容積莫耳濃度

1、目錄Reactions in Aqueous Solution1 Precipitation Reactions 沈澱反應4 Acid-base reactions6 Oxidation-reduction Reactions Redox reactions12Reactions in Aqueous SolutionAbstract： 4-1： Solute Concentration： Molarity (M)： 容積莫耳濃度 Normality (N)： 當量濃度 4-2： Precipitation reactions fig 4.4 4-3： Acid-base reactions

2、Acid Base Titration 4-4： oxidation-reduction reactions oxidation number 氧化數 Balancing half-equations (oxidation or reduction) Balancing Redox equations 4-1 Solute Concentrations；Molarity (M) 容積莫耳濃度 M = =符號： 【 】例： A solution containing 1.20 mol of substance A in 2.50 L of solution. 【A】= = Fig 4.1 Pre

3、paring one liter of 0.100M potassium chromate鉻酸鉀 a) 取1個1 L之定量瓶volumetric flask，置入19.4 g b) 加入少量水，完全溶解c) 加足量水主頸線，搖晃均勻Ex 4-1： A dilute nitric acid 6.0 mol per liter of solution 6.0 Ma) How many moles of are in 75 ml of this solution?b) What volume of dilute nitric acid must be taken to contain one mol

4、e of?Sol： a) 6.0 mole ： 1 L = n ： 75 ml n = b) 6.0 mole ： 1 L = n ： x L x = An ionic solid dissolves in water, the cations and anions separate from each other + molarity = molarity 【】 = 【】 molarity = 2 molarity 【】 = 2【】例： + molarity = 3 molarity molarity = 1 molarity + molarity = 2 molarity molarity

5、 = 3 molarity Ex 4-2： Give the concentration, in moles per liter, of each ion in a) 0.08 M b) 0.40 M Sol： a) + 【】= 2 0.08 = 0.16 M 【】= 1 0.08 = 0.08 M b) + 【】 = 1 0.40 = 0.40 M 【】 = 3 0.40 = 1.2 M Precipitation Reactions 沈澱反應 Sometimes when water solutions of two different ionic compounds are mixed,

6、 an insoluble solid separate out of solution. precipitate Ksp solubility product 溶解度積Fig 4.3Group I cations Group II cations Transition metal cations Fig 4.2 Ex 4-3： 下列二組溶液混合是否生成沈澱？ (用fig 4.3) a) and b) and Sol： a) NO. NO. NO Precipitate b) NO. Yes. 沈澱 Not ionic equation + + Reactants： ； Product： ；

7、+ + + + + Not ionic equation + Ex 4-4： Write a net ionic equation of the followings：a) and b) and c) and Sol：a) No. Yes. + b) Yes. + Yes. + c) No. No. No equation.此章中，都用 net ionic equation 表示之 Stoichiometry 化學計量 Mole-mass calculationEx 4-5： 與混合，生成紅色沈澱 a) Net ionic equation b) 若使用 30.0 ml 0.125 M 生成若

8、干沈澱？ c) 50.00 ml 0.200 M 與 30.0 ml 0.125 M 反應生成沈澱若干Sol： a) + b) = 0.125 = = = c) + n 0.01 1 ： 3 = n ： 0.0100 為 limiting reactant = Acid-base reactions Acidic solution： 1. sour taste 2. Litmus(石蕊) turns from blue to red. Basic solution： 1. slippery feeling 2. Litmus turns from red to blue.Svcmte Arrh

9、enius： Acid-base definitions： An acid is a species that produces ions in water solution. An base is a species that produces ions in water solution.-： (Ch. 13) 酸：反應中提供者 鹼：反應中接受者 共軛酸鹼對 鹼2 酸2 + + 酸1 鹼1：酸：反應中接受共用電子對者 鹼：反應中提供共用電子對者 鹼 酸 Strong and Weak Acids and Bases.Strong base： Ex. + 開始 0.10 mol 平衡 0.1

10、0 mol 0.10 mol 0.10 mol ionize completely.Weak acid： Ex. + 開始 0.10 mol 平衡 0.09 mol 0.01 mol 0.01 mol partially ionized.Strong base： Ex. + + ionize completely.Weak base： Ex. + + 開始 0.10 mol 平衡 0.09 mol 0.01 mol 0.01 mol partially ionized.Table 4-1 Common Strong acids and basesAcidBase氫氯酸Hydrochloric

11、acidHydrobromic acidHydroiodic acidNitric acid過氯酸Porchloric acidStrontium hydroxideSulfuric acidAmine 弱鹼 + + Strong electrolytes 強電解質 completely ionizedWeak electrolytes partially ionized Equations for Acid-Base reactions.1. Strong acid-Strong base + + neutralization： + spectator ions 適合所有強酸強鹼反應2. W

12、eak acid-Strong base + +) + net reaction： + + Ex： + + + 3. Strong acid-Weak base + + +) + + Ex： + + Table 4.2： Types of acid-base reactions.ReactantsReacting SpeciesNet ionic reactionS acid S base, + W acid S base, + + S acid W base, + Ex 4-7 net ionic equationa) Hyopchlorous acid次氯酸 and calcium hyd

13、roxideb) Ammonia with perchloric acid 過氯酸c) Hydriodic acid () with sodium hydroxideAns： a) weak acid strong base + + b) weak base strong acid + c) strong acid strong base + Acid-base titrationTitration： Measuring the volume of a standard solution (a solution of known concentration) required to react

14、 with a measured amount of sample.Fig 4.7 titration of vinegar () with sodium hydroxide Step 1. 三角錐瓶加入已知量之 + 指示劑 (1-2 drop) Step 2. 以滴定管滴入 酸管：全玻璃 鹼管：玻璃 + 塑膠軟件 Step 3. 滴定至滴定泈點 () + + Equivalence point：當量點The number of equivalent mole of base equal to the number of equivalent mole of acid. 1M = 1N 1M

15、= 1N 1M = 2NEnd point：滴定泈點The point of the indicator change color.Neutralization point：中和點 pH = 7 Ex 4-8： 以25.0 ml 0.500M titrate：a) 15.0 ml , 求之molarity (M) ?b) 15.0 ml of weak acid , 之molarity (M) ? + 2 + 2c) 2.5 aspirin, 求 aspirin中acetylsalicylic acid 乙醯水楊酸之含量? + + Sol： a) + 1 M = 1 N 15.0 = 0.50

16、0 25.0 0.500 M = 0.500 N = 0.833 N 1 M = 1 N = 0.833 M b) + 2 + 2 15.0 = 0.500 25.0 = 0.833 N 1 M = 2 N or：1 ： 2 = ： 1 ： 2 = ： = 15.0 ： 0.500 = = 0.417 M c) + + = = 0.500 = 0.0125 mol = 0.0125 180.15 = 2.25 g Purity (%) = 100 = 90.0 % Oxidation-reduction Reactions Redox reactions involves a transfer

17、 of electrons between two species. 例：金屬與酸之反應 Zn 與 HCl 反應Oxidation： a specie loses electrons + The ion or molecule donates electrons Reducing agentReduction： a specie gains electrons + The ion or molecule accepts electrons Oxidizing agentRedox reaction： + + For a redox reaction：1. Oxidation and reduc

18、tion occur together.2. There is no net change in the number of electrons in a redox reaction. Oxidation number 氧化數 Oxidation number Ex.1. An element in an elementary substance 0 0 0 K 02. Monoatomic ion the charge of the ion -1 +3 -23. IA element in compound +1 +1 +1 IIA +2 +2 +2 VIIA -1 -1 -1 Oxyge

19、n in ordinary compound -2 O：-2 例外： O：-1 Hydrogen in ordinary compound +1 H：+1 例外： H：-1 H：-14. The sum of the oxidation numbers in a neutral species is O, in a polyatomic ion, it is equal to the charge of that ion.Ex 4-8 求氧化數 磷酸鈉中之P. 亞磷酸根中之P.Sol： Oxidation and Reduction (陽極)Oxidation： an increase in

20、oxidation number 氧化數增加，失去電子，還原劑Reduction： a decrease in oxidation number 氧化數減少，獲得電子，氧化劑 (陰極) +1 reduction 0 + + 0 oxidation +2 +2 reduction 0 + + Redox reaction 0 oxidation +3 0 + + +4 0 +4 Balancing half-equation (Oxidation or Reduction)Oxidation 左、右元素一致 +2 +1 +3 + Reduction 左、右元素一致 0 -12 -1 + 2 左、

21、右元素不一致平衡步驟：1. Balance the atoms of the element being oxidized or reduced.2. Balance the oxidation number by adding electrons. 氧化數多的一邊加電子3. Balance charge by adding ions in acidic solution, ions in basic solution. 酸溶液加，鹼溶液加4. Balance hydrogen by adding molecules H 不夠一邊加5. Balance Oxygen numberEx 4-9 Balance the following half-equationsA) B) A) a) +7 Reduction +2 b) 氧化數多的一邊加 左：+5 + 5 c) 酸性溶液加 左：