Acid Base the Water Constant and PH Presentation

User Generated

ncenqrrc92

Science

Description

2 labs and 1 assignment needs to be done as per instructions in the attachments.

Unformatted Attachment Preview

module 5 lab 3: equilibrium name: In this virtual experiment you will investigate how equilibrium can be shifted in a chemical reaction by changes in solute concentration or temperature. Start by going to the Cobalt Chloride Lab and Le Chatliers Principle site within the Chemistry Collective Lab Page (http://chemcollective.org/equilib) the instructions for this lab are found by clicking on the Cobalt Lab link; however they are listed on the following slides for your convenience. module 5 lab 3: equilibrium Go to the Cobalt Chloride Lab and Le Chatliers Principle site, Add 25 mL of [Co[H2O]6]+2 to an empty Erlenmeyer flask. Now add 12M HCl in 1 mL increments until the equilibrium color has changed. Hint: Type in “1” for the volume to be transferred, and then keep clicking “Pour” until you see a change, counting the number of 1 mL clicks to determine the total volume added. 1. What was the minimum mL of 12M HCl needed for the color change? name: Now, remove some of the free chloride ions by adding some silver nitrate. Hint: add 1 mL amounts of the silver nitrate successively until the equilibrium has been shifted instead of adding a whole bunch at once. Take a screenshot after the equilibrium has changed 3.Take a screenshot after the equilibrium has changed and place it in this box.. Include a screenshot after the color change in this box. Predict the effect of removing chloride ions based on Le Chatelier’s Principle. Here are the two equilibria involved. Note how Ag+ scavenges free chloride ions by tying them up in a precipitate, and thus removes them from solution: AgNO3 D Ag+(aq) + NO3- (aq) Ag+(aq) + Cl-(aq) D AgCl(s) ___________________________ AgNO3(aq) + Cl- D AgCl(s) + NO3- (aq) 2. I predict that removing chloride ions will: Predict the effect of adding HCl to this reaction mixture: 4. I predict that adding HCl will: module 5 lab 3: equilibrium Add some HCl in small increments as you have learned. until the equilibrium shifts 5. Did the equilibrium shift in the direction you predicted? Explain. name: Next, lets see how this reaction equilibrium is shifted by a change in temperature. Right click on the flask and choose thermal properties. You can now change the temperature between 0 and 99 OC. Heat or cool the system until you have perturbed the equilibrium 6. Is this reaction exothermic or endothermic?Explain based on what happened when you changed the temperature, and include a screenshot. Insert your screenshot that shows the reaction mixture after adding HCl module 5 lab 3: equilibrium name: Finally, let’s calculate the value of the equilibrium constant at two different temperatures- this should show us if the reaction is endothermic or exothermic. Allow the system to equilibrate (constant temperature). Use the concentration values to determine K. Now go to the thermal properties, change the temperature, and click on the thermally isolated system option. Determine the new K at the new temperature. From the new K at the new temperature, determine if the system is endothermic or exothermic. 7. Calculate your K value at your first temperature showing your work below and include a screenshot: 8. Calculate your K value at your second temperature showing your work below and include a screenshot: 9. What do those two K values tell you about this reaction and why?. module 5 lab 3: equilibrium name: Post-lab: please answer these 3 questions: 10. Based on the colors of the reactants and products, which do you think is in a higher concentration in the flask of 0.1M cobalt chloride? 11. Describe what happens when you add concentrated HCl to the flask of cobalt chloride. Did you add a product or reactant when you added the HCl? 12. Describe what happens when you add silver nitrate to the blue cobalt chloride solution. What does adding the silver nitrate do to the reaction? name: module 5 lab 2: acid base chemistry In these activites you will explore acids and bases using the chemistry collective, which are a series of onlne labs and activities that are part of the National Science Digital Laboratory. This lab is worth 15 points (1 point per question) Go to lab page: http://chemcollective.org/acid-base where you will see 4 activities. We will be performing the first activity: Strong acids and bases. You will perform a virtual titration, with several follow up questions, and a concept test. important help: You will really want to watch this video to help you do the titration!! https://www.you tube.com/watch ?v=gVe4y6o85R w&feature=yout u.be module 5 lab 2: acid base chemistry name: Go to lab page: http://chemcollective.org/acid-base (no sign-in required) 1. Following the instructions for the strong acids and bases activity. Perform the virtual titration, and determine the concentration of the unknown strong acid , showing your “correct” screenshot below: Next, answer the follow-up questions below. 2. Briefly describe the equivalence point in a titration. What species are present? 3. Phenolphthalein is an indicator that will change color at pH close to 8. Why would we use phenolphthalein to determine when the equivalence point is reached? 4. At what point do you know that you have added enough base from the buret into the acid solution? What happens to your calculations for the concentration of the acid if you accidentally add in more base than you need to? Acid Base Chemistry and Buffers Concept Test module 5 lab 2: acid base chemistry answer the questions below in the space provided name: 5. A comic book villain is holding you at gun point and is making you drink a sample of acid. She gives you a beaker with 100ml of a strong acid with pH=5. She also gives you a beaker of a strong base with a pH=10. You can add as much of the strong base to the strong acid as you want, and you must then drink the solution. You'd be best off trying to make the solution neutral before drinking it. How much of the base should you add? 1.1 ml 2.10 ml 3.100 ml 4.1000 ml answer (choose number): show your work here: s 6. Consider an exceptionally weak acid, HA, with a Ka = 1x10 -20 . You make a 0.1M solution of this weak acid. answer (choose show your work here: 1.What is the pH? number): 2.1 3.2 4.12 5.13 7. You have a powder that dissolves in water to form a strong acid. You dissolve 2g of the powder in 100ml of water, and measure the pH to be 4. You would like to form 100ml of a solution with pH 5. How many grams of the powder do you dissolve in 100ml of water? 1.0.2 g 2.0.4 g answer (choose show your work here: 3.20 g number): 4.40 g module 5 lab 2: acid base chemistry name: Acid Base Chemistry and Buffers Concept Test answer the questions below in the space provided 8. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is: 1.1 2.2 answer (choose show your work here: 3.4 number): 4.5 9. You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger, [F-]/[HF] or [CN-]/[HCN]? 1.[F-]/[HF] answer (choose show your work here: 2.[CN-]/[HCN] number): 3.can.t tell from available information 10. A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to: 1.a 0.05M solution of NaCOOH 2.a 0.025M solution of NaCOOH 3.a solution that is 0.05M in HCOOH and 0.05M in NaCOOH 4.a solution that is 0.025M in HCOOH and 0.025M in NaCOOH answer (choose number): show your work here: module 5 lab 2: acid base chemistry name: Acid Base Chemistry and Buffers Concept Test answer the questions below in the space provided 11. Which do you think is a better buffer (more resistant to addition of acid and base)? 1.a solution that is 0.1M in HA and 0.01M in A answer (choose 2.a solution that is 0.01M in HA and 0.1M in A number): 3.a solution that is 0.1M in HA and 0.1M in A 4.a solution that is 0.05M in HA and 0.05M in A show your explanation here: 12. Consider a protein with the acidic side chains shown in the table below. Given that the pH of blood is about 7.3, how many of the above side chains would be mostly in their ionic form (A-) in blood? 1.2 Amino Acid side-chain: pKa 2.3 Arginine pKa = 12.48 3.4 4.5 answer (choose show your explanation here: Aspartic Acid PKa = 3.90 number): Cysteine pKa = 8.33 Glutamic acid pKa = 4.07 Histidine pKa = 6.04 Lysine pKa =10.79 Tyrosine pKa =10.13 13. You are titrating an acid with a pKa of 7. You are given a choice of indicators with the following pKa's, which do you use? 1.3 2.5 answer (choose show your work here: 3.7 number): 4.9 module 5 lab 2: acid base chemistry Acid Base Chemistry and Buffers Concept Test answer the questions below in the space provided 14. CO2 dissolves in water to form carbonic acid: 1.10-2 2.10-4 3.10-6 4.10-8 H2CO3 D H+ + HCO3- Ka1 = 4.3 x l0-7 HCO3- H+ + CO3-2 Ka2 = 4.8 x l0-11 In an acid solution with pH =4.35 ([H+ ]=4.5x10-5), what is the ratio [CO3-2] /[H2CO3]? answer (choose number): show your work here: name: module 5 assignment name: Chemical Equations: show work for credit 1. How many moles of NO2 forms when 0.356 moles of N2O5 decomposes?: 2N2O5(g) → 4 NO2(g) + O2(g) .356 g mol N2O5 x 4 mol NO2/2mol N2O5 = 0.712 mol NO2 page 1 of 3 Properties of Liquids and Solids Instructions: Answer the following questions using the following information: ∆Hfus = 6.02 kJ/mol; ∆Hvap= 40.7 kJ/mol; 2. For the following reaction, calculate how many moles of each product are formed when 0.356 moles of PbS completely react. Assume there is an excess of oxygen. 2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g) This will make 0.356 moles of PbO and 2 moles of SO2. specific heat of water is 4.184 J/g∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of water vapor is 2.03 J/g∙˚C. 6. How much heat is required to vaporize 25 g of water at 100˚C? 3. For the following reaction, calculate how many grams of each product are formed when 4.05 g of water is used. 2H2O(g) → 2H2(g) + O2(g) 4. Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation: 4P + 5O2 → 2P2O5 5. Determine the percent yield of the following reaction when 2.80 g of P reacts with excess oxygen. The actual yield of this reaction is determined to by 3.89 g of P2O5. 4P + 5O2 → 2P2O5 7. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your answer to three significant figures)? 8. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)? module 5 assignment name: page 2 of 3 Use the following image to answer the following questions. 1. A solution contains 40 g of NaCl per 100 g of water at 100˚C. Is the solution unsaturated, saturated or supersaturated? Explain. Acid/Base: The Water Constant and pH Instructions: Determine if each solution is acidic, basic, or neutral. Explain each answer. For all calculations show your work for credit. 1. 2. A solution contains 50 g of KCl per 100 g of water at 25˚C. Is the solution unsaturated, saturated or supersaturated? Explain. 3. A solution contains 10 g of KNO3 per 100 g of water at 30˚C. Is the solution unsaturated, saturated or supersaturated? Explain. [H3O+] = 1 x 10-10 M; [OH-] = 1 x 10-4 M 2. [H3O+] = 1 x 10-7 M; [OH-] = 1 x 10-7 M 3. [H3O+] = 1 x 10-1 M; [OH-] = 1 x 10-13 M 4. An 8 oz bottle of Dr. Pepper made with imperial cane sugar (C12H22O11) contains 27 g of sugar in 236.6 mL of water. What is the molarity of the solution with respect to cane sugar? 4. [H3O+] = 1 x 10-13 M; [OH-] = 1 x 10-1 M 5. Calculate the mass of NaCl in a 55 mL sample of a 3.5 M saline solution. 6. What volume of water should be added to 2.6 g of KCl to make a 4 M solution? module 5 assignment name: page 3 of 3 Equilibrium and Le Chateliers Principle Instructions: Write the equilibrium expression for each chemical equation. 1) Instructions: Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic. 1. [H3O+] = 2.6 x 10-3 M 2. [H3O+] = 2.6 x 10-8 M 3. [H3O+] = 1.6 x 10-2 M 4. [H3O+] = 4.3 x 10-3 M Instructions: Calculate the [H3O+] of solutions 9 and 10; calculate the [OH-] solutions 11 and 12. 5. pH = 2.76 2H2S(g) D 2H2(g) + S2(g) 2) NH4HS(s) D NH3(g) + H2S(g) 3) Oxidation-Reduction Reactions Instructions: Assign an oxidation state to each element, ion, or molecule. 1. Cl- -1 2. Br2 0 3. K 0 4. Ca2+ +2 Instructions: Assign an oxidation state to each atom in each compound. 1. 2. 3. 4. NaCl Na – (+1) CH4 SbCl5 Al2O3 Cl- (-1) Instructions: Assign an oxidation state to each atom in each polyatomic ion. 6. pH = 3.65 7. pOH = 3.65 1. 2. 3. 4. NH4+ ClO4Cr2O72MnO4- Instructions: Identify the oxidation state of P in each ion. 8. pOH = 6.87 1. 2. 3. 4. PO43HPO42H2PO4H3PO4
Purchase answer to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Explanation & Answer

Attached.

module 5 assignment

name:

Chemical Equations: show work for credit
1.

How many moles of NO2 forms when 0.356 moles of N2O5 decomposes?:
2N2O5(g) → 4 NO2(g) + O2(g)
.356 g mol N2O5 x 4 mol NO2/2mol N2O5 = 0.712 mol NO2

page 1 of 3

Properties of Liquids and Solids
Instructions: Answer the following questions using the following information:
∆Hfus = 6.02 kJ/mol;
∆Hvap= 40.7 kJ/mol;

2. For the following reaction, calculate how many moles of each product are formed when 0.356
moles of PbS completely react. Assume there is an excess of oxygen.
2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g)
This will make 0.356 moles of PbO and 0.356 moles of SO2.

specific heat of water is 4.184 J/g∙˚C;
specific heat of ice is 2.06 J/g∙˚C;
specific heat of water vapor is 2.03 J/g∙˚C.
6. How much heat is required to vaporize 25 g of water at 100˚C?

3. For the following reaction, calculate how many grams of each product are formed when 4.05 g of
water is used.
2H2O(g) → 2H2(g) + O2(g)
4.05g of water = 0.225 moles of water.
0.225 moles of water produces 0.225 moles of H2 = 0.45 g of H2
0.225 moles of water produces 0.1125 moles of O2 = 3.6 g of O2.
4. Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the
following chemical equation:
4P + 5O2 → 2P2O5
O2 required: (3.07g P / 31g P) x (5/4) moles of O2 x (32g O2) = 3.96 g of O2
Limiting Reagent = P. So, P2O5 theoretical yield = (3.07g P / 31g P) x (2/4) x (142 g
P2O5) = 7.03 g
5. Determine the percent yield of the following reaction when 2.80 g of P reacts with excess oxygen.
The actual yield of this reaction is determined to by 3.89 g of P2O5.
4P + 5O2 → 2P2O5

O2 required: (2.80g P / 31g P) x (5/4) moles of O2 x (32g O2) = 3.61g of O2
Theoretical yield = 2.80 + 3.61 g = 6.41g
Percentage yield = (3.89/6.41) x 100 = 60.7%

25g of water = 25/18 = 1.389 moles
Q = ∆Hvap x 1.389 moles = 40.7 x 1.389 = 56.5 kJ

7. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report
your answer to three significant figures)?
25g ice = 1.389 moles
Heat required = (2.06 x 25 x 4) + (1.389 x 6020) + (4.184 x 25 x 100) + (1.389 x 40700) +
(2.03 x 25 x 5) = 75.8 kJ

8. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If
no heat is lost to the surroundings, what is the final temperature of the entire water sample after
all the ice is melted (report your answer to three significant figures)?

8.32g ice = 8.32/18 = 0.462 moles
Let, final temperature is T.
(6020 x 0.462) + (4.184 x 8.32 x T) = (4.184 x 55 x (25 - T))
So, T = 11.2 ˚C

module 5 assignment

name:

page 2...

Related Tags