9/24/12. Chemistry Second Edition Julia Burdge. Reactions in Aqueous Solutions

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1 Chemistry Second Edition Julia Burdge 4 Reactions in Aqueous Solutions Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 4 Reactions in Aqueous Solutions 4.1 General Properties of Aqueous Solutions Electrolytes and Nonelectrolytes Strong Electrolytes and Weak Electrolytes 4.2 Solubility Guidelines for Ionic Compounds in Water Molecular Equations Ionic Equations Net Ionic Equations 4.3 Acid-Base Reactions Strong Acids and Bases Brønsted Acids and Bases Acid-Base Neutralization 4.4 Oxidation Numbers Oxidation of Metals in Aqueous Solutions Balancing Simple Redox Equations Other Types of Redox Reactions 1

2 4 Reactions in Aqueous Solutions 4.5 Molarity Dilution Solution Stoichiometry 4.6 Aqueous Reactions and Chemical Analysis Gravimetric Analysis Acid-Base Titrations 4.1 General Properties of Aqueous Solutions A solution is a homogenous mixture of two or more substances. The substance present in the largest amount (moles) is referred to as the solvent. The other substances present are called the solutes. A substance that dissolves in a particular solvent is said to be soluble in that solvent. General Properties of Aqueous Solutions An electrolyte is a substance that dissolves in water to yield a solution that conducts electricity. NaCl(s) H 2 O Na + (aq) + Cl (aq) An electrolyte undergoes dissociation and breaks apart into its constituent ions. 2

3 General Properties of Aqueous Solutions A non-electrolyte is a substance that dissolves in water to yield a solution that does not conduct electricity. C 12 H 22 O 11 (s) H 2 O C 12 H 22 O 11 (aq) The sucrose molecules remain intact upon dissolving. General Properties of Aqueous Solutions An electrolyte that dissociates completely is known as a strong electrolyte. Water soluble ionic compounds NaCl(s) H 2O Na + (aq) + Cl (aq) Strong Acids HCl(g) H 2O H + (aq) + Cl (aq) Strong Bases NaOH(s) H 2O Na + (aq) + OH (aq) Aqueous Solutions 3

4 General Properties of Aqueous Solutions A weak electrolyte is a compound that produces ions upon dissolving but exists in solution predominantly as molecules that are not ionized. Weak Acids HC 2 H 3 O 2 (l) H + (aq) + C 2 H 3 O 2 (aq) Weak Bases NH 3 (g) + H 2 O(l) + NH 4 (aq) + OH (aq) General Properties of Aqueous Solutions The double arrow, directions., denotes a reaction that occurs in both NH 3 (g) + H 2 O(l) + NH 4 (aq) + OH (aq) When both the forward and reverse reactions occur at the same rate, the reaction is in a state of dynamic equilibrium. 4.1 General Properties of Aqueous Solutions 4

5 General Properties of Aqueous Solutions Classify each of the following as a nonelectrolyte, a weak electrolyte, or a strong electrolyte. Citric acid, H 3 C 6 H 5 O 7 Potassium phosphate, K 3 PO 4 Glucose, C 6 H 12 O An insoluble product that separates from a solution is called a precipitate. 2NaI(aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + 2NaNO 3 (aq) A chemical reaction in which a precipitate forms is called a precipitation reaction. 5

6 Water is a good solvent for ionic compounds because it is a polar molecule. The polarity of water results from electron distributions within the molecule. The oxygen atom has an attraction for the hydrogen atoms electrons and is therefore partially negative compared to hydrogen. The oxygen atom is δ partially negative δ + δ + The hydrogen atoms are partially positive Hydration occurs when water molecules remove the individual ions from an ionic solid surround them so the substances dissolves. Solubility is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. 6

7 Classify the following as soluble or insoluble in an aqueous solution. Ba(NO 3 ) 2 AgI Mg(OH) 2 Classify the following as soluble or insoluble in an aqueous solution: Ba(NO 3 ) 2 AgI Mg(OH) 2 Identify the precipitate in the following reaction: Pb(NO 3 ) 2 (aq) + 2NaI(aq) 2NaNO 3 (?) + PbI 2 (?) Copyright McGraw-Hill

8 Identify the precipitate in the following reaction: Pb(NO 3 ) 2 (aq) + 2NaI(aq) 2NaNO 3 (?) + PbI 2 (?) Pb(NO 3 ) 2 (aq) + 2NaI(aq) 2NaNO 3 (aq) + PbI 2 (s) In a molecular equation compounds are represented by chemical formulas as though they exist in solution as molecules or formula units Na 2 SO 4 (aq) + Ba(OH) 2 (aq) 2NaOH(aq) + BaSO 4 (s) In the reaction between aqueous Na 2 SO 4 and Ba(OH) 2 Na 2 SO 4 (aq) + Ba(OH) 2 (aq) 2NaOH(aq) + BaSO 4 (S) the aqueous species are represented as follows: Na 2 SO 4 (aq) 2Na + 2 (aq) + SO 4 (aq) Ba(OH) 2 (aq) Ba 2+ (aq) + 2OH (aq) NaOH(aq) Na + (aq) + OH (aq) In an ionic equation compounds that exist completely or predominately as ions in solution are represented as those ions. 2Na + 2 (aq) + SO 4 (aq) + Ba 2+ (aq) + 2OH (aq) 2Na + (aq) + 2OH (aq) + BaSO 4 (s) 8

9 An equation that includes only the species that are actually involved in the reaction is called a net ionic equation. Ions that appear on both sides of the equation are called spectator ions. Spectator ions do not participate in the reaction. Ba 2+ 2 (aq) + SO 4 (aq) BaSO 4 (s) To determine the molecular, ionic and net ionic equations: 1) Write and balance the molecular equation, predicting the products by assuming that the cations trade anions. 2) Write the ionic equation by separating strong electrolytes into their constituent ions. 3) Write the net ionic equation by identifying and canceling spectator ions on both sides of the equation. 4) If both the reactants and products are all strong electrolytes, there is no net ionic equation; no reaction takes place. Write the molecular, ionic, and net ionic equations for the combination of Fe(NO 3 ) 2 (aq) with Na 2 CO 3 (aq). Step 1: Write and balance the molecular equation, predicting the products by assuming that the cations trade anions; checking the solubility of each product. 9

10 Write the molecular, ionic, and net ionic equations for the combination of Fe(NO 3 ) 2 (aq) with Na 2 CO 3 (aq). Step 2: Write the ionic equation by separating strong electrolytes into their constituent ions. Write the molecular, ionic, and net ionic equations for the combination of Fe(NO 3 ) 2 (aq) with Na 2 CO 3 (aq). Step 3: Write the net ionic equation by identifying and canceling spectator ions on both sides of the equation. 4.3 Acid-Base Reactions Acids can be either strong or weak. A strong acid is a strong electrolyte. 10

11 Acid-Base Reactions A weak acid is a weak electrolyte; it does not dissociate completely. Acetic acid, HC 2 H 3 O 2, is an example. HC 2 H 3 O 2 (l) H + (aq) + C 2 H 3 O 2 (aq) acidic proton Most acids are weak acids. Acid-Base Reactions Strong bases are strong electrolytes. Strong bases are the hydroxides of Group 1A and heavy Group 2A Sodium hydroxide, NaOH, is an example. NaOH(s) H 2O Na + (aq) + OH (aq) Acid-Base Reactions An Arrhenius acid is one that ionizes in water to produce H + ions. HCl(g) H 2 O H + (aq) + Cl (aq) An Arrhenius base is one that dissociates in water to produce OH ions. NaOH(s) H 2 O Na + (aq) + OH (aq) 11

12 Acid-Base Reactions A Brønsted acid is a proton donor. A Brønsted base is a proton acceptor. In these definitions, a proton refers to a hydrogen atom that has lost its electron also known as a hydrogen ion (H + ). NH 3 (g) + H 2 O(l) + NH 4 (aq) + OH (aq) NH 3 is a Brønsted base: accepts a proton to become NH 4 + H 2 O is a Brønsted acid: donates a proton to become OH Acid-Base Reactions Brønsted acids donate protons to water to form the hydronium ion (H 3 O + ). hydrogen ion (H + ) proton All refer to the same aqueous species hydronium ion (H 3 O + ) HF(aq) + H 2 O(l) H 2O H 3 O + (aq) + F (aq) Acid-Base Reactions A monoprotic acid has one proton to donate. Hydrochloric acid is an example: HCl(g) H + (aq) + Cl (aq) one equivalent of solvated hydrogen ion 12

13 Acid-Base Reactions A polyprotic acid has more than one acidic hydrogen atom. Sulfuric acid, H 2 SO 4, is an example of a diprotic acid; there are two acidic hydrogen atoms. Polyprotic acids lose protons in a stepwise fashion: Step 1: H 2 SO 4 (aq) H + (aq) + HSO 4 (aq) In H 2 SO 4, the first ionization is strong. Step 2: HSO 4 (aq) H + 2 (aq) + SO 4 (aq) In H 2 SO 4, the second ionization occurs only to a very small extent. Acid-Base Reactions Bases that produce only one mole of hydroxide per mole of compound are called monobasic. Sodium hydroxide is an example: NaOH(s) H 2O Na + (aq) + OH (aq) one equivalent of hydroxide Acid-Base Reactions Some strong bases produce more than one hydroxide per mole of compound. Barium hydroxide is an example of a dibasic base. Ba(OH) 2 (s) H 2 O Ba 2+ (aq) + 2OH (aq) two equivalents of hydroxide 13

14 Acid-Base Reactions A neutralization reaction is a reaction between an acid and a base. Generally, a neutralization reaction produces water and a salt. HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) acid base water salt The net ionic equation of many acid-base reactions is: H + (aq) + OH (aq) H 2 O(l) 4.4 Reactions in which electrons (e ) are transferred from one substance to another Oxidation loss of electrons (e ) by one substance Zn Zn e (oxidation of zinc) Reduction gain of electrons (e ) by another substance Zn metal loses 2 electrons and is oxidized to Zn 2+ Zn 2+ is called the reducing agent Cu e Cu (reduction of copper) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Cu 2+ gains 2 electrons and is reduced to Cu metal Cu is called the oxidizing agent Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) 14

15 A redox reaction is the sum of an oxidation half-reaction and a reduction half-reaction. Oxidation (lose 2e ) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Reduction (gain 2e ) Oxidation half-reaction: Zn(s) Zn 2+ (aq) + 2e Reduction half-reaction: Overall redox reaction: Cu 2+ (aq) + 2e Cu 2+ (aq) + Zn(s) Cu(s) Zn 2+ (aq) + Cu(s) The oxidation number is the charge an atom would have if electrons were transferred completely. H 2 (g) + F 2 (g) 2HF(g) Oxidation number: Total contribution to charge: N 2 (g) + 3H 2 (g) 2NH 3 (g) Oxidation number: Total contribution to charge: The oxidation number is sometimes called the oxidation state. To assign oxidation numbers: 1) The oxidation number of an element, in its elemental form, is zero. 2) The oxidation numbers in any chemical species must sum to the overall charge on the species. must sum to zero for any molecule must sum to the charge on any polyatomic ion the oxidation number of a monoatomic ion is equal to the charge on the ion 15

16 To assign oxidation numbers: 3) Know the elements that nearly always have the same oxidation number. Assign the oxidation numbers to the elements in the compound KMnO 4. Step 1: Start with the oxidation numbers you know: Oxidation number: Total contribution to charge: K Mn O Step 2: The numbers in the boxes (total contribution to charge) must sum to zero (KMnO 4 is a neutral compound). Assign the oxidation numbers to the elements in the compound H 2 SO 4. Step 1: Start with the oxidation numbers you know: Oxidation number: Total contribution to charge: H 2 S O Step 2: The numbers in the boxes (total contribution to charge) must sum to zero (the chemical species is neutral). 16

17 Assign the oxidation numbers to the elements in the ion ClO 3. Step 1: Start with the oxidation numbers you know: Oxidation number: Total contribution to charge: Cl O Step 2: The numbers in the boxes (total contribution to charge) must sum to negative one (the chemical species is a 1 anion). In a displacement reaction, an atom or an ion in a compound is replaced by an atom of another element. Zn(s) + CuCl 2 (aq) ZnCl 2 (aq) + Cu(s) Zinc displaces, or replaces copper in the dissolved salt. Zn is oxidized to Zn 2+ Cu 2+ is reduced to Cu When a metal is oxidized by an aqueous solution, it becomes an aqueous ion. The activity series is a list of metals (and hydrogen) arranged from top to bottom in order of decreasing ease of oxidation. Metals listed at the top are called active metals. Metals listed at the bottom are called noble metals. An element in the series will be oxidized by the ions of any element that appears below it in the table. Increasing ease of oxidation Element Oxidation Half-Reaction Zinc Zn Zn e Iron Fe Fe e Nickel Ni Ni e Hydrogen H 2 2H + + 2e Copper Cu Cu e Silver Ag Ag + + e Gold Au Au e Zn(s) + CuCl 2 (aq) ZnCl 2 (aq) + Cu(s) Cu(s) + ZnCl 2 (aq) no reaction 17

18 Which of the following reactions will occur? Co(s) + BaI 2 (aq)? Element Oxidation Half-Reaction Sodium Na Na + + 1e Sn(s) + CuBr 2 (aq) Ag(s) + NaCl(aq)?? Increasing ease of oxidation Barium Ba Ba e Cobalt Co Co e Tin Sn Sn e Hydrogen H 2 2H + + 2e Copper Cu Cu e Silver Ag Ag + + e Co(s) + BaI 2 (aq) Sn(s) + CuBr 2 (aq) Ag(s) + NaCl(aq) No reaction. Cobalt is below barium. Cu(s) + Sn 2+ (aq) No reaction. Silver is below sodium. Redox reactions must have both mass balance and charge balance. Cr(s) + Ni 2+ (aq) Cr 3+ (aq) + Ni(s) Oxidation half-reaction: Cr(s) Cr 3+ (aq) + 3e Reduction half-reaction: Ni 2+ (aq) + 2e Ni(s) Before adding half-reactions, the electrons must balance. Prior to adding the two half-reactions, balance the electrons. Step 1: Multiply the oxidation half-reaction by 2 Oxidation half-reaction: 2 Cr(s) Cr 3+ (aq) + 3e Step 3: Multiply the reduction half-reaction by 3 Reduction half-reaction: 3 Ni 2+ (aq) + 2e Ni(s) Oxidation half-reaction: 2Cr(s) 2Cr 3+ (aq) + 6e Reduction half-reaction: 3Ni 2+ (aq) + 6e 3Ni 2+ (aq) + 2Cr(s) 2Ni(s) 2Ni(s) + 2Cr 3+ (aq) 18

19 Give the overall balanced equation for the following reactions. Cu(s) + H + (aq)? Element Oxidation Half-Reaction Barium Ba Ba e Cd(s) + Ag + (aq)? Increasing ease of oxidation Cadmium Cd Cd e Tin Sn Sn e Hydrogen H 2 2H + + 2e Copper Cu Cu e Silver Ag Ag + + e Cu(s) + H + (aq) Cd(s) + 2Ag + (aq) No reaction. Copper is below hydrogen. 2Ag(s) + Cd 2+ (aq) Combination reactions can involve oxidation and reduction. N 2 (g) + 3H 2 (g) 2NH 3 (g) Hydrogen is oxidized from 0 to +1 Nitrogen is reduced from 0 to 3 Decomposition can also be a redox reaction. NaH(s) 2Na(s) + 3H 2 (g) Na + is reduced to Na H is oxidized to H 2 19

20 Disproportionation reactions occur when one element undergoes both oxidation and reduction. reduction oxidation 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) Oxygen in H 2 O 2 (and other peroxides) has an oxidation number of 1. Combustion is also a redox process. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution. moles solute molarity = liters solution Other common rearrangements: mol L = M mol = M L 20

21 Determine the molarity of a 5.00 L solution that contains 235 g of sucrose, C 12 H 22 O 11. Step 1: Determine the molar mass of sucrose. ( g/mol) Step 2: Determine the moles of sucrose. ( moles sucrose) Step 3: Determine the molarity of the solution: moles solute molarity = liters solution mol sucrose molarity = = M 5.00 L How many millilitres of 3.50 M NaOH can be prepared from grams of solid NaOH? Step 1: Convert grams to moles. ( moles NaOH) Step 2: Use the molarity equation to find liters; convert to ml: moles solute molarity = liters solution mol NaOH liters of solution = = L = 536 ml 3.50 M Dilution is the process of preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution 21

22 In an experiment, a student needs ml of a M CuCl 2 solution. A stock solution of 2.00 M CuCl 2 is available. How much of the stock solution is needed? Use the relationship that moles of solute before dilution = moles of solute after dilution. M c x L c = M d x L d (2.00 M CuCl 2 )(L c ) = (0.100 M CuCl 2 )( L) L c = L or 12.5 ml To make the solution: 1) Pipet 12.5 ml of stock solution into a ml volumetric flask. 2) Carefully dilute to the calibration mark. What volume of 6.0 M H 2 SO 4 is needed to prepare ml of a solution that is 0.25 M H 2 SO 4? Use the relationship that moles of solute before dilution = moles of solute after dilution. M c x L c = M d x L d (6.0 M H 2 SO 4 )(L c ) = (0.25 M H 2 SO 4 )( L) L c = L or 21 ml What are the concentrations of ions in a solution that is M in barium nitrate? [Ba(NO 3 ) 2 ] Step 1: Determine if barium nitrate is a strong electrolyte using solubility rules and, if so, write the equation for the dissociation. Ba(NO 3 ) 2 (s) H 2 O Ba 2+ (aq) + 2NO 3 (aq) Step 2: Determine the concentration of Ba 2+ (aq) ions using the stoichiometric ratio. 2+ [Ba ] = [Ba(NO 3) 2] 1 mol Ba mol Ba(NO 3) 2 = 1 mol Ba = M Ba mol Ba(NO 3) 2 L 1 mol Ba(NO 3) 2 Square brackets around a chemical species indicates concentration. 22

23 What are the concentrations of ions in a solution that is M in barium nitrate? [Ba(NO 3 ) 2 ] Ba(NO 3 ) 2 (s) H 2 O Ba 2+ (aq) + 2NO 3 (aq) Step 3: Determine the concentration of NO 3 (aq) ions using the stoichiometric ratio. [NO 3] = [Ba(NO 3) 2] 2 mol NO mol Ba(NO 3) 2 = 2 mol NO3 = 1.50 M NO 1 mol Ba(NO 3) 2 L 1 mol Ba(NO 3) 2 3 Spectrophotometry An Absorption Spectrophotometer 23

24 Spectrophotometry Amount of light absorbed by a sample depends on path length and solute concentration. Different concs of Cu 2+ Same concentrations but different path lengths Concentration Path Length Spectrophotometry BEER-LAMBERT LAW relates amount of light absorbed and the path length and solute concentration. A = absorbance = path length ε = molar absorptivity c = concentration There is a linear relation between A and c for a given path length and compound. This means you can determine solution concentration if A is measured. Spectrophotometry To use the Beer-Lambert law you must first calibrate the instrument. The calibration plot can be used to find the unknown concentration of a solution from a measured A. 24

25 Problem: The nitrite ion is involved in the biochemical nitrogen cycle. You can analyze for the nitrite ion content of a sample using spectrophotometry by first using several organic compounds to create a colored compound from the ion. The following data were collected. [NO 2 ] 550 nm M M M M M Unknown solution a) Construct a calibration plot, and determine the slope and intercept b) What is the nitrite ion concentration in the unknown solution? y = 33375x E E E E E E E E E E E E E E E E E E E E Aqueous Reactions and Chemical Analysis Gravimetric analysis is an analytical technique based on the measurement of mass. Gravimetric analysis is highly accurate. Applicable only to reactions that go to completion or have nearly 100 % yield. 25

26 Aqueous Reactions and Chemical Analysis One common type of gravimetric analysis involves the isolation of a precipitate. Typical steps involve: 1) Mass an unknown solid. 2) Dissolve the unknown in water. 3) React the unknown with an excess amount of a substance that is known to form a precipitate. 4) Filter, dry and weigh the precipitate. 5) Use the formula mass and the mass of the precipitate to find the % mass of the unknown ion. Aqueous Reactions and Chemical Analysis A g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample? Step 1: Find the percent by mass chlorine in AgCl g Cl % Cl = 100 = 24.7 % g AgCl Step 2: Multiply the percent of Cl by the mass of the precipitate to obtain the mass of Cl in the sample x g AgCl = g Cl in the sample Aqueous Reactions and Chemical Analysis A g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample? Step 3: Divide the mass of Cl in sample by the total mass of sample; multiply by 100 to determine percent g Cl % Cl in unknown = 100 = 51.7% Cl g sample 26

27 Aqueous Reactions and Chemical Analysis Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as a titration. A titration is a volumetric technique that uses burets. The point in the titration where the acid has been neutralized is called the equivalence point. Aqueous Reactions and Chemical Analysis The equivalence point is usually signalled by a color change. The color change is brought about by the use of an indicator. Indicators have distinctly different colors in acidic and basic media. The indicator is chosen so that the color change, or endpoint, is very close to the equivalence point. Phenolphthalein is a common indicator. Aqueous Reactions and Chemical Analysis Typical steps of a titration include: 1) Prepare an unknown solution 2) Add an appropriate indicator to visualize the endpoint. 3) Carefully add a standard solution with a buret until the endpoint is reached. 4) Using solution stoichiometry, calculate the molarity of the unknown solution. 27

28 Aqueous Reactions and Chemical Analysis Sodium hydroxide solutions are commonly used in titrations. NaOH solutions must be standardized as the concentrations change over time. (NaOH reacts with CO 2 that slowly dissolves into the solution forming carbonic acid.) The acid potassium hydrogen phthalate (KHP) is frequently used to standardize NaOH solutions. acidic proton of KHP; KHP is a monoprotic acid Aqueous Reactions and Chemical Analysis A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required ml of NaOH to reach the endpoint. What is the concentration of the NaOH solution? Step 1: Use the molar mass of KHP to determine the moles of KHP g KHP moles of KHP = = moles KHP g/mol Aqueous Reactions and Chemical Analysis A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required ml of NaOH to reach the endpoint. What is the concentration of the NaOH solution? Step 2: Using the net ionic equation, convert to moles of NaOH KHC 8 H 4 O 4 (aq) + NaOH(aq) KNaC 8 H 4 O 4 (aq) + H 2 O(l) 1 mol NaOH moles of NaOH = mol KHP = mol NaOH 1 mol KHP 28

29 Aqueous Reactions and Chemical Analysis A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required ml of NaOH to reach the endpoint. What is the concentration of the NaOH solution? Step 3: Use the molarity equation to calculate molarity of the sodium hydroxide solution. mol NaOH moles NaOH molarity of NaOH = = = M NaOH liters of solution L solution Aqueous Reactions and Chemical Analysis How many milliliters of a 1.42 M H 2 SO 4 solution are needed to neutralize 95.5 ml of a M KOH solution? Step 1: Write and balance the chemical equations that corresponds to the neutralization reaction: H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) Step 2: Use the molarity and volume given to determine the moles of KOH. moles of KOH = M KOH x L = moles KOH Aqueous Reactions and Chemical Analysis How many milliliters of a 1.42 M H 2 SO 4 solution are needed to neutralize 95.5 ml of a M KOH solution? Step 3: Using the moles of KOH and the stoichiometric ratios from the balanced equation, convert to moles of H 2 SO 4. H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) 1 mol H SO 2 mol KOH mol KOH = mol H2SO4 29

30 Aqueous Reactions and Chemical Analysis How many millilitres of a 1.42 M H 2 SO 4 solution are needed to neutralize 95.5 ml of a M KOH solution? Step 4: Using the moles of H 2 SO 4 and the concentration given, calculate the millilitres of solution mol H2SO 4 = L or 11.3 ml 1.42 M solution 4 Chapter Summary: Key Points Properties of Aqueous Solutions Electrolytes and Nonelectrolytes Strong Electrolytes and Weak Electrolytes Solubility of Ionic Compounds in Water Molecular Equations Ionic Equations Net Ionic Equations Acid-Base Reactions Strong Acids and Bases Brønsted Acids and Bases Acid-Base Neutralization Oxidation Numbers Oxidation of Metals in Aqueous Solutions Balancing Simple Redox Equations Other Types of Redox Reactions Molarity Dilution Solution Stoichiometry Aqueous Reactions and Chemical Analysis Gravimetric Analysis Acid-Base Titrations 30

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