CHAPTER 5 INTRODUCTION TO REACTIONS IN AQUEOUS SOLUTIONS PRACTICE EXAMPLES
|
|
- Carol Griffith
- 6 years ago
- Views:
Transcription
1 CHAPTER 5 INTRODUCTION TO REACTIONS IN AQUEOUS SOLUTIONS PRACTICE EXAMPLES 1A In determining total Cl, we recall the definition of molarity: moles of solute per liter of solution. mol NaCl mol Cl from NaCl, Cl = =0.8M Cl 1 L soln 1 mol NaCl mol MgCl mol Cl from MgCl, Cl = = 0.10 M Cl 1 L soln 1 mol MgCl Cl total = Cl from NaCl Cl from MgCl = 0.8 M 0.10 M = 0.50 M Cl 1B (a) mg F L 1 g F 1000 mg F 1 mol F g F = M F - (b) L mol F 1 mol CaF 1L mol F g CaF 1 mol CaF 1 kg 1000 g =.1 kg CaF A In each case, we use the solubility rules to determine whether either product is insoluble. The ions in each product compound are determined by simply switching the partners of the reactant compounds. The designation (aq) on each reactant indicates that it is soluble. (a) Possible products are potassium chloride, KCl, which is soluble, and aluminum hydroxide, Al aq OH aq Al OH s Al OH, which is not. Net ionic equation: (b) Possible products are iron(iii) sulfate, Fe SO, and potassium bromide, KBr, both of which are soluble. No reaction occurs. (c) Possible products are calcium nitrate, Ca(NO ), which is soluble, and lead(ii) iodide, Pb aq I aq PbI s PbI, which is insoluble. The net ionic equation is: B (a) Possible products are sodium chloride, NaCl, which is soluble, and aluminum phosphate, Al aq PO aq AlPO s AlPO, which is insoluble. Net ionic equation: (b) Possible products are aluminum chloride, AlCl, which is soluble, and barium sulfate, BaSO, which is insoluble. Net ionic equation: Ba aq SO aq BaSO s 97
2 (c) Possible products are ammonium nitrate, NHNO, which is soluble, and lead (II) carbonate, Pb aq CO aq PbCO s PbCO, which is insoluble. Net ionic equation: A B A Propionic acid is a weak acid, not dissociated completely in aqueous solution. Ammonia similarly is a weak base. The acid and base react to form a salt solution of ammonium propionate. NH aq HC H O aq NH aq C H O aq 5 5 Since acetic acid is a weak acid, it is not dissociated completely in aqueous solution (except at infinite dilution); it is misleading to write it in ionic form. The products of this reaction are the gas carbon dioxide, the covalent compound water, and the ionic solute calcium acetate. Only the latter exists as ions in aqueous solution. CaCO s HC H O aq CO g H O l Ca aq C H O aq (a) This is a metathesis or double displacement reaction. Elements do not change oxidation states during this reaction. It is not an oxidation reduction reaction. (b) The presence of O (g) as a product indicates that this is an oxidation reduction reaction. Oxygen is oxidized from O.S. = - in NO - to O.S. = 0 in O (g). Nitrogen is reduced from O.S. = 5 in NO - to O.S. = in NO. B 5A 5B 6A Vanadium is oxidized from O.S. = in VO to an O.S. = 5 in VO while manganese is reduced from O.S. = 7 in MnO - to O.S. = in Mn. Aluminum is oxidized (from an O.S. of 0 to an O.S. of ), while hydrogen is reduced (from an O.S. of 1 to an O.S. of 0). Oxidation : Al s Al aq e Reduction: H aq e H g Net equation : Als 6 H aq Al aq H g Bromide is oxidized (from 1 to 0), while chlorine is reduced (from 0 to 1). Oxidation : Br aq Br l e Cl g e Cl aq Reduction: Net equation : Br aq Cl g Br l Cl aq Step 1: Write the two skeleton half reactions. MnO aq Mn aq and Fe aq Fe aq Step : Balance each skeleton half reaction for O (with HO ) and for H atoms (with H ). MnO aq 8 H aq Mn aq H O(l) and Fe aq Fe aq 98
3 Step : Balance electric charge by adding electrons. and MnO aq 8 H aq 5 e Mn aq H O(l) Fe aq Fe aq e Step : Combine the two half reactions Fe aq Fe aq e 5 MnO aq 8 H aq 5 e Mn aq H O(l) MnO aq 8 H aq 5 Fe aq Mn aq H O(l) 5 Fe aq 6B 7A Step 1: Uranium is oxidized and chromium is reduced in this reaction. The skeleton half-equations are: UO aq UO aq and CrO 7 (aq) Cr (aq) Step : First, balance the chromium skeleton half-equation for chromium atoms: CrO 7 aq Cr aq Next, balance oxygen atoms with water molecules in each half-equation: UO aq HO(l) UO aq and CrO 7 (aq) Cr (aq) 7HO(l) Then, balance hydrogen atoms with hydrogen ions in each half-equation: UO aq H O(l) UO aq H aq 7 Cr O (aq) 1H (aq) Cr (aq) 7H O(l) Step : Balance the charge of each half-equation with electrons. UO aq H O(l) UO aq H aq e CrO7 aq 1 H aq 6 e Cr aq 7 HO(l) Step : Multiply the uranium half-equation by and add the chromium half-equation to it. UO aq H O(l) UO aq H aq e Step 5: 7 Cr O aq 1 H aq 6 e Cr aq 7 H O(l) - UO (aq)cr O (aq)1 H (aq) H O(l) UO (aq) Cr (aq)7 H O(l)6 H (aq) 7 Simplify. Subtract HO (l) and 6 H (aq) from each side of the equation. UO aq Cr O aq 8 H aq UO aq Cr aq H O(l) 7 Step 1: Write the two skeleton half-equations. S(s) SO ( aq ) and OCl ( aq ) Cl ( aq ) Step : Balance each skeleton half-equation for O (with H O ) and for H atoms (with H ). HO(l) Ss SO aq 6 H OCl (aq) H Cl (aq) HO(l) Step : Balance electric charge by adding electrons. H O(l) S s SO aq 6 H (aq) e OCl (aq) H (aq) e Cl (aq) H O(l) 99
4 Step : Change from an acidic medium to a basic one by adding OH to eliminate H. H O(l) S s 6 OH (aq) SO aq 6 H (aq) 6 OH (aq) e OCl aq H (aq) OH (aq) e Cl aq H O(l) OH (aq) Step 5: Simplify by removing the items present on both sides of each half-equation, and combine the half-equations to obtain the net redox equation. {S s 6 OH (aq) SO aq H O(l) e } 1 {OCl aq H O(l) e Cl aq OH (aq)} - S s 6 OH (aq) OCl aq H O(l) SO aq H O(l) Cl aq OH Simplify by removing the species present on both sides. S s OH aq OCl aq SO aq H O(l) Cl aq Net ionic equation: 7B Step 1: Write the two skeleton half-equations. MnO aq MnOs and SO (aq) SO (aq) Step : Balance each skeleton half-equation for O (with H O ) and for H atoms (with H ). MnO aq H aq MnO s H O(l) Step : SO (aq) H O(l) SO (aq) H (aq) Balance electric charge by adding electrons. MnO aq H aq e MnO s H O(l) SO aq H O(l) SO aq H aq e Step : Change from an acidic medium to a basic one by adding OH to eliminate H. MnO aq H aq OH aq e MnO s H O(l) OH aq SO aq H O(l) OH aq SO aq H aq OH aq e Step 5: Simplify by removing species present on both sides of each half-equation, and combine the half-equations to obtain the net redox equation. {MnO aq H O(l) e MnO s OH aq } - MnO s SO aq H O(l)8 OH aq {SO aq OH aq SO aq H O(l) e } MnO aq SO aq 6 OH (aq) H O(l) Simplify by removing species present on both sides. Net ionic equation: MnO aq SO aq H O(l) MnO s SO aq OH aq 100
5 8A Since the oxidation state of H is 0 in H (g) and is 1 in both NH (g) and H O(g), hydrogen is oxidized. A substance that is oxidized is called a reducing agent. In addition, the oxidation state of N in NO (g) is, while it is innh ; the oxidation state of the element N decreases during this reaction, meaning that NO (g) is reduced. The substance that is reduced is called the oxidizing agent. 8B In Au CN aq, gold has an oxidation state of 1; Au has been oxidized and, thus, Au(s) (oxidization state = 0), is the reducing agent. In OH - (aq), oxygen has an oxidation state of -; O has been reduced and thus, O (g) (oxidation state = 0) is the oxidizing agent. 9A We first determine the amount of NaOH that reacts with g KHP. 1 mol KHP 1 mol OH 1 mol NaOH n NaOH = g KHP = mol NaOH 0. g KHP 1 mol KHP 1 mol OH mol NaOH 1000 ml [NaOH] = = M.0 ml soln 1 L 9B The net ionic equation when solid hydroxides react with a strong acid is OH - H H O. There are two sources of OH - : NaOH and Ca(OH). We compute the amount of OH - from each source and add the results. moles of OH from NaOH: 9.5 g NaOH 1 mol NaOH 1 mol OH = 0.5 g sample = mol OH g sample g NaOH 1 mol NaOH : moles of OH from Ca OH 7.5 g Ca OH 1 mol Ca OH mol OH = 0.5 g sample = mol OH g sample 7.09 g Ba OH 1 mol Ca OH - total amount OH = mol from NaOH mol from Ca OH = mol OH mol OH 1 mol H 1 mol HCl 1000 ml soln [HCl] = = 0.10 M 5.6 ml HCl soln 1 mol OH 1 mol H 1 L soln 10A First, determine the mass of iron that has reacted as Fe with the titrant. The balanced chemical equation provides the essential conversion factor to answer this question. Namely: 5 Fe aqmno aq 8 H aq 5 Fe aq Mn aq HOl mol MnO 5 mol Fe g Fe mass Fe = L titrant = 0.6 g Fe 1 L titrant 1 mol MnO 1 mol Fe Then determine the % Fe in the ore. % Fe = g Fe g ore 100% = 65.% Fe 10B The balanced equation provides us with the stoichiometric coefficients needed for the solution. - Namely: 5 C O aq MnO aq16 H aq10 CO g Mn aq8 H Ol 101
6 1 mol Na C O 1 mol C O mol MnO amount MnO = 0.8 g Na C O 1.00 g Na C O 1 mol Na C O 5 mol C O = mol MnO mol MnO 1000 ml 1 mol KMnO [KMnO ] = = M KMnO.68 ml soln 1 L 1 mol MnO INTEGRATIVE EXAMPLE A. First, balance the equation. Break down the reaction of chlorate and ferrous ion as follows: ClO 6H 6e Cl H O 6 Fe Fe e Net reaction: ClO 6Fe 6H Cl 6Fe H O The reaction between Fe and Ce is already balanced. To calculate the moles of Fe that remains after the reaction with ClO -, determine the moles of Ce that react with Fe : mol Ce = L M = mol = mol of excess Fe total mol of Fe = L = mol Therefore, the moles of Fe reacted = = mol. To determine the mass of KClO, use the mole ratios in the balanced equation in conjunction with the molar mass of KClO. 1 mol ClO 1 mol KClO 1.5 g KClO mol Fe 6 mol Fe 1 mol ClO 1 mol KClO = g KClO g %KClO = 100% = 9.89% 0.1 g B. First, balance the equation. Break down the reaction of arsenous acid and permanganate as follows: 5 H AsO H O H AsO e H - - MnO 8H 5e Mn HO Net reaction: 5H AsO MnO 6H 5H AsO Mn H O moles of MnO - = L 0.01 M = mol 10
7 To calculate the mass of As, use the mole ratios in the balanced equation in conjunction with the molar mass of As: 5 mol HAsO 1 mol As 7.9 g As mol MnO mol MnO 1 mol H AsO 1 mol As = g As g mass% As = 100% = 1.% 7.5 g EXERCISES Strong Electrolytes, Weak Electrolytes, and Nonelectrolytes 1. (a) Because its formula begins with hydrogen, HC H O 6 5 is an acid. It is not listed in Table 5-1, so it is a weak acid. A weak acid is a weak electrolyte. (b) (c) LiSOis an ionic compound, that is, a salt. A salt is a strong electrolyte. MgI also is a salt, a strong electrolyte. (d) CHCH O is a covalent compound whose formula does not begin with H. Thus, it is neither an acid nor a salt. It also is not built around nitrogen, and thus it does not behave as a weak base. This is a nonelectrolyte. (e) Sr OH is a strong electrolyte, one of the strong bases listed in Table 5-.. HCl is practically 100% dissociated into ions. The apparatus should light up brightly. A solution of both HCl and HC H O will yield similar results. In strongly acidic solutions, the weak acid HC H O is molecular and does not contribute to the conductivity of the solution. However, the strong acid HCl is practically dissociated into ions and is unaffected by the presence of the weak acid HC H O. The apparatus should light up brightly. 5. (a) Barium bromide: strong electrolyte (b) Propionic acid: weak electrolyte (c) Ammonia: weak electrolyte Ion Concentrations 7. (a) (b) K mol KNO = L soln 1 mol K 1 mol KNO = 0.8 M K mol Ca NO mol NO NO = = 0. M NO 1 L soln 1 mol Ca NO 10
8 (c) (d) 0.08 mol Al SO mol Al Al = = M Al 1 L soln 1 mol Al SO Na mol Na PO = L soln mol Na 1 mol Na PO = 0.67 M Na 9. Conversion pathway approach: 0.1 g Ba OH 8H O 1 mol Ba OH 8H O 1000 ml mol OH OH = 75 ml soln 1 L 15.5g Ba OH 8H O 1 mol Ba OH 8 H O =.010 M OH Stepwise approach: 0.1 g Ba OH 8H O 1000 ml = 0.80 g/l 75 ml soln 1 L 0.80 g 1 mol Ba OH 8H O mol Ba OH 8H O = L 15.5g Ba OH 8H O L mol Ba OH 8H O mol OH =.010 M OH L 1 mol Ba OH 8 H O (a) (b) (c) 1. mg Ca 1 g Ca 1 mol Ca [Ca ] =.5 10 M Ca 1 L solution 1000 mg Ca g Ca.8 mg K 1 g K 1000 ml solution 1 mol K [K ] = M K 100 ml solution 1000 mg K 1 L solution g K 5 g Zn 1 g Zn 1000 ml solution 1 mol Zn [Zn ] =.10 M Zn 6 1 ml solution 110 g Zn 1 L solution 65.9 g Zn 1 In order to determine the solution with the largest concentration of K, we begin by converting each concentration to a common concentration unit, namely, molarity of K M KSO mol K 0.17 M K 1 L solution 1 mol K SO 1.5 g KBr 1000 ml solution 1 mol KBr 1 mol K M K 100 ml solution 1 L solution g KBr 1 mol KBr 8.1 mg K 1000 ml solution 1 g K 1 mol K 0.07 M K 1 ml solution 1 L solution 1000 mg K g K Clearly, the solution containing 8.1 mg K per ml gives the largest K of the three solutions. 10
9 15. Determine the amount of I in the solution as it now exists, and the amount of I in the solution of the desired concentration. The difference in these two amounts is the amount of I that must be added. Convert this amount to a mass of MgI in grams. 1 L mol I moles of I in final solution = 50.0 ml 1000 ml 1 L soln = mol I 1 L mol KI 1 mol I moles of I in KI solution = 50.0 ml = mol I 1000 ml 1 L soln 1 mol KI 1 mol MgI g MgI 1000 mg mass MgI required = mol I mol I 1 mol MgI 1 g =. 10 mg MgI 17. moles of chloride ion 0.65 mol KCl 1 mol Cl 0.85 mol MgCl mol Cl = 0.5 L L 1 L soln 1 mol KCl 1 L soln 1 mol MgCl = 0.11 mol Cl 0.7 mol Cl = mol Cl mol Cl Cl = = 0.7 M 0.5 L L Predicting Precipitation Reactions 19. In each case, each available cation is paired with the available anions, one at a time, to determine if a compound is produced that is insoluble, based on the solubility rules of Chapter 5. Then a net ionic equation is written to summarize this information. (a) Pb aq Br aq PbBr s (b) No reaction occurs (all are spectator ions). Fe aq OH aq Fe OH s (c) 1. Mixture Result (Net Ionic Equation) (a) HIa Zn NO (aq): No reaction occurs. (b) CuSOaq Na COaq : Cu aq CO aq CuCOs (c) Cu NO aq Na PO aq : Cu aq PO aq Cu PO s. (a) Add KSO aq; BaSO s will form and MgSO will not precipitate. BaCl s K SO aq BaSO s KClaq 105
10 (b) Add HOl; NaCO water Na CO s Na aq CO aq s dissolves, but MgCO (s) will not dissolve (appreciably). (c) Add KCl(aq); AgCl(s) will form, while Cu(NO ) (s) will dissolve. AgNO s KCl aq AgCl s KNO aq 5. Mixture Net Ionic Equation Sr NO aq K SO aq : Sr aq SO aq SrSO s (a) (b) Mg NO aq NaOH aq : Mg aq OH aq Mg OH s BaCl aq K SO aq : Ba (aq) SO (aq) BaSO (s) (c) (upon filtering, KCl (aq) is obtained) Acid Base Reactions 7. The type of reaction is given first, followed by the net ionic equation. (a) Neutralization: OH aq HC H O aq H Ol C H O aq (b) No reaction occurs. This is the physical mixing of two acids. FeS s H aq H S g Fe aq (c) Gas evolution: (d) Gas evolution: HCO aq H aq "HCOaq " HOl CO g (e) Redox: Mg s H aq Mg aq H g 9. As a salt: NaHSO aq Na aq HSO aq As an acid: HSO aq OH aq H Ol SO aq 1. Use (b) NH (aq): NH affords the OH - ions necessary to form Mg(OH) (s). Applicable reactions: {NH (aq) H O(l) NH (aq) OH - (aq)} MgCl (aq) Mg (aq) Cl - (aq) Mg (aq) OH - (aq) Mg(OH) (s) 106
11 Oxidation Reduction (Redox) Equations. (a) The O.S. of H is 1, that of O is, that of C is, and that of Mg is on each side of this equation. This is not a redox equation. (b) (c) (d) The O.S. of Cl is 0 on the left and 1 on the right side of this equation. The O.S. of Br is 1 on the left and 0 on the right side of this equation. This is a redox reaction. The O.S. of Ag is 0 on the left and 1 on the right side of this equation. The O.S. of N is 5 on the left and on the right side of this equation. This is a redox reaction. On both sides of the equation the O.S. of O is, that of Ag is 1, and that of Cr is 6. Thus, this is not a redox equation. 5. (a) Reduction: (b) Reduction: NO aq 10 H aq 8 e NOg 5 HOl (c) Oxidation: Als OH aq AlOH aq e 7. (a) Oxidation: { SO aq 6 H aq e S O aq H O(l) I aq I s e } 5 Reduction: { MnO aq 8 H aq 5 e Mn aq HOl } Net: 10 I aq MnO aq 16 H aq 5 I s Mn aq 8 H O l (b) Oxidation: { Reduction: { BrO aq 6 H aq 6 e Br aq HOl N H l N g H aq e } } Net: N H l BrO aq N g Br aq 6 H Ol (c) Oxidation: Fe aq Fe aq e Reduction: VO aq 6 H aq e VO aq HOl Net: Fe aq VO aq 6 H aq Fe aq VO aq H Ol (d) Oxidation: { UO aq HO l UO aq H aq e } Reduction: { NO aq H aq e NOg HOl } Net: UO aq NO aq H aq UO aq NOg H Ol 9. (a) Oxidation: { Reduction: ClO aq H O(l) 6 e Cl aq 6 OH aq MnO s OH aq MnO aq H O(l) e } Net: MnO s ClO aq OH aq MnO aq Cl aq H O(l) 107
12 (b) Oxidation: { - Reduction: { OCl aq H O(l) e Cl aq OH aq Fe OH s 5 OH aq FeO aq HO(l) e } } Net: Fe OH s OCl aq OH aq FeO aq Cl aq 5 H O(l) (c) Oxidation: { ClO (aq) OH aq ClO aq HOl e Reduction: ClO (aq) HOl 5 e Cl (aq) OH aq Net: } 5 6 ClO (aq) 6 OH aq 5ClO aq Cl aq H O (l) (d) Oxidation: (Ag (s) Ag (aq) 1 e ) - Reduction: H O(l) CrO e Cr(OH) (s) 5 OH - - Net: Ag(s) CrO H O(l) Ag(aq) Cr(OH) (s) 5 OH - 1. (a) Oxidation: Reduction: { Cl g e Cl aq Cl g 1 OH aq ClO aq 6 HO(l) 10 e } 5 Net: Or: 6 Cl g 1 OH aq 10 Cl aq ClO aq 6 H O(l) Cl g 6 OH aq 5 Cl aq ClO aq H O(l) (b) Oxidation: Reduction: S O aq H O(l) HSO aq H aq e S O aq H aq e S O aq H O (l) Net: S O aq H O(l) HSO aq S O aq. (a) Oxidation: { NO aq HOl NO aq H aq e } 5 Reduction: { MnO aq 8 H aq 5 e Mn aq HOl } Net: 5 NO aq MnO aq 6 H aq 5 NO aq Mn aq H Ol (b) Oxidation: { Mn (aq) OH - (aq) MnO (s) H O (l) e - } Reduction: { MnO - (aq) H O (l) e - MnO (s) OH - (aq) } Net: Mn (aq) MnO - (aq) OH - (aq) 5 MnO (s) H O (l) (c) Oxidation: { CH5OH CHCHO H aq e Reduction: Cr O7 aq 1 H aq 6 e Cr aq 7 HOl Net: } Cr O aq 8 H aq C H OH Cr aq 7 H O l CH CHO
13 5. For the purpose of balancing its redox equation, each of the reactions is treated as if it takes place in acidic aqueous solution. (a) H O(g) CH (g) CO (g) 8 H (g) 8 e - { e - H (g) NO(g) ½ N (g) H O(g) } CH (g) NO(g) N (g) CO (g) H O(g) (b) {H S(g) 1/8 S 8 (s) H (g) e - } e - H (g) SO (g) 1/8 S 8 (s) H O(g) H S(g) SO (g) /8 S 8 (s) H O(g) or 16 H S(g) 8 SO (g) S 8 (s) 16 H O(g) (c) {Cl O(g) NH (aq) H (aq) e - NH Cl(s) H O(l) } { NH (g) N (g) 6 e - 6 H (aq) } 6 NH (g) 6 H (aq) 6 NH (aq) 10 NH (g) Cl O(g) 6 NH Cl(s) N (g) H O(l) Oxidizing and Reducing Agents 7. The oxidizing agents experience a decrease in the oxidation state of one of their elements, while the reducing agents experience an increase in the oxidation state of one of their elements. (a) SO aq MnO aq is the reducing agent; the O.S. of S = in SO and =6 in SO. is the oxidizing agent; the O.S. of Mn =7 in MnO and in Mn. (b) H gis the reducing agent; the O.S. of H = 0 in H g and=1 in NO g is the oxidizing agent; the O.S. of N = in NO g and HOg. in NH (g). (c) Fe CN aq 6 is the reducing agent; the O.S. of Fe = in FeCN and = in FeCN. H O aq in HO and= 6 6 is the oxidizing agent; the O.S. of O = 1 in HO. Neutralization and Acid Base Titrations 9. The problem is most easily solved with amounts in millimoles. V 0.18 mmol HCl 1 mmol H 1 mmol OH NaOH = ml HClaq 1 ml HCl aq 1 mmol HCl 1 mmol H 109
14 1 mmol NaOH 1 ml NaOH aq 1 mmol OH mmol NaOH = 1. ml NaOH aq soln 51. The net reaction is OH aq HC H O aq H O(l) C H O aq. 5 5 Conversion pathway approach: mmol HC H O 1 mmol KOH 1 ml base 5 V base = 5.00 ml acid 1 ml acid 1 mmol HC H O.155 mmol KOH 5 =.56 ml KOH solution Stepwise approach: mmol HC H O ml acid = 7.6 mmol HC H O 1 ml acid 1 mmol KOH 7.6 mmol HCH5O = 7.6 mmol KOH 1 mmol HC H O 5 1 ml base 7.6 mmol KOH =.56 ml KOH solution.155 mmol KOH 5 5. NaOHaq HClaq NaClaq HO(l) is the titration reaction mol HCl 1mol NaOH 0.08 L 1Lsoln 1 molhcl [NaOH] = = 0.10 M NaOH L sample 55. The mass of acetylsalicylic acid is converted to the amount of NaOH, in millimoles, that will react with it. 0. g HC9H7O 1 mol HC9H7O 1 mol NaOH 1000 mmol NaOH NaOH = ml NaOH aq 180. g HC H O 1 mol HC H O 1 mol NaOH = M NaOH The equation for the reaction is HNO aq KOH aq KNO aq H O 1. This equation shows that equal numbers of moles are needed for a complete reaction. We compute the amount of each reactant. mmol HNO = 5.00 ml acid mmol HNO ml acid =.0 mmol HNO 0.18 mmol KOH mmol KOH = ml acid =.18 mmol KOH 1 ml base There is more acid present than base. Thus, the resulting solution is acidic. 110
15 V 1.01 g vinegar.0 g HC H O 1 mol HC H O = 5.00 ml vinegar 59. base 1 ml g vinegar 60.0 g HCHO 1 mol NaOH 1 L base 1 mol HC H O mol NaOH ml L = ml base 61. Answer is (d): 10 % of necessary titrant added in titration of NH 5 NH 5 HCl 1 HCl required for equivalence point 0 % excess 5 NH 6 Cl - H O (depicted in question's drawing ) Stoichiometry of Oxidation Reduction Reactions 6. Conversion pathway approach: [MnO ]= 1mol As O mol MnO 1mol KMnO g As O g As O 5 mol As O 1molMnO 1L.15 ml 1000 ml = M KMnO Stepwise approach: mol KMnO [KMnO ]= L solution 1mol As O g As O = mol As O g AsO mol MnO mol As O =.5910 mol MnO - - 5mol AsO 1mol KMnO mol MnO = molMnO 1L.15 ml L solution 1000 ml mol KMnO.5910 mol KMnO mol KMnO - - [ KMnO ]= = = M L solution L solution 111
16 65. First, we will determine the mass of Fe, then the percentage of iron in the ore. 1 L mol CrO 7 6 mol Fe g Fe mass Fe = 8.7 ml 1000 ml 1 L soln 1 mol Cr O 7 1 mol Fe 0.861g Fe mass Fe = g Fe % Fe = 100% 5. % Fe 0.91 g ore 67. First, balance the titration equation: Oxidation: { C O aq CO g e } 5 } Reduction: { MnO aq 8 H aq 5 e Mn aq H Ol Net: 5 C O aq MnO aq 16 H aq 10 CO g Mn aq 8 H Ol mass mass =1.00 L satd soln Na C O NaCO 1000 ml 5.8 ml satd soln KMnO mol KMnO 1 L 5.00 ml satd soln Na C O 1000 ml KMnO 1 mol MnO 5 molc O 1 mol Na C O 1.0 g Na C O 1 mol KMnO molmno 1 molc O 1 mol Na C O = 7.0 g Na C O NaCO Integrative and Advanced Exercises 71. A possible product, based on solubility rules, is Ca (PO ). We determine the % Ca in this compound. molar mass g Ca 0.97 g P g O 10. g Ca g P g O g 10. g Ca % Ca 100% 8.76% g Ca (PO ) Thus, Ca (PO ) is the predicted product. The net ionic equation follows. Ca (aq) HPO (aq) Ca (PO ) (s) H (aq) 7. Let us first determine the mass of Mg in the sample analyzed. Conversion pathway approach: 1 mol Mg P O mol Mg.05 g 7 mass Mg g MgPO g Mg.55 g MgP O7 1 mol MgPO 7 1 mol Mg g Mg ppm Mg 10 g sample 108 ppm Mg g sample 11
17 Stepwise approach: 1 mol Mg P O g Mg P O =.710 mol Mg P O g MgP O7 mol Mg.710 mol Mg P O.910 mol Mg mol Mg PO 7.05 g 1 mol Mg mol Mg g Mg g Mg ppm Mg 10 g sample 108 ppm Mg g sample 75. Let V represent the volume of added 0.8 M CaCl that must be added. We know that [Cl ] [Cl 0.5 L ] = 0.50 M, but also, mol KCl 1L soln 1mol Cl V 1mol KCl 0.5 L V 0.8 mol CaCl 1L soln mol Cl 1mol CaCl (0.5 V ) V V V L (a) [FeS 8 H O Fe SO 16 H 15 e ] [O H e H O] 15 overall: FeS (s) 15 O (g) H O(l) Fe (aq) 8 SO (aq) H (aq) (b) One kilogram of tailings contains 0.0 kg (0 g) of S. We have 1molS 1molFeS moles of FeS = 0 g S 0.68 mol FeS.07 g S mol S moles of H = moles of CaCO = molh 0.68 mol FeS 0.67 mol H molfes 1 mol CaCO 0.67 mol H 0. mol CaCO molh g CaCO mass of CaCO = 0. mol CaCO. g CaCO 1molCaCO 11
18 8. Oxidation :{ Cl (aq) Cl (g) e } Reduction :CrO 7 (aq) 1 H (aq) 6 e Cr (aq) 7 H O Net :6 Cl (aq) CrO 7 (aq) 1 H (aq) Cr (aq) 7 H O Cl (g) We need to determine the amount of Cl (g) produced from each of the reactants. The limiting reactant is the one that produces the lesser amount of Cl.. amount Cl amount Cl 1.15 g 0.1g HCl 5 ml 1mL 100. g soln 1.5 mol Cl 1mol HCl 6.6 g HCl 1mol Cl 1mol HCl mol Cl 6 mol Cl 98.5 g K CrO 7 1mol K CrO 7 1mol CrO 7 mol Cl 6.6 g 100. g sample 9. g K CrO 7 1mol K CrO 7 1mol CrO mol Cl, the amount produced from the limiting reactant g Cl Then we determine the mass of Cl (g) produced. = 0.69 mol Cl =.6 g Cl 1 mol Cl 85. Cl (g) NaClO (aq) NaCl(aq) ClO (g) (not balanced) Cl (g) NaClO (aq) NaCl(aq) ClO (g) amount ClO g ClO (g) 100 g ClO calculated.785 L.0 mol NaClO mol ClO 67.5 g ClO 1 gal 1 gal 1 L soln mol NaClO 1 mol ClO 97 g ClO produced 88. (a) First, balance the redox equations needed for the calculation. Oxidation: {HSO - (aq) H O(l) SO - (aq) H (aq) e - } Reduction: {IO - (aq) 6 H (aq) 6 e - I - (aq) H O(l) } 1 Net: HSO - (aq) IO - (aq) SO - (aq) H (aq) I - (aq) The solution volume of 5.00 L contains 9.0 g NaIO. This represents 9.0 g/197.9g/mol NaIO = 0.17 mol NaIO. (b) From the above equation, we need times that molar amount of NaHSO, which is (0.17 mol) = 0.1 mol NaHSO ; the molar mass of NaHSO is g/mol. The required mass then is 0.1(10.06) = 5.9 g. 11
19 For the second process: Oxidation: { I - (aq) I (aq) e - } 5 Reduction: { IO - (aq) 1 H (aq) 10 e - I (aq) 6 H O(l) } 1 Net: 5 I - (aq) IO - (aq) 6 H (aq) I (aq) H O(l) In Step 1, we produced 1 mol of I - for every mole of IO - reactant; therefore we had 0.17 mol I -. In step, we require 1/5 mol IO - for every mol of I -. We require only 1.00 L of the solution in the question instead of the 5.00 L in the first step. 89. Mg(OH) (aq) HCl(aq) MgCl (aq) H O(l) (1) Al(OH)(aq) HCl(aq) AlCl(aq) HO(l) () HCl(aq) NaOH(aq) NaCl(aq) HO(l) () mol L = mol initial moles of HCl = 1 L moles of HCl that reacted with NaOH = moles of HCl left over from reaction with active ingredients = 0.77 mol NaOH 1 mol HCl 1 L 1 mol NaOH L = 6.10 mol moles of HCl that react with active ingredients = mol mol = mol # moles HCl that react with Mg(OH) # moles HCl that = total moles of HCl reacted/used react with Al(OH) # moles HCl that react with Mg(OH) = X grams Mg(OH) 1 mol Mg(OH) mol HCl 58. g Mg(OH) 1 mol Mg(OH) # moles HCl that react with Al(OH) = X grams Al(OH) 1 mol Al(OH) mol HCl g Al(OH) 1 mol Al(OH) 115
20 X (0.500 X) X = 0.108, therefore the mass of Mg(OH) in the sample is grams. % Mg(OH) = (0.108/0.500) 100 = 1.6 %Al = 100 %Mg(OH) = mol AgI 1 mol CHI 1 mol C H O 08. g C H O g AgI.77 g AgI mol AgI 1 mol CHI 1 mol C H O g C H O % C19H16O = g % 1.96 g 9. (a) CaO(s) H O(l) Ca (aq) OH (aq) H PO (aq) OH (aq) PO (aq) H O(l) HPO (aq) OH (aq) PO (aq) H O(l) 5 Ca (aq) PO (aq) OH (aq) Ca 5 (PO ) OH(s) (b) g P 1 mol P 1molPO 5 mol Ca L L 0.97 g P 1 mol P mol PO 1 mol CaO g CaO = g CaO = 0 g = 0.0 kg 1molCa 1molCaO FEATURE PROBLEMS 9. From the volume of titrant, we can calculate both the amount in moles of NaC 5 H 5 and (through its molar mass of g/mol) the mass of NaC 5 H 5 in a sample. The remaining mass in a sample is that of CH8 O (7.11 g/mol), whose amount in moles we calculate. The ratio of the molar amount of CH8 O in the sample to the molar amount of NaC 5 H 5 is the value of x. 116
21 Conversion pathway approach: mol HCl 1 mol NaOH 1 mol NaC H moles of NaC5H 5 = L 1 L soln 1 mol HCl 1 mol NaOH = mol NaC H 5 5 mass of C H O = 0. g sample mol NaC H = g C H O molCH8O 0.110g CH8O 7.11g C H 8 O x = = mol NaC H g NaC H 1 mol NaC H Stepwise approach: mol HCl L = mol HCl 1 L soln - 1 mol NaOH mol HCl = mol NaOH 1 mol HCl - 1 mol NaC5H mol NaOH mol NaC H 1 mol NaOH g NaC5H mol NaC5H g NaC5H5 1 mol NaC H mass of C H O = 0. g sample g NaC H = g C H O molC H O 0.111g C H O = mol C H O g CH8O mol NaC H mol CH8O = For the second sample, parallel calculations give mol NaC 5 H 5, 0.09 g C H x = 1.1. There is rounding error in this second calculation because it is limited to two significant figures. The best answer is from the first run x ~1.0 or 1. The formula is NaC 5 H 5 (THF) First, we balance the two equations. H C O aq CO g H aq e Oxidation: Reduction: MnO s H aq e Mn aq H Ol Net: H C O aq MnO s H aq CO g Mn aq H Ol 8, 117
22 Oxidation: { HCO aq CO g H aq e } 5 Reduction: { MnO aq 8 H aq 5 e Mn aq HOl } Net: 5 H C O aq MnO aq 6 H aq 10 CO g Mn aq 8 H Ol Next, we determine the mass of the excess oxalic acid mol KMnO 1mol MnO 5mol HCO mass HCOHO L 1L 1mol KMnO mol MnO 1 mol HCO HO g HCO HO = 0.97 g HCO HO 1 mol HCO 1 mol HCO HO The mass of H C O H O that reacted with MnO = g 0.97 g = 0.70 g HCO HO 1 mol H C O 1 mol MnO 86.9 g MnO mass MnO = 0.70 g H C O H O g H C O H O 1 mol H C O 1 mol MnO %MnO = 0.85 g MnO 0.85g MnO 100% 91.0% MnO 0.5g sample 97. The molecular formula for CH CH OH is C H 6 O and for CH COOH is C H O. The first step is to balance the oxidation reduction reaction. Oxidation: [C H 6 O H O C H O H e ] Reduction: [Cr O 7 1 H 6e Cr 7 H O] Overall: C H 6 O Cr O 7 16 H C H O Cr 11 H O Before the breath test: 0.75 mg K Cr O 1 g 1 mol 1000 ml 7 = M ml 1000 mg 9.19 g 1 L = M (to 1 sig fig) For the breath sample: 0.05 g CH6O 1 ml blood BrAC = = 100 ml blood 100 ml breath g CH6O ml breath mass C H 6 O = g CH6O ml breath 500. ml breath = g C H 6 O 118
23 Calculate the amount of K Cr O 7 that reacts: 1 mol CH6O mol CrO 7 1 mol K CrO g CH6O mol CH6O 1molCrO g C H O 6 = mol KCrO 7 # mol K Cr O 7 remaining = moles K Cr O 7 before moles K Cr O 7 that reacts moles K Cr O 7 before = 1 g 1 mol 1000 mg 9.19 g mg KCrO 7 =.5 10 mol # mol K Cr O 7 remaining = mol mol = mol concentration of K Cr O 7 after the breath test = mol/0.00 L =.6 10 mol/l = 10 mol/l (to 1 sig fig) 10. The answer is (b). Conversion pathway approach: mol Ba(OH) mol OH 0.00 L = mol 1 L 1mol Ba(OH) Stepwise approach: mol Ba(OH) L = mol Ba(OH) 1 L - mol OH mol Ba(OH) = mol 1mol Ba(OH) 10. The answer is (d), because H SO is a strong diprotic acid and theoretically yields 0.0 mol of H for every 0.10 mol of H SO. 10. The answer is (c). Based on the solubility guidelines in Table 5-1, carbonates (CO - ) are insoluble The answer is (a). Reaction with ZnO gives ZnCl (soluble) and H O. There is no reaction with NaBr and Na SO, since all species are aqueous. By the process of elimination, (a) is the answer Balanced equation: KI Pb(NO ) - Net ionic equation: I Pb PbI (s) KNO PbI 119
24 107. Balanced equation: Na CO HCl NaCl H O CO Net ionic equation: CO H H O (l) CO (g) 108. (a) Balanced equation: Na PO Zn(NO ) 6NaNO Zn (PO ) Net ionic equation: Zn PO Zn (PO ) (s) (b) (c) Balanced equation: NaOH Cu(NO ) Cu(OH) NaNO Net ionic equation: Cu OH Cu(OH) (s) Balanced equation: NiCl Na CO NiCO NaCl Net ionic equation: Ni CO NiCO (s) 109. (a) Species oxidized: N in NO (b) Species reduced: O (c) Oxidizing agent: O (d) Reducing agent: NO (e) Gains electrons: O (f) Loses electrons: NO 110. The answer is (b). The charges need to be balanced on both sides. Using a coefficient of, the charges on both sides of the reaction becomes The answer is (d), 5 ClO - to 1 I. The work to balance the half-reactions is shown below: Reduction: 5ClO H e Cl HO Oxidation: I 6HO IO 10e 1H To combine the above reactions, the oxidation reaction should be multiplied by 5. The combined equation is: Combined: 5ClO I H O 5Cl IO H 11. The answer is (a). The balanced half-reaction is as follows: - NpO H e Np H O 11. (a) False. Based on solubility rules, BaCl dissolves well in water. Therefore, it is a strong electrolyte. (b) True. Since H - is a base, H O is by necessity an acid. It also reduces H - (-1) to H (0). (c) False. The product of such a reaction would be NaCl and H CO, neither of which precipitates out. (d) False. HF is among the strongest of weak acids. It is not a strong acid, because it doesn t completely dissociate. 10
25 (e) True. For every mole of Mg(NO ), there are moles of ions, in contrast to moles of ions for NaNO. 11. (a) No. Oxidation states of C, H or O do not change throughout the reaction. (b) Yes. Li is oxidized to Li and H in H O is reduced from 1 to 0 in H. (c) Yes. Ag is oxidized and Pt is reduced. (d) No. Oxidation states of Cl, Ca, H, and O remain unchanged. 11
TYPES OF CHEMICAL REACTIONS
TYPES OF CHEMICAL REACTIONS Precipitation Reactions Compounds Soluble Ionic Compounds 1. Group 1A cations and NH 4 + 2. Nitrates (NO 3 ) Acetates (CH 3 COO ) Chlorates (ClO 3 ) Perchlorates (ClO 4 ) Solubility
More informationed. Brad Collins Aqueous Chemistry Chapter 5 Some images copyright The McGraw-Hill Companies, Inc. Sunday, August 18, 13
ed. Brad Collins Aqueous Chemistry Chapter 5 Some images copyright The McGraw-Hill Companies, Inc. A solution is a homogenous mixture of 2 or more substances at the molecular level The solute(s) is(are)
More informationChapter 4. Reactions in Aqueous Solution
Chapter 4 Reactions in Aqueous Solution Topics General properties of aqueous solutions Precipitation reactions Acid base reactions Oxidation reduction reactions Concentration of solutions Aqueous reactions
More information1) What is the volume of a tank that can hold Kg of methanol whose density is 0.788g/cm 3?
1) Convert the following 1) 125 g to Kg 6) 26.9 dm 3 to cm 3 11) 1.8µL to cm 3 16) 4.8 lb to Kg 21) 23 F to K 2) 21.3 Km to cm 7) 18.2 ml to cm 3 12) 2.45 L to µm 3 17) 1.2 m to inches 22) 180 ºC to K
More informationReactions in Aqueous Solutions
Reactions in Aqueous Solutions Chapter 4 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A solution is a homogenous mixture of 2 or more substances. The solute
More informationChapter 4 Suggested end-of-chapter problems with solutions
Chapter 4 Suggested end-of-chapter problems with solutions a. 5.6 g NaHCO 1 mol NaHCO 84.01 g NaHCO = 6.69 10 mol NaHCO M = 6.69 10 mol 50.0 m 1000 m = 0.677 M NaHCO b. 0.1846 g K Cr O 7 1 mol K 94.0 g
More informationChapter 4 Reactions in Aqueous Solutions. Copyright McGraw-Hill
Chapter 4 Reactions in Aqueous Solutions Copyright McGraw-Hill 2009 1 4.1 General Properties of Aqueous Solutions Solution - a homogeneous mixture Solute: the component that is dissolved Solvent: the component
More informationAqueous Reactions. The products are just the cation-anion pairs reversed, or the outies (A and Y joined) and the innies (B and X joined).
Aqueous Reactions Defining Aqueous Reactions Aqueous reactions are reactions that take place in water. To understand them, it is important to understand how compounds behave in water. Some compounds are
More informationHonors Unit 4 Homework Packet
1 Honors Homework Packet Reactions in Aqueous Solutions Part I: Aqueous Solns. Part II: Acid/Base Chemistry Part III: Redox Reactions Name: 2 Molarity of Solutions (pg. 2 & 3) Directions: Solve each of
More informationChemistry 101 Chapter 4 STOICHIOMETRY
STOICHIOMETRY Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation. Stoichiometry allows chemists to predict how much of a reactant is necessary
More informationChem 110 General Principles of Chemistry
Chem 110 General Principles of Chemistry Chapter 3 (Page 88) Aqueous Reactions and Solution Stoichiometry In this chapter you will study chemical reactions that take place between substances that are dissolved
More informationHomework #3 Chapter 4 Types of Chemical Reactions and Solution Stoichiometry
Homework #3 Chapter 4 Types of Chemical Reactions and Solution Stoichiometry 13. Determine the concentrations of the solutions Solution A 4 particles 1.0 L Solution B 6 paticles 4.0 L Solution C 4 particles
More informationChapter 4. Reactions In Aqueous Solution
Chapter 4 Reactions In Aqueous Solution I) General Properties of Aqueous Solutions Homogeneous mixture on a molecular level - prop. same throughout - separable by physical means - variable composition
More informationSolution Stoichiometry
Chapter 8 Solution Stoichiometry Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the
More informationPage 1. Exam 2 Review Summer A 2002 MULTIPLE CHOICE. 1. Consider the following reaction: CaCO (s) + HCl(aq) CaCl (aq) + CO (g) + H O(l)
Page 1 MULTIPLE CHOICE 1. Consider the following reaction: CaCO (s) + HCl(aq) CaCl (aq) + CO (g) + H O(l) The coefficient of HCl(aq) in the balanced reaction is. a) 1 b) 2 c) 3 d) 4 e) 0 2. Given the information
More informationDuring photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H 12 O 6 ) according to the reaction:
Example 4.1 Stoichiometry During photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H 12 O 6 ) according to the reaction: Suppose that a particular plant consumes 37.8 g of CO 2
More informationCHEMISTRY - CLUTCH CH.4 - CHEMICAL QUANTITIES & AQUEOUS REACTIONS
!! www.clutchprep.com CONCEPT: MOLARITY Molarity (M) can serve as the connection between the interconversion of to and vice versa. For example, a 5.8 M NaCl solution really means per. ( Molarity = MolesSolute
More informationChapter 4. The Major Classes of Chemical Reactions 4-1
Chapter 4 The Major Classes of Chemical Reactions 4-1 The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions
More informationReactions in Aqueous Solution
1 Reactions in Aqueous Solution Chapter 4 For test 3: Sections 3.7 and 4.1 to 4.5 Copyright The McGrawHill Companies, Inc. Permission required for reproduction or display. 2 A solution is a homogenous
More informationCh 7 Chemical Reactions Study Guide Accelerated Chemistry SCANTRON
Ch 7 Chemical Reactions Study Guide Accelerated Chemistry SCANTRON Name /80 TRUE/FALSE. Write 'T' if the statement is true and 'F' if the statement is false. Correct the False statments by changing the
More informationChapter 8 Chemical Reactions
Chemistry/ PEP Name: Date: Chapter 8 Chemical Reactions Chapter 8: 1 7, 9 18, 20, 21, 24 26, 29 31, 46, 55, 69 Practice Problems 1. Write a skeleton equation for each chemical reaction. Include the appropriate
More information**The partially (-) oxygen pulls apart and surrounds the (+) cation. The partially (+) hydrogen pulls apart and surrounds the (-) anion.
#19 Notes Unit 3: Reactions in Solutions Ch. Reactions in Solutions I. Solvation -the act of dissolving (solute (salt) dissolves in the solvent (water)) Hydration: dissolving in water, the universal solvent.
More informationConcentration of Solutions
Solutions We carry out many reactions in solutions Remember that in the liquid state molecules move much easier than in the solid, hence the mixing of reactants occurs faster Solute is the substance which
More informationH H H H H O H O. Role of Water. Role of Water. Chapter 4. Chemical Reactions in Aqueous Solution H 2 H H H 2 O. Role of H 2 O(l) as solvent.
Role of Water Role of Water Chemical Reactions in Aqueous Solution Role of H 2 O(l) as solvent The polar nature of water molecule Two key features: 1. The distribution of bonding electrons O H covalent
More informationChemistry 150/151 Review Worksheet
Chemistry 150/151 Review Worksheet This worksheet serves to review concepts and calculations from first semester General Chemistry (CHM 150/151). Brief descriptions of concepts are included here. If you
More informationChapter 4 Notes Types of Chemical Reactions and Solutions Stoichiometry A Summary
Chapter 4 Notes Types of Chemical Reactions and Solutions Stoichiometry A Summary 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen s electronegativity is high (3.5) and hydrogen s is low (2.1)
More informationChapter 4: Reactions in Aqueous Solutions
Chapter 4: Reactions in Aqueous Solutions Water 60 % of our bodies heat modulator solvent for reactions covers 70% of Earth Chapter 4 3 types of reactions that occur in H 2 O 1. precipitation 2. acid-base
More informationCHAPTER 4 THREE MAJOR CLASSES OF CHEMICAL REACTIONS
CHAPTER 4 THREE MAJOR CLASSES OF CHEMICAL REACTIONS END OF CHAPTER PROBLEMS 4.1 Plan: Review the discussion on the polar nature of water. Water is polar because the distribution of its bonding electrons
More informationChapter 4 Reactions in Aqueous Solution
Chapter 4 Reactions in Aqueous Solution Homework Chapter 4 11, 15, 21, 23, 27, 29, 35, 41, 45, 47, 51, 55, 57, 61, 63, 73, 75, 81, 85 1 2 Chapter Objectives Solution To understand the nature of ionic substances
More informationChapter 4. Properties of Aqueous Solutions. Electrolytes in Aqueous Solutions. Strong, weak, or nonelectrolyte. Electrolytic Properties
Chapter 4 Reactions in Aqueous Solution Observing and Predicting Reactions How do we know whether a reaction occurs? What observations indicate a reaction has occurred? In your groups, make a list of changes
More informationUnit 4a: Solution Stoichiometry Last revised: October 19, 2011 If you are not part of the solution you are the precipitate.
1 Unit 4a: Solution Stoichiometry Last revised: October 19, 2011 If you are not part of the solution you are the precipitate. You should be able to: Vocabulary of water solubility Differentiate between
More information7/16/2012. Chapter Four: Like Dissolve Like. The Water Molecule. Ionic Compounds in Water. General Properties of Aqueous Solutions
General Properties of Aqueous Solutions Chapter Four: TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY A solution is a homogeneous mixture of two or more substances. A solution is made when one substance
More informationElectrolytes do conduct electricity, in proportion to the concentrations of their ions in solution.
Chapter 4 (Hill/Petrucci/McCreary/Perry Chemical Reactions in Aqueous Solutions This chapter deals with reactions that occur in aqueous solution these solutions all use water as the solvent. We will look
More informationChapter 4; Reactions in Aqueous Solutions. Chapter 4; Reactions in Aqueous Solutions. V. Molarity VI. Acid-Base Titrations VII. Dilution of Solutions
Chapter 4; Reactions in Aqueous Solutions I. Electrolytes vs. NonElectrolytes II. Precipitation Reaction a) Solubility Rules III. Reactions of Acids a) Neutralization b) Acid and Carbonate c) Acid and
More informationChapter 04. Reactions in Aqueous Solution
Chapter 04 Reactions in Aqueous Solution Composition Matter Homogeneous mixture Contains One visible distinct phase Uniform properties throughout Two or more substances that are mixed together Substances
More informationChapter 4. Types of Chemical Reactions and Solution Stoichiometry
Chapter 4 Types of Chemical Reactions and Solution Stoichiometry Chapter 4 Table of Contents 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition
More informationChapter 4. Aqueous Reactions and Solution Stoichiometry
Sample Exercise 4.1 (p. 127) The diagram below represents an aqueous solution of one of the following compounds: MgCl 2, KCl, or K 2 SO 4. Which solution does it best represent? Practice Exercise 1 (4.1)
More informationCHEMISTRY - ZUMDAHL 2E CH.6 - TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY
!! www.clutchprep.com CONCEPT: MOLARITY Molarity (M) can serve as the connection between the interconversion of to and vice versa. For example, a 5.8 M NaCl solution really means per. ( Molarity = MolesSolute
More informationTypes of chemical reactions
PowerPoint to accompany Types of chemical reactions Chapters 3 & 16.1 M. Shozi CHEM110 / 2013 General Properties of Aqueous Solutions Solutions are mixtures of two or more pure substances. The solvent
More informationCHM152LL Solution Chemistry Worksheet
Name: Section: CHM152LL Solution Chemistry Worksheet Many chemical reactions occur in solution. Solids are often dissolved in a solvent and mixed to produce a chemical reaction that would not occur if
More informationBalancing Equations Notes
. Unit 6 Chemical Equations and Reactions What is a Chemical Equation? A Chemical Equation is a written representation of the process that occurs in a chemical reaction. A chemical equation is written
More informationBalancing Equations Notes
. Unit 9 Chemical Equations and Reactions What is a Chemical Equation? A Chemical Equation is a written representation of the process that occurs in a chemical reaction. A chemical equation is written
More informationThe solvent is the dissolving agent -- i.e., the most abundant component of the solution
SOLUTIONS Definitions A solution is a system in which one or more substances are homogeneously mixed or dissolved in another substance homogeneous mixture -- uniform appearance -- similar properties throughout
More informationCH 4 AP. Reactions in Aqueous Solutions
CH 4 AP Reactions in Aqueous Solutions Water Aqueous means dissolved in H 2 O Moderates the Earth s temperature because of high specific heat H-bonds cause strong cohesive and adhesive properties Polar,
More informationCHEM 1413 Chapter 4 Homework Questions TEXTBOOK HOMEWORK
CHEM 1413 Chapter 4 Homework Questions TEXTBOOK HOMEWORK Chapter 3 3.68 Calculate each of the following quantities: (a) Mass (g) of solute in 185.8 ml of 0.267 M calcium acetate (b) Molarity of 500. ml
More informationReactions in Aqueous Solution
Reactions in Aqueous Solution Chapter 4 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A solution is a homogenous mixture of 2 or more substances The solute
More informationPart One: Ions in Aqueous Solution
A. Electrolytes and Non-electrolytes. CHAPTER FOUR: CHEMICAL REACTIONS Part One: Ions in Aqueous Solution 1. Pure water does not conduct electric current appreciably. It is the ions dissolved in the water
More informationReactions in Aqueous Solutions
Reactions in Aqueous Solutions 1 Chapter 4 General Properties of Aqueous Solutions (4.1) Precipitation Reactions (4.2) Acid-Base Reactions (4.3) Oxidation-Reduction Reactions (4.4) Concentration of Solutions
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) When the following equation is balanced, the coefficients are. 1) NH3 (g) + O2 (g) NO2
More informationChapter 6. Types of Chemical Reactions and Solution Stoichiometry
Chapter 6 Types of Chemical Reactions and Solution Stoichiometry Chapter 6 Table of Contents (6.1) (6.2) (6.3) (6.4) (6.5) (6.6) (6.7) (6.8) Water, the common solvent The nature of aqueous solutions: Strong
More informationAP Chemistry Note Outline Chapter 4: Reactions and Reaction Stoichiometry:
AP Chemistry Note Outline Chapter 4: Reactions and Reaction Stoichiometry: Water as a solvent Strong and Weak Electrolytes Solution Concentrations How to Make up a solution Types of Reactions Introduction
More informationChapter 4 Types of Chemical Reaction and Solution Stoichiometry
Chapter 4 Types of Chemical Reaction and Solution Stoichiometry Water, the Common Solvent One of the most important substances on Earth. Can dissolve many different substances. A polar molecule because
More informationCHAPTER 4 TYPES OF CHEMICAL REACTIONS & SOLUTION STOICHIOMETRY
Advanced Chemistry Name Hour Advanced Chemistry Approximate Timeline Students are expected to keep up with class work when absent. CHAPTER 4 TYPES OF CHEMICAL REACTIONS & SOLUTION STOICHIOMETRY Day Plans
More informationSolubility & Net Ionic review
Solubility & Net Ionic review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which of the following statements is/are correct? 1. All ionic compounds
More informationChapter 4. Reactions in Aqueous Solution
Chapter 4. Reactions in Aqueous Solution 4.1 General Properties of Aqueous Solutions A solution is a homogeneous mixture of two or more substances. A solution is made when one substance (the solute) is
More informationReaction Writing Sheet #1 Key
Reaction Writing Sheet #1 Key Write and balance each of the following reactions and indicate the reaction type(s) present: 1. zinc + sulfur zinc sulfide 8 Zn (s) + S 8 (s) 8 ZnS (s) synthesis 2. potassium
More informationEXPERIMENT 10: Precipitation Reactions
EXPERIMENT 10: Precipitation Reactions Metathesis Reactions in Aqueous Solutions (Double Displacement Reactions) Purpose a) Identify the ions present in various aqueous solutions. b) Systematically combine
More informationChapter 4. Reactions in Aqueous Solution. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO
Lecture Presentation Chapter 4 in Solution 2012 Pearson Education, Inc. John D. Bookstaver St. Charles Community College Cottleville, MO Properties of Solutions Solute: substance in lesser quantity in
More informationRevision of Important Concepts. 1. Types of Bonding
Revision of Important Concepts 1. Types of Bonding Electronegativity (EN) often molecular often ionic compounds Bonding in chemical substances Bond energy: Is the energy that is released when a bond is
More information11/3/09. Aqueous Solubility of Compounds. Aqueous Solubility of Ionic Compounds. Aqueous Solubility of Ionic Compounds
Aqueous Solubility of Compounds Not all compounds dissolve in water. Solubility varies from compound to compound. Chapter 5: Chemical Reactions Soluble ionic compounds dissociate. Ions are solvated Most
More informationReactions in Aqueous Solutions
Chapter 4 Reactions in Aqueous Solutions Some typical kinds of chemical reactions: 1. Precipitation reactions: the formation of a salt of lower solubility causes the precipitation to occur. precipr 2.
More informationCH 221 Chapter Four Part II Concept Guide
CH 221 Chapter Four Part II Concept Guide 1. Solubility Why are some compounds soluble and others insoluble? In solid potassium permanganate, KMnO 4, the potassium ions, which have a charge of +1, are
More informationChapter 4 Electrolytes and Aqueous Reactions. Dr. Sapna Gupta
Chapter 4 Electrolytes and Aqueous Reactions Dr. Sapna Gupta Aqueous Solutions Solution - a homogeneous mixture of solute + solvent Solute: the component that is dissolved Solvent: the component that does
More informationChemical reactions describe processes involving chemical change
1.1 Chemical Reactions 1.2 Chemical Equations Chemical reactions describe processes involving chemical change The chemical change involves rearranging matter Converting one or more pure substances into
More informationCh 4-5 Practice Problems - KEY
Ch 4-5 Practice Problems - KEY The following problems are intended to provide you with additional practice in preparing for the exam. Questions come from the textbook, previous quizzes, previous exams,
More informationGeneral Chemistry 1 CHM201 Unit 2 Practice Test
General Chemistry 1 CHM201 Unit 2 Practice Test 1. Which statement about the combustion of propane (C 3H 8) is not correct? C 3H 8 5O 2 3CO 2 4H 2O a. For every propane molecule consumed, three molecules
More informationAP Chapter 4 Study Questions
Class: Date: AP Chapter 4 Study Questions True/False Indicate whether the statement is true or false. 1. Ca(OH) 2 is a strong base. 2. The compound HClO 4 is a weak acid. 4. The compound NH 4 Cl is a weak
More informationCHEM134- Fall 2018 Dr. Al-Qaisi Chapter 4b: Chemical Quantities and Aqueous Rxns So far we ve used grams (mass), In lab: What about using volume in lab? Solution Concentration and Solution Stoichiometry
More informationNet Ionic Reactions. The reaction between strong acids and strong bases is one example:
Net Ionic Reactions Model 1 Net Ionic Reactions. Net ionic reactions are frequently used when strong electrolytes react in solution to form nonelectrolytes or weak electrolytes. These equations let you
More informationReactions in Aqueous Solution
Reading Assignments: Reactions in Aqueous Solution Chapter 4 Chapter 4 in R. Chang, Chemistry, 9 th Ed., McGraw-Hill, 2006. or previous editions. Or related topics in other textbooks. Consultation outside
More information1. Hydrochloric acid is mixed with aqueous sodium bicarbonate Molecular Equation
NAME Hr Chapter 4 Aqueous Reactions and Solution Chemistry Practice A (Part 1 = Obj. 1-3) (Part 2 = Obj. 4-6) Objective 1: Electrolytes, Acids, and Bases a. Indicate whether each of the following is strong,
More information2H 2 (g) + O 2 (g) 2H 2 O (g)
Mass A AP Chemistry Stoichiometry Review Pages Mass to Mass Stoichiometry Problem (Review) Moles A Moles B Mass B Mass of given Amount of given Amount of unknown Mass of unknown in grams in Moles in moles
More informationUNIT (4) CALCULATIONS AND CHEMICAL REACTIONS
UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. Atomic mass
More informationAssignment 04 (A) a) ii and iii b) i, ii, and iii c) i, iv, and v d) iii e) ii (These are molecular compounds.)
Assignment 04 (A) 1- Which of the following are nonelectrolytes in water? (i) HF (ii) ethanol, C 2 H 5 OH (iii) C 12 H 22 O 11 (iv) KClO 3 (v) Cu(NO 3 ) 2 a) ii and iii b) i, ii, and iii c) i, iv, and
More informationChapter 4 Aqueous Reactions and Solution Stoichiometry
Chapter 4 Aqueous Reactions and Solution Stoichiometry 4.1 General Properties of Aqueous Solutions What is a solution? How do you identify the following two? Solvent. Solute(s). Dissociation. What is it?
More informationSOLUBILITY REVIEW QUESTIONS
Solubility Problem Set 1 SOLUBILITY REVIEW QUESTIONS 1. What is the solubility of calcium sulphate in M, g/l, and g/100 ml? 2. What is the solubility of silver chromate? In a saturated solution of silver
More informationGeneral Chemistry. Contents. Chapter 5: Introduction to Reactions in Aqueous Solutions. Electrolytes. 5.1 The Nature of Aqueous Solutions
General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 5: Introduction to Reactions in Aqueous Solutions Philip Dutton University of Windsor, Canada N9B 3P4
More information15.0 g Fe O 2 mol Fe 55.8 g mol Fe = g
CHAPTER Practice Questions.1 1 Mg, O, H and Cl (on each side).. BaCl (aq) + Al (SO ) (aq) BaSO (s) + AlCl (aq).5 0.15 mol 106 g mol 1 = 1. g 15.0 g Fe O mol Fe 55.8 g mol Fe = 10.9 g 1 159.7 g mol FeO
More informationTYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY
CHAPTER TYPES OF CHEMICA REACTIONS AND SOUTION STOICHIOMETRY Questions 17. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that
More information9/24/12. Chemistry Second Edition Julia Burdge. Reactions in Aqueous Solutions
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
More informationGeneral Chemistry. Chapter 5: Introduction to Reactions in Aqueous Solutions. Principles and Modern Applications Petrucci Harwood Herring 8 th Edition
General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 5: Introduction to Reactions in Aqueous Solutions Philip Dutton University of Windsor, Canada N9B 3P4
More informationChapter Four: Reactions in Aqueous Solution
Chapter Four: Reactions in Aqueous Solution Learning Outcomes: Identify compounds as acids or bases, and as strong, weak, or nonelectrolytes Recognize reactions by type and be able to predict the products
More informationSolubility Rules See also Table 4.1 in text and Appendix G in Lab Manual
Ch 4 Chemical Reactions Ionic Theory of Solutions - Ionic substances produce freely moving ions when dissolved in water, and the ions carry electric current. (S. Arrhenius, 1884) - An electrolyte is a
More informationChemistry deals with matter and its changes CHEMICAL REACTIONS
Chemistry deals with matter and its changes CHEMICAL REACTIONS CHEMICAL EQUATIONS N 2 + 3 H 2 2 NH 3 2 N 6 H 2 N 6 H reactants products balanced means equal numbers of atoms of each element on each side
More informationAP Chemistry Honors Unit Chemistry #4 2 Unit 3. Types of Chemical Reactions & Solution Stoichiometry
HO AP Chemistry Honors Unit Chemistry #4 2 Unit 3 Chapter 4 Zumdahl & Zumdahl Types of Chemical Reactions & Solution Stoichiometry Students should be able to:! Predict to some extent whether a substance
More informationChem 1A Dr. White Fall Handout 4
Chem 1A Dr. White Fall 2014 1 Handout 4 4.4 Types of Chemical Reactions (Overview) A. Non-Redox Rxns B. Oxidation-Reduction (Redox) reactions 4.6. Describing Chemical Reactions in Solution A. Molecular
More informationReview 7: Solubility Equilibria
Review 7: Solubility Equilibria Objectives: 1. Be able to write dissociation equations for ionic compounds dissolving in water. 2. Given Ksp, be able to determine the solubility of a substance in both
More informationChemical Reaction Defn: Chemical Reaction: when starting chemical species form different chemicals.
Chemistry 11 Notes on Chemical Reactions Chemical Reaction Defn: Chemical Reaction: when starting chemical species form different chemicals. Evidence to indicate that a chemical reaction has occurred:
More informationUnit 1 - Foundations of Chemistry
Unit 1 - Foundations of Chemistry Chapter 2 - Chemical Reactions Unit 1 - Foundations of Chemistry 1 / 42 2.1 - Chemical Equations Physical and Chemical Changes Physical change: A substance changes its
More information2. The reaction of carbon monoxide and diiodine pentoxide as represented by the equation
1. The complete combustion of phenylhydrazine, C 6 H 5 NHNH 2, with the oxidizer dinitrogen tetraoxide is shown in the equation C 6 H 5 NHNH 2 + N 2 O 4 CO 2 + H 2 O + N 2 When balanced, the sum of all
More informationReactions in aqueous solutions Precipitation Reactions
Reactions in aqueous solutions Precipitation Reactions Aqueous solutions Chemical reactions that occur in water are responsible for creation of cenotes. When carbon dioxide, CO2, dissolves in water, the
More informationReview Questions (Exam II)
Announcements Exam tonight, 7-8:15pm (locations posted on website) Conflict Exam, 5:15-6:30pm (114 Transportation Bldg) No lab this week! Start new material on Thursday (read chapter 10!) Review Questions
More information7.01 Chemical Reactions
7.01 Chemical Reactions The Law of Conservation of Mass Dr. Fred Omega Garces Chemistry 100 Miramar College 1 Chemical Reactions Making Substances Chemical Reactions; the heart of chemistry is the chemical
More informationChapter 7 Chemical Reactions
Chapter 7 Chemical Reactions Evidence of Chemical Change Release or Absorption of Heat Color Change Emission of Light Formation of a Gas Formation of Solid Precipitate Tro's "Introductory 2 How Do We Represent
More informationPractice questions for Chapter 4
Practice questions for Chapter 4 1. An unknown substance dissolves readily in water but not in benzene (a nonpolar solvent). Molecules of what type are present in the substance? A) neither polar nor nonpolar
More informationPractice Worksheet - Answer Key. Solubility #1 (KEY)
Practice Worksheet - Answer Key Solubility #1 (KEY) 1 Indicate whether the following compounds are ionic or covalent a) NaCl ionic f) Sr(OH) 2 ionic b) CaBr 2 ionic g) MgCO 3 ionic c) SO 2 covalent h)
More informationReactions in Aqueous Solutions
Copyright 2004 by houghton Mifflin Company. Reactions in Aqueous Solutions Chapter 7 All rights reserved. 1 7.1 Predicting if a Rxn Will Occur When chemicals are mixed and one of these driving forces can
More informationTypes of Reactions: Reactions
1 Reactions On the A.P. Test there will be one question (question #4) that will say: Give the formulas to show the reactants and the products for the following chemical reactions. Each occurs in aqueous
More informationReaction Classes. Precipitation Reactions
Reaction Classes Precipitation: synthesis of an ionic solid a solid precipitate forms when aqueous solutions of certain ions are mixed AcidBase: proton transfer reactions acid donates a proton to a base,
More informationName AP Chemistry September 30, 2013
Name AP Chemistry September 30, 2013 AP Chemistry Exam Part I: 40 Questions, 40 minutes, Multiple Choice, No Calculator Allowed Bubble the correct answer on the blue side of your scantron for each of the
More information