents.

Balancing Oxidation-Reduction Rxns Sometimes it is difficult to balance redox rxns in aqueous solutions. We can use a method called the half reaction method to make balancing easier

Half Reactions We separate the reaction into two half reactions One involves oxidation while the other involves reduction

Half Reactions Example: Ce4+(aq) + Sn2+(aq)--> Ce3+(aq) + Sn4+(aq) We separate the unbalanced reaction into two half reactions: Ce4+(aq)--> Ce3+ (reduced) Sn2+(aq)--> Sn4+ (oxidized)

Half Reactions You will generally write the two half reactions, balance them, and then add them together. However, the procedure differs slightly depending on whether you are in an acidic or basic solution

Half Reactions in Acidic Solutions Write the two half reactions For each half reaction: Balance all elements except hydrogen and oxygen Balance Oxygen using H2O Balance Hydrogen using H+ Balance charges using electrons

Half Reactions in Acidic Solutions If needed, multiply one or both balanced half reactions by an integer to equalize the number of electrons transferred in the two half reactions Add the half reactions and cancel identical species Check that atoms and charges are balanced

Half Reactions in Acidic Solutions Follow the steps for balancing the reaction MnO4-(aq)+Fe2+--acid-->Fe3+(aq)+Mn2+(aq)

Sample Ex. 4.19 Potassium dichromate (K2Cr2O7) is a bright orange compound that can be reduced to a blue-violet solution of Cr3+ ions. Under certain conditions, K2Cr2O7 reacts with ethyl alcohol (C5H5OH) as follows: H+(aq)+Cr2O72-(aq)+C2H5OH(l)-->Cr3+(aq)+CO2(g)+H2O(l) Use half reaction method to balance the equation.

Half Reactions in Basic Solutions Follow the same steps as an acidic solution to find a balanced equation Add OH- to both sides to cancel out any H+ ions Form H2O with the H+ and OH- and eliminate any H2O appearing on both sides of the equation Check that elements and charges balance

Sample Ex. 4.20 Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ore. An aqueous solution containing cyanide ions is often used to extract the silver using the following reaction that occurs in a basic solution: Ag(s)+CN-(aq)+O2--basic-->Ag(CN)2-(aq)