Complete the following table: Experiment 10. Red Cabbage Chemistry...

Complete the following table: Experiment 10. Red Cabbage Chemistry

[H3O+]	[OH-]	pH	pOH	Acidic/Basic/Neutral
2.35 ´ 10-3
	4.93 ´ 10-8
		8.32
			10.27
3.72 ´ 10-10
	1.00 ´ 10-7
		2.58
			5.26
1.00 ´ 10-3
	4.27 ´ 10-2
		8.00
			2.04

 

 

 

Experiment 11. Buffer pH and Property (Worksheet)

 

Student name (Print):_______________________
I. Suggestion of Use (1) This is a dry lab.  (2) Instructors may assign different problems for different students in the same group to work on in the lab. (3) If preferred, instructors may randomize those numbers in the given problems and then assign to students.
II. Objectives (1) Identify necessary components for the constitution of a buffer; (2) Calculate buffer pH using Henderson-Hasselbalch equation; (3) Calculate buffer pH after the addition of strong acid/base.
III. Principles, Procedures, and Practices
Buffers are widely used in chemical reaction where solution pH is required to be constant. With the addition of limited amount of acid/base, buffer solution pH does not change significantly.   III.1 Constitution of a buffer To constitute a buffer, a pair of conjugate weak acid and weak base is required. The conjugate weak acid and the conjugate weak base has a difference of only one H+. For example, HC2H3O (acetic acid) and C2H3O- (acetate) will be t constitute a buffer.
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Question: Identify which of the following combination will be able to constitute a buffer.
Problem 1		Problem 2				Problem 3
HCN (aq) + HSCN (aq)	yes;  no.	HNO3 (aq) + NO3- (aq)		yes;  no.		HClO3 (aq) + ClO3- (aq)	yes;  no.
HC2H3O (aq) + C2H3O- (aq)	yes;  no.	HNO2 (aq) +NaNO2(aq)		yes;  no.		HClO (aq) + KClO (aq)	yes;  no.
Cl- (aq) + HCl (aq)	yes;  no.	NH4+(aq) + NH3 (aq)		yes;  no.		NaH2PO4 (aq) + Na2HPO4 (aq)	yes;  no.
H2S (aq) + Na2S (aq)	yes;  no.	H2SO4 (aq) +Na2SO4(aq)		yes;  no.		H2CO3 (aq) +NaHCO3(aq)	yes;  no.
III.2 Buffer solution pH calculation For buffer solution pH-related calculation, Henderson-Hasselbalch equation is often used.    However, one must always keep in mind that Henderson-Hasselbalch equation is good for use only when,     Example: What is the pH of a buffer solution composed of 0.12 M NaNO2 and 0.15 M HNO2 at 25°C? It is known Ka for HNO2 is 4.5×10-4 at 25°C. Solution: HNO2 is a weak acid. The NO2- derived from NaNO2 is a weak base. HNO2 and NO2- are a conjugate pair. [base]ini=[NO2-]ini=0.12M; [acid]ini=[HNO2]ini=0.15M   Attention: Sometimes, mass may be given instead of mole or molarity. In that case, you have to calculate molarity first.
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Question: Calculate the buffer solution pH.
Problem 1		Problem 2				Problem 3
What is the pH of a buffer solution composed of 0.0200 M boric acid and 0.020 M sodium borate at 25 °C? Boric acid Ka=5.8×10-10.		A 500-ml buffer solution contains 15.0 g NH4Cl and 6.15 g 30% (by mass) NH3 solution at 25 °C. What is the pH of this buffer? NH4+ Ka=5.8×10-10.  MMNH3=17.03g/mol, MMNH4Cl=53.49g/mol.				What is the pH of a 250-ml buffer containing 0.15 mol NaH2PO4 and 0.22 mol Na2HPO4 at 25 °C. Ka for H2PO4- is 6.2×10-8.
Solution:
III.3 Buffer solution Preparation for a Required pH Buffer preparation has two common ways: (1) direct mixing conjugate acid and base; and (2) partial neutralization.
Way 1 - Direct mixing conjugate acid and base				Way 2- Partial neutralization
Example: It is known nitrous acid (HNO2) pKa = 3.35. If you intend to use HNO2 and NaNO2 to buffer solution at pH = 4.00, what is molar ration of NaNO2 and HNO2 should be? To a solution containing 15.0 g HNO2, what mass of NaNO2 you need to add? Solution:     Hence,         = 6.9				Example: It is known nitrous acid (HNO2) pKa = 3.35. If you have 200 ml 0.100 M HNO2 solution, what volume of 0.500 M NaOH you need to add in order to have a buffer with pH at 3.00? Solution:   With partial neutralization, the solution will include HNO2 and NO2-, which can constitute a buffer.     Hence,     Assuming ml of 0.100 M NaOH is required to add.   Based on above chemical equation, moles of NaOH quals to moles of NaNO2 generated.
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Question:
Problem 1			Problem 2		Problem 3
It is known H2PO4- pKa=7.20. If you intend to use NaH2PO4 and Na2HPO4 to make a buffer with a pH at 7.00, to a solution containing 10.0 g NaH2PO4, what mass of Na2HPO4 should be added?  MMNaH2PO4=120.0 g/mol, MMNa2HPO4=142.0 g/mol.			It is known H2PO4- pKa=7.20. If you have 150 ml 0.150 M NaH2PO4 solution, what volume of 0.200 M NaOH you need to add to make a buffer with pH at 8.00? NaH2PO4(aq) + NaOH(aq)                      à Na2HPO4(aq) + H2O(l)		It is known H2PO4- pKa=7.20. If you intend to use NaH2PO4 and Na2HPO4 to make a buffer with a pH at 7.50, to a solution containing 22.0 g NaH2PO4, what mass of Na2HPO4 should be added? MMNaH2PO4=120.0 g/mol, MMNa2HPO4=142.0 g/mol.
Solution:
III.4 Buffer property Buffer solution can react with both acid and base to keep pH relative stable, while water does not have this property.
Adding strong acid to pure water Example: Pure water pH=7.00 at 25 °C. If 1.00 ml 0.100 M HCl is added into 100 ml pure water, what is the pH? Solution: HCl is a strong acid. It 100% ionize after being added into water.
Adding strong base to pure water Example: Pure water pH=7.00 at 25 °C. If 1.00 ml 0.100 M NaOH is added into 100 ml pure water, what is the pH? Solution: NaOH is a strong acid. It 100% ionize after being added into water.
Adding strong acid into buffer Example: A 100-ml buffer solution containing 0.100 M HNO2 and 0.100 M NaNO2. What is the pH before and after adding 1.00 ml 0.100 M HCl? HNO2 pKa = 3.35. Solution: Before adding HCl     After adding HCl     Reaction between NO2- and HCl is 1:1 molar ratio reaction. Moles of HCl reacted equal to moles of NO2- consumed and moles of HNO2 generated.
Adding strong base into buffer Example: A 100-ml buffer solution containing 0.100 M HNO2 and 0.100 M NaNO2. What is the pH after adding 1.00 ml 0.100 M NaOH? HNO2 pKa = 3.35. Solution:   Reaction between HNO2 and NaOH is 1:1 molar ratio reaction. Moles of NaOH reacted equal to moles of HNO2 consumed and moles of NO2- generated.
Summary Adding 1.00 ml 0.100 M HCl into 100ml water results a pH change from 7.00 to 3.00.                          (DpH = -4.00) Adding 1.00 ml 0.100 M NaOH into 100ml water results a pH change from 7.00 to 11.00.                    (DpH = +4.00) Adding 1.00 ml 0.100 M HCl into 100ml buffer results a pH change from 3.35 to 3.34.                          (DpH = -0.01) Adding 1.00 ml 0.100 M NaOH into 100ml water results a pH change from 3.35 to 3.36.                       (DpH = +0.01) Attention: If too much acid/base is added resulting in [base]ini/[acid]ini <0.1 or >10, one can't use Henderson-Hasselbalch equation to calculate buffer pH any more. Instead, common ion effect with I.C.E. table should be used for pH calculation.
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Questions:
Problem 1			Problem 2		Problem 3
A 100-ml buffer solution containing 0.200 M HC2H3O2 (acetic acid) and 0.300 M NaC2H3O2. What is the pH before and after adding 2.00 ml 0.100 M HCl? Acetic acid pKa = 4.74.			A 100-ml buffer solution containing 0.200 M HC2H3O2 (acetic acid) and 0.150 M NaC2H3O2. What is the pH before and after adding 3.00 ml 0.200 M HCl? Acetic acid pKa = 4.74.		A 100-ml buffer solution containing 0.250 M HC2H3O2 (acetic acid) and 0.200 M NaC2H3O2. What is the pH before and after adding 5.00 ml 0.200 M NaOH? Acetic acid pKa = 4.74.
Solution:

 

 

Experiment 12. Solubility Product and Common Ion Effect

 

 

Student Name (Print):__________________________   Pre-lab Study Questions (Submit upon arrival for experiment)

1. Write the solubility product Ksp for the following compounds. a. FeS                                      b. AgNO3                                        c. Mg(OH)2     d. Ca3(PO4)2     2. Consider a saturated solution of magnesium hydroxide Mg(OH)­2. Answer the following questions. a. What would happen to the solubility of Mg(OH)2 if the solution were made up in 0.01 M Mg(NO3)2 instead of pure water? Explain.       b. What would happen to the solubility of Mg(OH)2 if the solution were made up in 0.01 M NaOH instead of pure water? Explain.          c. What would happen to the solubility of Mg(OH)2 if the solution were made up in 0.01 M NaCl instead of pure water? Explain.

 

 

 

 

Experiment 12. Solubility Product and Common Ion Effect Student Name (Print):__________________________
I. Objectives
(1) Measure the solubility of calcium hydroxide Ca(OH)2 by titration (2) Determine the common ion effect by measuring the solubility of Ca(OH)2 with and without the presence of excess Ca2+
II. Materials
500-mL beaker 250-mL Erlenmeyer flask Buret Solid Ca(OH)2 1 M HCl Phenolphthalein indicator
III. Safety Information and Requirements
(1) Wear goggles or safety glasses;  (2) Dump waste according to instruction;
IV. Procedure   and Data Collection
IV.1 Information An ionic compound is formed between a metal cation and a nonmetal anion or a polyatomic ion. When an ionic compound dissolves in water, the compound dissociates into a cation and an anion. For example: NaCl(s) ® Na+(aq) + Cl-(aq). However not all ionic compounds can completely dissolve in water. Some ionic compounds have low solubility in water and are considered as insoluble compounds. Review solubility rules in General Chemistry I. Calcium hydroxide Ca(OH)2 has a low solubility in water. The solubility equilibrium of calcium hydroxide Ca(OH)2 in water is shown below: Ca(OH)2(s)  ⇌ Ca2+(aq) + 2 OH-(aq) The equilibrium constant of the above reversible equation, aka solubility product of Ca(OH)2 is given as:  Ksp = [Ca2+][OH-]2 (Note that Ca(OH)2, as a solid, does not appear in the expression of Ksp, in keeping with the rules of equilibrium constant expressions for heterogeneous systems.) Many factors can affect the solubility of a solid, including the presence of other ions in solution. If an ion present in the solution in which the solid is dissolved is also one of the ions from the solid, this ion is called a common ion. The presence of a common ion shifts the equilibrium to the reactant side of the solubility equilibrium, i.e. the common ion decreases the solubility of the solid. This is known as the Common Ion Effect. In this experiment, we will examine the solubility of calcium hydroxide Ca(OH)2, a slightly soluble solid and the common ion effect.
IV. 2 Procedure Preparation of Saturated Ca(OH)2 Solutions (1) Measure 2 grams of solid Ca(OH)2 to a 250-mL Erlenmeyer flask. Fill the flask about half full with 0.010 M Ba(NO3)2 solution.  (2) Close the flask with a cork or a stopper. Shake the solution for 5 minutes. Label the solution as "Ca(OH)2 Solution 1".  (3) Measure 2 grams of solid Ca(OH)2 to another Erlenmeyer flask, repeat steps 1-2 using 0.010 M Ca(NO3)2 solution. Label the solution as "Ca(OH)2 Solution 2".  (4) Filter the above two solutions using filter paper. Label the two filtered solutions as "Saturated Ca(OH)2 Solution 1" and "Saturated Ca(OH)2 Solution 2".        Titration of the Ca(OH)2 Solutions (1) Record the molarity of the standardized HCl solution provided by the instructor.  (2) Use a pipet to transfer 25.00 mL "Saturated Ca(OH)2 Solution 1" to a clean Erlenmeyer flask. Add 2-3 drops of the indicator to the flask.  (3) Titrate the 25.00 mL of "Saturated Ca(OH)2 Solution 1" with the HCl solution to the end-point, twice. Record the mL of the HCl solution used in titrations. Take the average of the volumes and convert the average volume to L. (4) Repeat Steps (2) and (3) with "Saturated Ca(OH)2 Solution 2". Record the mL of the HCl solution used in titrations. Take the average of the volumes and convert the average volume to L.	Think: Ca(OH)2 has low solubility in water. In which solution,  Ba(NO3)2 or Ca(NO3)2, do you think Ca(OH)2 has lower solubility than in pure water? (Hint: which solution has a common ion? How does a common ion affect the solubility of an insoluble compound?)       Record Molarity of the standard HCl solution  MHCl = _________________M   Saturated Ca(OH)2 Solution 1 Saturated Ca(OH)2 Solution 2   Trial 1 Trial 2 Trial 1 Trial 2 VHCl (mL)         Average VHCl (L)
V. Calculation and Conclusions
V. 1 Information In this experiment, acid HCl is used to titrate base Ca(OH)2. With the known molarity M of HCl and the measured volume V of HCl to titrate 25.00 mL Ca(OH)2, we can calculate the molarity M of Ca(OH)2. Since 1 mol Ca(OH)2 dissociating in water produces 1 mol Ca2+, the calculated molarity M of Ca(OH)2 is equal to the molarity of Ca2+ in "Saturated Ca(OH)2 Solution 1" where no common ion is present:  Ca2+ in "Saturated Ca(OH)2 Solution 1" = M of Ca(OH)2 calculated from the titration equation However, when the saturated solution is prepared in 0.01 M Ca(NO3)2, the molarity of Ca2+ is equal to the sum of Ca2+ from Ca(OH)2 and the Ca2+ from 0.01 M Ca(NO3)2, which is 0.01 M: Ca2+ in "Saturated Ca(OH)2 Solution 2" = M of Ca(OH)2 calculated from the titration equation + 0.01 M V.2 Calculation Plug the average volumes of the HCl solution from Data Collection into the titration equation to calculate the molarities of Ca2+ in the two saturated solutions. Complete the following table. Saturated Ca(OH)2 Solution 1 Average HCl volume VHCl (L)   Moles of H+ (mol) = MHCl´VHCl   [OH-] = Moles of H+/0.025 (M)   [Ca2+] = 2 ´ [OH-] (M)   Ksp = [Ca2+][OH-]2   Saturated Ca(OH)2 Solution 1 Average HCl volume VHCl (L)   Moles of H+ (mol) = MHCl´VHCl   [OH-] = Moles of H+/0.025 (M)   [Ca2+] = 2 ´ [OH-] + 0.01 (M)   Ksp = [Ca2+][OH-]2   V.3 Conclusions Compare the [OH-] in two saturated solutions, which one is smaller? What can you conclude?     Compare the two Ksp values, are they close? What can you conclude?

 

 

    

 

Experiment 13. Heat of Reactions

 

 

Student Name (Print):__________________________   Pre-lab Study Questions (Submit upon arrival for experiment)

1. Calculate the enthalpy of reaction DHrxno for each of the following reactions using the tabulated standard enthalpy of formation DHfo. (Show calculation) a. H2O(g) ® H2O(l)     b. CaCO3(s) ® CaO(s) + CO2(g)       c. CH4(g) + 2 O2(g)  ® CO2(g) + 2 H2O(g)      2. Consider the following reaction and the given standard enthalpy of formation DHfo.                             NH4NO3(s)  ® NH4+(aq) + NO3-(aq) DHfo (kJ/mol) -365.6 -132.0 -205.0   a. Calculate the enthalpy of reaction DHrxno for the above reaction using the given standard enthalpy of formation DHfo. (Show calculation)       b. The above reaction equation represents the dissolving process of 1 mol of solid ammonium nitrate NH4NO3 in water. This process is endothermic or exothermic? This process will cause the temperature of the surroundings to rise or d  op? Explain.        c. If 2.00 g of solid NH4NO3 is dissolved in 50.0 mL of water in a coffee-cup calorimeter, how much heat is released or absor  ed? Show calculation. (Hint: the calculation in a gives the heat transferred when 1 mol of NH4NO3 is dissolved in wa    er. If there are 2.00 g of NH4NO3, how many moles of NH4NO3 are there?)           d. Calculate the temperature change Dt of the 50.0 mL of water in the calorimeter if 2.00 g of solid NH4NO3 is dissolved in 50.0 mL of water. Show calculation.

 

 

 

 

Experiment 13. Heat of Reactions Student Name (Print):__________________________
I. Objectives
(1) Understand reaction enthalpy: endothermic and exothermic and relate reaction enthalpy to the sign of the temperature change of the surroundings (2) Determine enthalpy change DH of dissolving solid NH4NO3 in water (3) Determine enthalpy change DH of acid-base neutralization reactions
II. Materials
(1) Coffee-cup calorimeter (2) Ammonium nitrate NH4NO3 (3) 1 M hydrochloric acid, HCl (4) 1 M sodium hydroxide, NaOH
II. Safety Information and Requirements
(1) Wear goggles or safety glasses (2) Dump waste according to instruction
IV. Procedure   and Data Collection
IV.1 Information The First Law of Thermodynamics (aka Law of Energy Conservation) states that the total energy of the universe must remain constant. Therefore, the heat (q) lost by a system is equal to the heat gained by the surroundings. The only difference between these two values is the sign:  qsystem = -qsurroundings         ..................................................Equation 1 The standard S.I. unit for heat is the joule (J). The specific heat Cs of a substance is the heat it takes to raise the temperature of one gram of the substance by 1 oC. The specific heat of water is 4.18 J/g oC.
In this experiment, a coffee-cup calorimeter is used to study the heat transfer involved in chemical reactions. The calorimeter contains two-nested Styrofoam coffee cups. Water is often used in the calorimeter to make thermal contact with the process under investigation (i.e. water is the surroundings). If the process releases heat, the heat transferred to the surroundings causes the temperature of the water to rise and vice versa. The temperature change (Dt) can be measured as the difference between the final and initial temperatures of the water in the calorimeter. The heat transfer in a calorimeter may be expressed with the following equation:  qwater = m ´Cs ´ Dt                     ............................. Equation 2 where m is the mass of water in the calorimeter, Cs of water is 4.18 J/g oC and Dt is the temperature difference of water before and after the process occurs. Note the water is the surroundings to the process and one has to reverse the sign of qwater to find the heat transfer of the system (Equation 1).			Coffee-cup calorimeter
To determine the temperature difference Dt, the temperatures over a certain time period will be recorded and plotted on y­-axis versus time on x-axis. The temperature difference Dt will be calculated as  Dt = tafter,average - tbefore,average ............Equation 3
In this experiment, two reactions will be studied. The first reaction is the dissolving process of solid ammonium nitrate NH4NO3.  Dissolving NH4NO3 is an endothermic reaction. This is the reaction often used in cold packs sold at drug stores. When NH4NO3 is dissolved in water in a coffee-cup calorimeter, the temperature of water will drop.  The second type of reaction that will be studied in this experiment is acid-base netrulization reactions. Two acid-base reactions can be studied. The first reaction is between a sodium hydroxide NaOH solution and a hydrochloric HCl solution: NaOH(aq) + HCl(aq) ® NaCl(aq) + H2O(l) The second reaction is a NaOH solution reacting with an acetic acid CH3COOH solution: NaOH(aq) + CH3COOH(aq) ® CH3COONa(aq) + H2O(l) Neutralization reactions are exothermic. Mixing of a base solution and an acid solution in a coffee-cup calorimeter will make the temperature of the mixed solutions higher than those of the original base and acid solutions.
IV. 2 Procedure and Data Collection Heat of Solution (1) Add 50 mL of deionized water into the calorimeter and allow the temperature of the water to equilibrate with the calorimeter. Record the temperature.  (2) Weigh approximately 2 grams of ammonium nitrate NH4NO3, Record the exact mass and covert the grams to moles.  (3) Add the ammonium nitrate to the calorimeter and quickly  swirl to dissolve. Record the temperature every 15 seconds for 5 minutes.                       Heat of Neutralization Reaction NaOH - HCl neutralization reaction (1) Add 25 mL of 1.0 M NaOH to the calorimeter (Caution:  NaOH is a strong base!). Cover it with the lid and let the solution temperature equilibrate for 5 minutes. Record the temperature after 5 minutes. (2) Measure 25 mL of 1.0 M HCl in a clean graduated cylinder (Caution: HCl is a strong acid!). Allow to sit for 5 minutes and then record the temperature. The temperature of the HCl solution should be approximately the same as the NaOH solution. (3) Quickly but carefully add the HCl solution to the calorimeter. Put the lid on and swirl the solution. Record the temperature every 15 seconds for 5 minutes.	Record Temperature of 50 mL water in the calorimeter ___________________ oC   Mass of NH4NO3 = ________________ grams                               = ________________ moles   Time  t (oC) Time (sec) t (oC) 0 min   45 sec   15 sec   3 min   30   15 sec   45   30   1 min   45   15 sec   4 min   30   15 sec   45   30   2 min   45   15 sec   5 min   30         Record Temperature of 1.0 M NaOH solution ____________ oC Temperature of 1.0 M HCl solution ____________ oC Time  t (oC) Time (sec) t (oC) 0 min   45 sec   15 sec   3 min   30   15 sec   45   30   1 min   45   15 sec   4 min   30   15 sec   45   30   2 min   45   15 sec   5 min   30
V. Calculations
Heat of Solution (1) Plot temperature (°C) on the y-axis vs time (min) on the x-axis and find the temperature difference Dt of water. Dt = _____________ oC This reaction is ________ (endothermic or exothermic).  (2) Calculate heat absorbed from or released to the water using Equation 2. qwater = _______________ J (3) Calculate heat of solution (heat absorbed or released by the dissolving process) using Equation 1. qsolution = _______________ J (4) Calculate molar heat of solution using the following equation:   DHsolution = ________________ kJ/mol (5) Compare the value calculated above to the theoretical value (Pre-Lab Study Question 2a.). Calculate the percent difference.
Heat of Neutralization NaOH - HCl neutralization (1) Plot temperature (°C) on the y-axis vs time (min) on the x-axis and find the temperature difference Dt of water. Dt = _____________ oC (2) Approximately 50 g of water H2O is present for this neutralization reaction. Calculate heat absorbed from or released to the water using Equation 2. qwater = _______________ J (3) Calculate heat of neutralization (heat absorbed or released by the neutralization reaction between NaOH, a base and HCl, an acid) using Equation 1. qneutralization = _______________ J (4) Approximately 0.025 mol (25 mL 1 M H+ with 25 mL 1 M OH-) of water H2O is produced from this reaction. Calculate DHneutralization using the following equation:   DHneutralization = ________________ kJ/mol
VI. Post-Laboratory Questions 1. In this experiment, reaction enthalpies DHrxn are studied. In the chapter of Thermodynamics, another property called entropy is also studied. Describe entropy of a substance. Which phase, solid, liquid or gas, has the greatest entropy?         2. Predict the entropy change DS of the following reactions and briefly explain. a. NH4NO3(s) ® NH4+(aq) + NO3-(aq)       b. H+(aq) + OH-(aq) ® H2O(l)       3. In the chapter of Thermodynamics, you have also learned the Gibbs free energy change  DG, which is a property that combines DH and DS:  DG = DH - TDS  where T is Kelvin temperature.  For a spontaneous process, DG < 0. Based on your experimental results on DH and your predications on DS of the two reactions, predict if the following reactions are always spontaneous, spontaneous only at high temperatures, spontaneous only at low temperatures, or always nonspontaneous. Briefly explain.  a. NH4NO3(s) ® NH4+(aq) + NO3-(aq)       b. H+(aq) + OH-(aq) ® H2O(l)

 

 

 

 

 

 

Experiment 14. Voltaic Cell and Redox Reaction

 

Student name (Print):_______________________
I. Objectives (1) Calculate standard cell potential and compare with experimental data; (2) Identify concentration effect on voltaic cell potential; (3) Write voltaic cell notation; (4) Write overall oxidation-reduction reaction equation.
II. Materials and Equipment
(1) 1.0 M CuSO4 solution; (2) 1.0 M ZnSO4 solution; (3) 0.10 M CuSO4 solution; (4) 0.010 M CuSO4 solution; (5) Cu plate; (6) Zn plate; (7) 1.0 M FeCl3 solution; (8) Iron plate; (9) 1.0 M KMnO4 + 1.0 M HCl + 1.0 M MnSO4 solution; (10) Graphite electrode; (11) 1.0 M H2O2 + 1.0 M HCl solution;			(12) 250-ml beaker (2 ea); (13) voltmeter; (14) Saturated KCl salt-bridge. (Can be substituted with KCl saturated filter paper stripe.); (15) Wires; (16) Fine sand paper.
III. Safety Information and requirements
(1) Wear goggles or safety glasses;  (2) Dump waste according to waste container; (3) Be careful of strong oxidant solutions. Immediately wash with plenty of water if any skin contact occurs. (4) Avoid breathing any chemical vapors.
IV. Principles, Procedures, and Practices
IV.1 Cu-Zn voltaic cell set-up and cell potential measurement Procedure: (1) Get two 250-ml clean beakers and pour about 100 ml 1.0 M ZnSO4 and 100 ml 1.0 M CuSO4 into these two beakers, respectively.  (2) Following the following chart to connect Cu-Zn voltaic cell. (3) Figure out the right wire connection. (4) Measure the cell potential and record. (5) Disconnect the cell wire connection. (Keep the solution for following operation.) Attention: (1) The electrode can't tough the salt-bridge; (2) You can switch the wire connection to "+" and "-" in the voltmeter. The right connection will result in a positive reading in the meter.
			Information: (1) With the right connection, the electrode connected to "+" position in the voltmeter is the cathode, and the electrode connected to "-" is the anode. (2) Oxidation half-reaction always happens in anode and reduction half-reaction always happens in cathode. (3) When oxidation happens, electrons are generated. When reduction happens, electrons are consumed.
Observation and record (1) In this Zn-Cu voltaic cell, the cell potential measured by the voltmeter is: __________V. (2) It is known Image transcription text O = Zn2+ -0.76V; E /Zn Cult /cu = +0.34V ... Show more . The theoretical cell potential = __________V. (3) Is the measured cell potential similar to the theoretical potential?  yes;  no. (Hint: 10% difference is believed to be similar.) (4) Under right connection, which electrode is connected to "+" in voltmeter?  Zn2+/Zn electrode;  Cu2+/Cu electrode. (5) In this voltaic cell, which electrode is anode?  Zn2+/Zn electrode;  Cu2+/Cu electrode. (6) What is the oxidation half-reaction on anode? _________________________________ (7) What is the reduction half-reaction on cathode? _________________________________ (8) Add half-reaction equations in (6) and (7) to get overall oxidation-reduction reaction balanced equation? ____________________________________________
IV.2 Effect of Cu2+ concentration on voltaic cell potential Procedure: (1) Keep the beaker with 100 ml 1.0 M ZnSO4; (2) Dump 1.0 M CuSO4 in the other beaker into the waste container and rinse the beaker with small portion of 0.10 M CuSO4 solution for 2-3 times. Pour about 100 ml 0.10 M CuSO4 solution in this beakers. (3) Reconnect the voltaic cell. Measure and record the cell potential.  (4) Disconnect the cell wire connection. (Keep the solution for following operation.)
Observation and record: (1) In this Zn-Cu voltaic cell, the cell potential measured by the voltmeter is: __________V.
Procedure: (5) Keep the beaker with 100 ml 1.0 M ZnSO4; (6) Dump 0.10 M CuSO4 in the other beaker into the waste container and rinse the beaker with small portion of 0.010 M CuSO4 solution for 2-3 times. Pour about 100 ml 0.010 M CuSO4 solution in this beakers. (7) Reconnect the voltaic cell. Measure and record the cell potential.  (4) Disconnect the cell wire connection.
Observation and record: (2) In this Zn-Cu voltaic cell, the cell potential measured by the voltmeter is: __________V.
Summary:
Cu2+ concentration (M)	1.0		0.10		0.010
Cell potential (V)	________		_________		________
Conclusion: (Check the right conclusion.) Cu2+ in the overall reaction equation is a product. The lower product concentration, the lower cell potential; Cu2+ in the overall reaction equation is a product. The lower product concentration, the higher cell potential; Cu2+ in the overall reaction equation is a react                                            nt. The lower reactant concentration, the lower cell potential; Cu2+ in the overall reaction equation is a reactant. The lower reactant concentration, the higher cell potential.
IV.3 Based on voltaic cells write oxidation-reduction reaction equations Different groups may be assigned to work on different voltaic cells.
Voltaic 1		Voltaic 2		Voltaic 3
Electrode 1: 1.0 M FeCl3 with Fe plate; Electrode 2: 1.0 M H2O2 + 1.0 M HCl solution with graphite piece.		Electrode 1: 1.0 M FeCl3 with Fe plate; Electrode 2: 1.0 M KMnO4 + 1.0 M HCl + 1.0 M MnSO4 solution with graphite piece.		Electrode 1: 1.0 M KMnO4 + 1.0 M HCl + 1.0 M MnSO4 solution with graphite piece; Electrode 2: 1.0 M H2O2 + 1.0 M HCl solution with graphite piece.
Procedure: (1) Obtain two clean 250-ml beakers and rinse each beaker with (small portion) the solution which will be contained. (2) Set up voltaic cell with right wire connection. (3) Record which electrode is connected to "+" in voltmeter. (4) Record cell potential. (5) Disconnect wire. (6) Write overall reaction equation.
Record: (1) The cathode of this cell is:  electrode 1;  electrode 2. (2) The cell potential is: _______V. (3) Oxidation half-reaction is: _____________________________________________ (4) Reduction half-reaction is: _____________________________________________ (5) Overall oxidation-reduction reaction equation is: ______________________________________________________________________ (Hint: You can obtain (5) by adding (3) and (4). However, before addition, you have to make the electron numbers are the same in (3) and (4) by multiplying a smallest whole number.)

 

 

 

Chapter 15

1. Give an example of a buffer in our bodies. 

2. How does it work. 

3. Draw four titration curves, including the equivalence point for each curve.

4. Define the endpoint