General Chemistry 2: Precipitation, Acid-Base Reactions - PDF

GENERAL CHEMISTRY 2  POWER COMPETENCY Apply the properties of solutions, solubility, and the stoichiometry of reactions in solutions in calculating stoichiometric calculations for reactions in solution and to apply the energy changes in chemical reaction in calculating enthalpy change. ...

GENERAL CHEMISTRY 2  POWER COMPETENCY Apply the properties of solutions, solubility, and the stoichiometry of reactions in solutions in calculating stoichiometric calculations for reactions in solution and to apply the energy changes in chemical reaction in calculating enthalpy change.  LEARNING TARGETS I can perform stoichiometric calculations for reactions in solution, STEM_GC11PP-IIId- f-112. Lesson 5.1 Precipitation Reactions General Chemistry 2 Science, Technology, Engineering, and Mathematics Have you ever wondered how X-ray images of the stomach are acquired? 5 What is a precipitation reaction? 6  Precipitation Reactions Precipitation Reaction a reaction in which an insoluble ionic compound is formed ○ ionic compounds: when a cation and an anion react with each other and form a chemical bond the insoluble product → precipitate 7 What is solubility? 8  Solubility the ability to dissolve in a given amount of solvent at a certain temperature the solubility values (s) of ionic compounds: quantitative values 9  Solubility e.g., NaCl → s = 35.89 g per 100 mL of water at 15℃ ○ for every 100 mL of water at 15℃, 35.89 g of NaCl will dissolve in it. e.g., BaSO4 → s = 0.0002448 g per 100 mL of water at the same temperature The simpler way is a qualitative assessment of ionic compounds through the use of solubility rules. 10  Solubility Rules Mixing an aqueous solution of NaCl and AgNO3 produces NaNO3 and AgCl, following a double displacement type of reaction. 11 Remember In a double displacement reaction, the two reactants exchange ions to form two new compounds. It takes the form: 12 Remember Take note that after the reaction, cation A is now bonded to anion Y while cation B is now bonded to anion X. You can think of the compounds in double displacement reactions as “switching partners”. This reaction will only occur if an insoluble precipitate or water is formed. 13 How can we predict if a precipitate will form from a given set of reactants? 14  Solubility Rules Soluble Compounds Exceptions alkali metals (Group 1), ammonium none ion (NH4+), nitrate, perchlorate, acetate chlorides, bromides, and iodides those of Pb2+, Ag+, and Hg22+ (which are insoluble) 15  Solubility Rules Soluble Compounds Exceptions sulfates those of Sr2+, Ba2+, Pb2+, and Hg22+ (which are insoluble) phosphates those of Group 1 metals and NH4+ (which are soluble) 16  Solubility Rules Insoluble Compounds Exceptions carbonates, phosphates those of Group 1 metals and NH4+, which are soluble sulfides those of Group 1 metals, Group 2 metals, and NH4+ (which are soluble) 17  Solubility Rules Insoluble Compounds Exceptions hydroxides those of Group 1 metals and NH4+ (which are soluble) those of Ca2+, Sr2+, and Ba2+ (which are slightly soluble) 18  Solubility Rules NaNO3: soluble compound ○ all nitrates are soluble → NaNO3(aq) AgCl: insoluble compound ○ Ag+ is an exception for soluble chloride compounds → AgCl(s) 19 Hard water is water that has high concentrations of Ca2+ and Mg2+ ions. How does washing soda (Na2CO3) remove the hardness of water? 20  Chemical Equations There are three types of writing chemical equations: ○ molecular equation ○ total ionic equation ○ net ionic equation 21  Molecular Equation An equation wherein all reactants and products are written as intact compounds 22  Total Ionic Equation All soluble ionic compounds are written as aqueous ions. Insoluble ionic compounds and molecular compounds are written as if they are intact molecules. 23  Total Ionic Equation Both 2K+(aq) and SO42-(aq) are present on both sides of the equation. ○ spectator ions - ions in the solution that do not participate in the formation of the precipitate 24 Why are ions that are present on both sides of the equation called spectator ions? 25 Net Ionic Equations The spectator ions in the total ionic equation are canceled out. 26  Let’s Practice! Predict the products and write the balanced molecular equation of the reaction between aqueous Mg(NO3)2 and aqueous NaOH. 27  Let’s Practice! Predict the products and write the balanced molecular equation of the reaction between aqueous Mg(NO3)2 and aqueous NaOH. The balanced chemical equation is shown below: 28  Let’s Practice! Write the total ionic equation of the reaction between aqueous Mg(NO3)2 and aqueous NaOH. 29  Let’s Practice! Write the total ionic equation of the reaction between aqueous Mg(NO3)2 and aqueous NaOH. The total ionic equation is shown below: 30  Let’s Practice! Write the net ionic equation of the reaction between aqueous Mg(NO3)2 and aqueous NaOH. 31  Let’s Practice! Write the net ionic equation of the reaction between aqueous Mg(NO3)2 and aqueous NaOH. The net ionic equation is shown below: 32 Try It! Write the net ionic equation of the reaction between aqueous Ba(CH3COO)2 and aqueous CaSO4. 33  Real-Life Applications of Precipitation Reactions BaSO4 as a contrast agent in X-ray and computed tomography scans. ○ nontoxic ○ coats the linings of the esophagus, stomach, and intestines 34  Real-Life Applications of Precipitation Reactions In photography, the off- white AgBr precipitate is used to develop a photographic image. AgBr grains are usually suspended in gelatin to create an emulsion. 35  Real-Life Applications of Precipitation Reactions When exposed to light, these grains undergo photolytic decomposition and become metallic silver. ○ allows AgBr to preserve a photographic image in black and white 36  Real-Life Applications of Precipitation Reactions washing soda or Na2CO3 as a water softener in many households ○ hard water = abundant in calcium and magnesium ions. 37  Real-Life Applications of Precipitation Reactions adding Na2CO3 to hard water ○ insoluble precipitates of CaCO3 and MgCO3 are formed ○ water softener 38  Check Your Understanding Determine whether the following compounds are soluble or insoluble. 1. AgNO3 2. PbS 3. Cu(OH)2 4. Mg(CH3COO)2 5. Al2(SO4)3 39  Check Your Understanding Write true if the following reactions are precipitation reactions. Otherwise, write false. 1. 40 Lesson 5.2 Quantitative Aspects of Precipitation Reactions General Chemistry 2 Science, Technology, Engineering, and Mathematics 42 How do we calculate the theoretical yield of a precipitation reaction? 43  Stoichiometry of Precipitation Reactions Precipitation Reactions double displacement reaction forms an insoluble product called the precipitate 44  Key Formula Concept Formula Description Use this formula to Moles solve for moles if molarity and volume where are given. n is number of moles (mol); M is molarity (M), and V is volume in liters (L). 45  Key Formula Concept Formula Description Use this formula Theoretical to solve for yield theoretical where yield provided mppt is the theoretical yield or mass of that the precipitate (g); balanced n, is the number of moles of limiting equation and reactant (mol); the limiting FWppt is the formula weight or molar mass of reactant have the precipitate (g/mol), and been stoichiometric ratio is the ratio of product to established. reactant in the balanced equation. 46  Key Formula Concept Formula Description Use this formula to Amount of solve for the Excess amount of excess Reactant where reactant that was Consumed nXR,consumed is the number of moles of consumed in the excess reactant that got consumed in the reaction provided reaction (mol); that the balanced nLR is the number of moles of limiting equation and the reactant (mol), and limiting reactant stoichiometric ratio is the ratio of have been reactant to reactant in the balanced established. equation. 47  Stoichiometry of Precipitation Reactions Suppose we mix 25.0 mL of 0.50 M AgNO3 and 10.0 mL of 1.0 M NaCl. Step 1: Write the Balanced Molecular Equation Step 2: Determine the Limiting Reactant Step 3: Calculate the Theoretical Yield Step 4: Calculate Amount of Excess Reactant Consumed 48 Remember The stoichiometric ratio is the mole ratio between the reactants and products in a given reaction. This is determined using the coefficients of the reactants and products in the balanced equation. 49 Remember The limiting reactant is the reactant that gets consumed first in a chemical reaction. Hence, it limits the amount of product formed. 50 Can two reactants be both limiting reactants? Why or why not? 51 Try It! Calculate the theoretical yield of the precipitate formed when 40.5 mL of 0.134 M Na2S is made to react with 27.8 mL of 0.163 M AgNO3. 52  Let’s Practice! Determine the amount of excess reactant remaining after 25.0 mL of 0.50 M AgNO3 is made to react with 10.0 mL of 1.0 M NaCl. 53  Let’s Practice! Determine the amount of excess reactant remaining after 25.0 mL of 0.50 M AgNO3 is made to react with 10.0 mL of 1.0 M NaCl. 0.0025 mol of AgNO3 is left after reaction. 54  Check Your Understanding Determine the stoichiometric ratio of the compounds asked in the given balanced chemical equations. 1. BaI2 and AgI 1. NaOH and Cu(OH)2 55  Check Your Understanding Determine the stoichiometric ratio of the compounds asked in the given balanced chemical equations. 3. KI and PbI2 56  Check Your Understanding Using the chemical equations in Part A, identify the limiting reactant(s) given the following starting amounts. 1. 5.0 mL of 0.10 M BaI2 and 5.0 mL of 0.10 M Ag2SO4 2. 10.0 mL of 0.10 M CuSO4 and 5.0 mL of 0.20 M NaOH 3. 25.0 mL of 1.5 M Pb(CH3COO)2 and 1.0 mL of 6.0 M KI 57 Challenge Yourself In calculating the theoretical yield, why is the stoichiometric ratio of reactant to product used, but in calculating the amount of excess reactant consumed by the reaction, the stoichiometric ratio of reactant to reactant is used? 58 Lesson 5.3 Acid-Base Reactions General Chemistry 2 Science, Technology, Engineering, and Mathematics The stomach is an important part of the human digestive system. It is a hollow, muscular organ located in the upper abdomen. 60 What is an acid and what is a base? 61  Acids Acid originates from the Latin word acidus→ sour can be identified by their sour taste defined as a substance that produces hydrogen ions (H+) in aqueous solutions 62  Acids Hydrogen chloride When it comes into contact with water vapor, forms aqueous hydrochloric acid that completely ionizes into H+ and Cl- ions. 63 Remember The H+ ion is actually a representation of the H3O+ ion, which is a hydrated proton. It is formed from the transfer of a proton from the acid to the H2O molecule. 64  Acids Ionizable Hydrogen Atoms typically written first in the molecular formula For example, the first H in acetic acid (HC2H3O2) is an ionizable hydrogen while the other three hydrogen atoms are not. 65  Acids Strong Acids completely ionizes into its component ions in aqueous solutions seven common strong acids HCl HNO3 HClO4 HBr HClO3 H2SO4* HI *H2SO4 ionizes in two distinct steps since it has two acidic hydrogens. It is only a strong acid in the first ionization. 66  Acids Weak Acids do not completely ionize into its component ions in aqueous solutions ionization does not go into completion → reversible a double harpoon (⇌) is used to indicate its reversibility 67  Bases Bases can be identified by their bitter taste, pungent odor, and slippery texture a substance that produces hydroxide ions (OH-) in aqueous solutions Example: sodium hydroxide (NaOH) 68  Bases Bases typically have a metal cation and an OH- anion Example: NaOH and Mg(OH) Other bases have no OH- in its formula but can form OH- in water. ○ Example: NH3, NaHCO3, Na2CO3 69  Bases Strong Bases completely ionizes into its component ions in aqueous solutions LiOH RbOH Ca(OH)2 NaOH CsOH Sr(OH)2 KOH Ba(OH)2 70  Bases Weak Bases do not completely ionize into its component ions in aqueous solutions ionization does not go into completion →reversible 71 What is a neutralization reaction? 72  Balancing Neutralization Reactions Neutralization Reactions can be classified as a double displacement reaction 73  Molecular Equations Example: ○ HCl → acid ○ NaOH → base ○ NaCl → salt 74 When a strong acid completely reacts with a strong base, will the resulting salt solution be acidic or basic? Explain. 75  Total Ionic Equations Strong acids and bases are written as aqueous ions. Weak acids and bases are written as molecular compounds. H2O (l) is written as is because it is an intact molecule. 76  Total Ionic Equations In the neutralization reaction of HCl(aq) and NaOH(aq), we acquire the following total ionic equation: 77  Net Ionic Equations Spectator ions are removed in writing the net ionic equation. 78  Net Ionic Equations shows the essential nature of neutralization reactions strong acid + strong base: ○ ○ Salt solutions produced are neutral — they are neither acidic nor basic. 79  Let’s Practice! Write the balanced molecular equation of the reaction between aqueous HC2H3O2 and aqueous Ba(OH)2. 80  Let’s Practice! Write the balanced molecular equation of the reaction between aqueous HC2H3O2 and aqueous Ba(OH)2. 81 Try It! Write the balanced molecular equation of the reaction between aqueous HNO3 and aqueous Ca(OH)2. 82  Let’s Practice! Write the total ionic equation of the reaction between aqueous HC2H3O2 and aqueous Ba(OH)2. 83  Let’s Practice! Write the total ionic equation of the reaction between aqueous HC2H3O2 and aqueous Ba(OH)2. 84 Try It! Write the total ionic equation of the reaction between aqueous HNO3 and aqueous Ca(OH)2. 85  Let’s Practice! Write the net ionic equation of the reaction between aqueous HC2H3O2 and aqueous Ba(OH)2. 86  Let’s Practice! Write the net ionic equation of the reaction between aqueous HC2H3O2 and aqueous Ba(OH)2. 87 Try It! Write the net ionic equation of the reaction between aqueous HNO3 and aqueous Ca(OH)2. 88  Real-Life Applications of Neutralization Reactions Antacids basic compounds that have the ability to neutralize stomach acid Mg(OH)2, Al(OH)3, CaCO3, NaHCO3 89  Real-Life Applications of Neutralization Reactions Lime Fertilizers added to soils to neutralize the acidity caused by the decay of organic matter CaCO3, CaO, Ca(OH)2 90  Real-Life Applications of Neutralization Reactions Slaked Lime Ca(OH)2 used to treat wastewater by neutralizing the acidity of industrial sludges increases the basicity of wastewater→prevent corrosion of pipes 91  Check Your Understanding Determine whether each of the following compounds is an acid or a base. 1. HNO3 2. H2S 3. Cu(OH)2 4. Mg(OH) 5. H2SO4 92  Check Your Understanding Write true if the following reactions are neutralization reactions. Otherwise, write false. 93 Challenge Yourself Is the reaction between H2SO4 and NH3 considered a neutralization reaction? Explain your answer. 94 Lesson 5.4 Quantitative Aspects of Acid-Base Reactions General Chemistry 2 Science, Technology, Engineering, and Mathematics Antacids are basic compounds used to treat heartburns and gastric hyperacidity, which are caused by excessive amounts of HCl in the stomach. Its common active ingredients include CaCO3, NaHCO3, Al(OH)3, and Mg(OH)2. 96 What is titration? 97  Titration Titration a technique for the determination of concentration of solutions done by reacting a solution with known concentration (called the titrant) to another solution with unknown concentration (called the analyte) 98  Titration Titration Setup titrant → burette analyte→ Erlenmeyer flask below the burette 99  Titration Titration carried out by adding the titrant slowly to the analyte until it reaches the equivalence point Equivalence point the condition in which both reactants have reacted completely with each other no reactant is in excess visualized using indicators 100  Titration Indicator in acid-base titrations change color when the titration reaches the endpoint a slight excess of the titrant must be present for an indicator to change color 101 For titrations with phenolphthalein as the indicator, why is the titration stopped when the color is faint pink and not when it is deep purple? 102  Stoichiometry of Acid-Base Titrations Step 1: Write the Balanced Molecular Equation For example, we want to determine the acetic acid content of a vinegar sample. Our analyte is acetic acid and our titrant is sodium hydroxide. 103 Remember The stoichiometric ratio is determined using the coefficients of the substances in the balanced equation. It may be between the two reactants or between a reactant and a product. In the case of acid-base titrations, we are more interested with the reactant to reactant stoichiometric ratio. 104  Stoichiometry of Acid-Base Titrations Step 2: Calculate the Moles of Titrant Let us say that a 5.00 mL sample of vinegar requires 37.5 mL of 0.100 M NaOH to reach the equivalence point. 105  Stoichiometry of Acid-Base Titrations Step 3: Calculate the Concentration of Analyte Let us say that a 5.00 mL sample of vinegar requires 37.5 mL of 0.100 M NaOH to reach the equivalence point. 106  Stoichiometry of Acid-Base Titrations Step 3: Calculate the Concentration of Analyte From this, we can calculate the molarity of HC2H3O2 to be: 107  Let’s Practice! A 25.0 mL sample of HCl requires 28.4 mL of 0.105 M NaOH to reach the equivalence point. What is the molarity of HCl in the sample? 108  Let’s Practice! A 25.0 mL sample of HCl requires 28.4 mL of 0.105 M NaOH to reach the equivalence point. What is the molarity of HCl in the sample? The molarity of HCl in the sample is 0.119M. 109 Try It! A 50.0 mL sample of H2SO4 requires 37.8 mL of 0.587 M NaOH to reach the equivalence point. What is the molarity of H2SO4 in the sample? Assume that the two hydrogens of H2SO4 are ionized completely. 110  Let’s Practice! What volume (in mL) of 2.105 M KOH is needed to titrate 100.0 mL of 0.3245 M HNO3? 111  Let’s Practice! What volume (in mL) of 2.105 M KOH is needed to titrate 100.0 mL of 0.3245 M HNO3? The volume of KOH to reach equivalence point is 15.42 mL. 112 Try It! What volume (in mL) of 2.345 M KOH is needed to titrate 250.0 mL of 0.1234 M H2SO4? Assume that the two hydrogens of H,SO4 are ionized completely. 113  Let’s Practice! A 0.500 g sample of KHP was dissolved in water and titrated with 24.3 mL of 0.100 M NaOH. What is the purity of KHP in the sample? KHP reacts with NaOH in a 1:1 stoichiometric ratio and has a formula weight of 204.22 g/mol. 114  Let’s Practice! A 0.500 g sample of KHP was dissolved in water and titrated with 24.3 mL of 0.100 M NaOH. What is the purity of KHP in the sample? KHP reacts with NaOH in a 1:1 stoichiometric ratio and has a formula weight of 204.22 g/mol. The percent purity of KHP in the sample is 99.2%. 115 Try It! A 1.30 g sample of an impure solid containing NaHCO3 was titrated with 37.5 mL of 0.396 M HCl. What is the purity of NaHCO3 in the sample? The formula weight of NaHCO3 is 84.0 g/mol. 116 Tips For a hypothetical equation: where a and t are the stoichiometric coefficient of the analyte and titrant, respectively, A is the analyte, and T is the titrant. The relationship between the moles of analyte and titrant, given the concentration (molarity) and volume (liters), can be obtained using the equation: 117  Check Your Understanding Identify the terms described in each of the following items. 1. It is the term used to describe the condition wherein an indicator changes color. 2. It is the term used to describe the condition wherein both reactants have completely neutralized one another. 3. It is the term used to refer to the careful and controlled addition of a titrant to an analyte. 118  Check Your Understanding Given the following acids and bases, determine their stoichiometric ratio when they undergo a neutralization reaction with each other. 1. NaOH and HClO4 1. Ba(OH)2 and HNO3 1. RbOH and H2SO4 119  Key Formula Concept Formula Description Moles Use this formula to where solve for moles if n is number of moles molarity and volume (mol); are given. M is molarity (M), and V is the volume in liters (L). 120  Key Formula Concept Formula Description Molarity Use this formula to solve for molarity if the number of where moles and volume M is molarity (M); are given. n is number of moles (mol), and V is the volume in liters (L). 121  Key Formula Concept Formula Description Volume Use this formula to solve for volume if the number of where moles and molarity V is the volume in liters (L); are given. n is number of moles (mol), and M is molarity (M). 122 Challenge Yourself A 1.30 g tablet of antacid claims to contain 38.5% (w/w) CaCO3. The tablet required 19.4 mL of 0.514 M HCl to complete its neutralization. Confirm or refute the claim. Show your calculations. The formula weight of CaCO3 is 100.1 g/mol. 123