CHEM 213 Chemical Analysis Exam 2 Tuesday May 11, 2004

Size: px
Start display at page:

Download "CHEM 213 Chemical Analysis Exam 2 Tuesday May 11, 2004"

Transcription

1 CHEM 213 Chemical Analysis Exam 2 Tuesday May 11, _ (of 10) 2 10_ (of 10) 3 10_ (of 10) 4 10_ (of 10) 5 10_ (of 10) 6 10_ (of 10) 7 20_ (of 20) 8 10_ (of 10) 9 10_ (of 10) Σ % KEY Name: (please print)

2 1. Calculate the solubility of PbI 2 (K sp = ) in a M solution of Mg(ClO 4 ) 2 using a. molar concentrations. (5 points) K sp = [Pb 2+ ][I - ] 2 = [Pb 2+ ] = s, [I - ] = 2s K sp = s(2s) 2 = s = solubility = ( /4) 1/3 = b. activities. (you find Kjelland s table of activity coefficients on page 11 of this exam). (5 points) µ = ½ Σc i Z i 2 = ½ [0.0333(2) (1) 2 ] = γ Pb2+ = 0.36, γ I- = 0.75 K sp = [Pb 2+ ][I - ] 2 γ Pb2+ ( γ I- ) 2 K sp = K sp /(γ Pb2+ ( γ I- ) 2 ) = /( ) = [Pb 2+ ] = s, [I - ] = 2s K sp = s(2s) 2 = s = solubility = ( /4) 1/3 =

3 2. A 40.0-mL solution of M Hg 2 (NO 3 ) 2 was titrated with 60.0 ml of M KI to precipitate Hg 2 I 2 (K sp = ). a. What volume of KI solution is needed to reach the equivalence point. (2 points) moles of I - = 2(moles of Hg 2 2+ ) (V c )(0.100 m) = 2(40.0 ml) M) V c = 32.0 ml b. Calculate the ionic strength (µ) of the solution when 60.0 ml of KI have been added. (4 points) Vitually all the Hg 2 2+ has precipitated, along with 3.20 mmol of I -. The ions remaining in solution are: [NO 3 - ] = 3.20 mmol/100 ml = M [I - ] = 2.80 mmol/100 ml = M [K + ] = 6.00 mmol/100 ml = M µ = ½ Σc i Z i 2 = ½ [0.032(1) (1) (1) 2 ] = M c. Using activities (activity coefficients on page 11), calculate phg 2 2+ (= -log a Hg2 2+). (4 points) a Hg2 2+ = K sp /(a I- ) 2 = /((0.81) 2 (0.028) 2 ) = phg 2 2+ =

4 3. (a) Find the ph of a solution prepared by dissolving g of tris (MW = g/mol) plus 4.67 g of tris hydrochloride (MW = g/mol, K a = ) in 1.00 L of water. (5 points) [B] = g L -1 / g mol -1 = M [BH + ] = 4.67 g L -1 / g mol -1 = M ph = pk a + log [B]/[BH + ] = log(0.1026/0.0296) = 8.61 (b) If we add 12.0 ml of 1.00 M HCl to the solution made in (a), what will be the new ph? (5 points) When a strong acid is added to a weak base, both react completely to give BH +. We are adding 12.0 ml of 1.00 M HCl, which contains ( L)(1.00 mol/l) = mol of H +. This much H + will consume mol of B to create mol of BH +, which is shown in the table below: B + H + BH + Initial moles Final moles ph = pk a + log[b]/[bh + ] = log(0.0906/0.0416) = 8.41 NOTE: The volume of the solution is in this case irrelevant (calculate the numbers with M instead of mol, and you will see), because volume cancels in the numerator and denominator of the log term: ph = pk a + log[(moles of B/L of solution)]/[(moles of BH + /L of solution)] 4

5 4. Calculate the concentration of CdS (K sp = ) in a solution in which the [H 3 O + ] concentration is held constant at M. Other equilibria that need to be considered to solve this problem are: H 2 S(aq) + H 2 O(l) HS - (aq) + H 3 O + (aq) K 1 = HS - (aq) + H 2 O(l) S 2- (aq) + H 3 O + (aq) K 2 = H 2 S(aq) + 2H 2 O(l) S 2- (aq) + 2H 3 O + K 1 K 2 = For full credit, I require more than a formula copied from your cheat sheet. I need to see how you obtained this formula. (10 points) S = solubility = [Cd 2+ ] = [S 2- ] + [HS - ] + [H 2 S] K sp = [Cd 2+ ][S 2- ] (1) K 2 = [H 3 O + ][S 2- ]/[HS - ] = (2) K 1 K 2 = [H 3 O + ] 2 [S 2- ]/[H 2 S] = (3) From mass-balance consideration [Cd 2+ ] = [S 2- ] + [HS - ] + [H 2 S] (4) Substituting equations (2) and (3) into (4) gives [Cd 2+ ] = [S 2- ] + [H 3 O + ][S 2- ]/K 2 + [H 3 O + ] 2 [S 2- ]/K 1 K 2 = [S 2- ](1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 ) [S 2- ] = [Cd 2+ ]/(1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 ) Substituting [S 2- ] in equation (1) yields [Cd 2+ ] = (K sp /[Cd 2+ ])(1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 ) [Cd 2+ ] = (K sp (1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 )) 1/2 Inserting the actual values for K sp, K 2, K 1 K 2, and [H 3 O + ] [Cd 2+ ] = ( ( / ( ) 2 / )) 1/2 [Cd 2+ ] = M 5

6 5. What is the ph of a solution that is prepared by dissolving 3.00 g of salicylic acid, C 6 H 4 (OH)COOH ( g/mol, K a = ) in 50.0 ml of M NaOH and diluting to ml? Hint: The simplified Henderson-Hasselbach equation will not work in this case. (10 points) Original amount HA = 3.00 g (mmol HA/ g) = mmol Original amount NaOH = 50.0 ml ( mol/ml) = 5.65 mmol c HA = ( )mmol/500 ml= M c NaA = 5.65 mmol/500 ml = M [H 3 O + ] = K a [HA]/[A - ] = x / = ph = However, [H 3 O + ] is not << c HA and c NaA [HA] = [H 3 O + ] + [OH - ] [A - ] = [H 3 O + ] - [OH - ] [OH - ] at this ph will be negligible = [H 3 O + ]( [H 3 O + ]0/( [H 3 O + ]) Rearranging gives [H 3 O + ] [H 3 O + ] = 0 [H 3 O + ] = ph =

7 6. Calculate the equilibrium concentration of undissociated HCOOH in a formic acid solution with an analytical formic acid concentration of and a ph of (K a (HCOOH) = ). (10 points) α 0 = [H 3 O + ]/([H 3 O + ] + K a ) = /( ) = [HCOOH]/C T = [HCOOH]/ = α 0 [HCOOH] = = M 7

8 7. Consider the titration of 50.0 ml of M malonic acid (K a1 = , K a2 = ) with M NaOH. The titration reaction occurring is: HO 2 CCH 2 CO 2 H + OH - - O 2 CCH 2 CO 2 H + H 2 O - O 2 CCH 2 CO 2 H + OH - - O 2 CCH 2 CO H 2 O Designate malonic acid as H 2 M and determine the ph after the following volumina of NaOH have been added: a. At 0.0 ml (3 points) H 2 M H + + HM x x x x 2 /(0.050 x) = K 1 x = ph = 2.11 b. At 8.0 ml (3 points) H 2 M + OH - HM - + H 2 O Initial: Final: 17 o 8 - ph = pk a1 + log([hm - ]/[H 2 M]) = log(8/17) = 2.52 c. At 12.5 ml (4 points) V b = ½ V e ph = pk a1 = 2.85 d. At 25.0 ml (4 points) At the first equivalence point, H 2 M has been converted to HM -. [H 3 O + ] = K a1 K a2 F + K a1 K w K a1 + F where F = (50/75)(0.050) = M [H 3 O + ] = M ph = 4.28 e. At 50.0 ml (3 points) At the second equivalence point, H 2 M has been converted to M 2- : M 2- + H 2 O HM - + OH - (50/100)(0.050) x x x x 2 /(0.025 x) = K b1 = K w /K a2 x = M ph = -log(k w /x) = 9.05 f. At 56.3 ml (3 points) There are 6.3 ml of excess NaOH [OH - ] = (6.3/106.3)(0.100) = M ph =

9 8. Consider the titration of ml of M HNO 2 (K a = ) with M NaOH solution. What is the ph after the addition of: a ml (1 point) c HA = M [H 3 O + ] = [A - ] [HA] = [H 3 O + ] K a = = [H 3 O + ][A - ]/[HA] = [H 3 O + ] 2 /( [H 3 O + ]) [H 3 O + ] [H 3 O + ] = 0 ph = 2.09 b ml (2 points) c NaA = ( )/ ) = c HA = ( ) )/65.00 = This is a buffer, but simplifying assumptions are not valid. K a = [H 3 O + ](c NaA + [H 3 O + ] [OH - ])/( c HA - [H 3 O + ] + [OH - ]), rearranged: [H 3 O + ] 2 + (K a + c NaA ) [H 3 O + ] - K a c HA = 0 [H 3 O + ] [H 3 O + ] = 0 ph = 2.83 c ml (2 points) c NaA = ( )/ ) = c HA = ( ) )/75.00 = This is still a buffer, use above equation. [H 3 O + ] 2 + (K a + c NaA ) [H 3 O + ] - K a c HA = 0 [H 3 O + ] [H 3 O + ] = 0 ph = 3.17 d ml ( 2 points) c NaA = ( )/ ) = c HA = ( ) )/85.00 = This is still a buffer, use above equation. [H 3 O + ] 2 + (K a + c NaA ) [H 3 O + ] - K a c HA = 0 [H 3 O + ] [H 3 O + ] = 0 ph = 3.53 e ml (2 points) This is the equivalence point and we have made a solution of NaNO 2. c NaNO2 = /100 ml = M K b = K w /K a = / = [OH - ] = (K b c NaNO2 ) 1/2 = poh = 6.08 ph = 7.92 f ml (1 point) We now have an excess of NaOH. [OH - ] c NaOH = )/101.0 = poh = 3.00 ph =

10 9. How would you prepare 1.00 L of a buffer with ph of 6.00 from Na 3 AsO 4 and M HCl? K a1 (H 3 AsO 4 ) = , K a2 (H 3 AsO 4 ) = , K a3 (H 3 AsO 4 ) = (10 points) ph = 6.00 [H 3 O + ] = antilog (-6.00) = [H 3 O + ][HAsO 4 2- ]/[H 2 AsO 4 - ] = [HAsO 4 2- ]/[H 2 AsO 4 - ] = / = 0.11 (1) Let V H3AsO4 and V NaOH be the volume in milliliters of the two reagents. Then: V H3AsO4 + V NaOH = 1000 ml (2) From mass-balance consideration, we may write that in the 1000 ml, no. mmol NaH 2 AsO 4 + no. mmol Na 2 HAsO 4 = V Na3AsO4 (3) 2 no. mmol NaH 2 AsO 4 + no. mmol Na 2 HAsO 4 = V HCl (4) Equation (1) can be written as: no. mmol Na 2 HAsO 4 /1000 no. mmol NaH 2 AsO 4 /1000 no. mmol Na 2 HAsO = 4 = 0.11 (5) no. mmol NaH 2 AsO 4 Thus we have four equations: (2), (3), (4), and (5) and four unknowns: V Na3AsO4, V HCl, no. mmol NaH 2 AsO 4 and no. mmol of Na 2 HAsO 4. Subtracting equation (3) from (4) yields: no. mmol NaH 2 AsO 4 = V HCl V Na3AsO4 (6) Substituting equation (6) into (3) gives no. mmol Na 2 HPO V HCl V Na3AsO4 = V Na3AsO4 no. mmol Na 2 HPO 4 = V HCl V Na3AsO4 (7) Substituting equations (6) and (7) into (5) gives ( V HCl V Na3AsO4 )/(0.400 V HCl V Na3AsO4 ) = 0.11 This equation rearranges to: V HCl = V Na3AsO4 Substituting equation (2) gives 0.444( V Na3AsO4 ) = V Na3AsO4 V Na3AsO4 = 444/1.499 = 296 ml V HCl = = 704 ml 10

11 11

Analyte: The substance whose concentration is not known in a titration. Usually the analyte is in the flask or beaker beneath the burette.

Analyte: The substance whose concentration is not known in a titration. Usually the analyte is in the flask or beaker beneath the burette. Key Worksheet 15 Acids & Base Equilibria: Acid Base Titrations Objectives To be able to calculate the ph, poh, and concentrations of all species present at any point of an acid base titration. Vocabulary

More information

Chem 222 #29 Review Apr 28, 2005

Chem 222 #29 Review Apr 28, 2005 Chem 222 #29 Review Apr 28, 2005 Announcement Please meet me after the class if you have any conflicts with the final exam schedule. You can expect similar questions with Quiz 6 in the final exam. If you

More information

Now, the excess strong base will react: HA + OH - A - + H 2 O Start mol End mol

Now, the excess strong base will react: HA + OH - A - + H 2 O Start mol End mol Chemistry Spring 016 Exam 3: Chapters 8-10 Name 80 Points Complete problem 1 and four of problems -6. CLEARLY mark the problem you do not want graded. You must show your work to receive credit for problems

More information

2] What is the difference between the end point and equivalence point for a monobasicmonoacid

2] What is the difference between the end point and equivalence point for a monobasicmonoacid 4 Titrations modified October 9, 2013 1] A solution of 0.100 M AgNO 3 is used to titrate a 100.00 ml solution of 0.100 M KCl. The K sp of AgCl is 1.8e-11 a) What is pag if 50.00 ml of the titrant is added

More information

HW 16-10: Review from textbook (p.725 #84, 87, 88(mod), 89, 95, 98, 101, 102, 110, 113, 115, 118, 120, SG#23,A)

HW 16-10: Review from textbook (p.725 #84, 87, 88(mod), 89, 95, 98, 101, 102, 110, 113, 115, 118, 120, SG#23,A) HW 6: Review from textbook (p.75 #84, 87, 88(mod), 89, 95, 98,,,, 3, 5, 8,, SG#3,A) 6.84 The pk a of the indicator methyl orange is 3.46. Over what ph range does this indicator change from 9 percent HIn

More information

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria and Solubility Equilibria Acid-Base Equilibria and Solubility Equilibria Acid-Base Equilibria and Solubility Equilibria Homogeneous versus Heterogeneous Solution Equilibria (17.1) Buffer Solutions (17.2) A Closer Look at Acid-Base

More information

Exam Practice. Chapters

Exam Practice. Chapters Exam Practice Chapters 16.6 17 1 Chapter 16 Chemical Equilibrium Concepts of: Weak bases Percent ionization Relationship between K a and K b Using structure to approximate strength of acids Strength of

More information

Copyright 2018 Dan Dill 1

Copyright 2018 Dan Dill 1 TP The expression for the equilibrium constant for the solubility equilibrium M 2 X 2 M X 2 is 1. sp 2 M X 2 / M 2 X 2. sp 2 M 2 X 2 / M 2 X 3. sp 2 M 2 X 2 4. sp M 2 X 2 Lecture 21 CH102 A1 (MWF 9:05

More information

Lecture Presentation. Chapter 16. Aqueous Ionic Equilibrium. Sherril Soman Grand Valley State University Pearson Education, Inc.

Lecture Presentation. Chapter 16. Aqueous Ionic Equilibrium. Sherril Soman Grand Valley State University Pearson Education, Inc. Lecture Presentation Chapter 16 Aqueous Ionic Equilibrium Sherril Soman Grand Valley State University The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze.

More information

Chapter 15. Acid-Base Equilibria

Chapter 15. Acid-Base Equilibria Chapter 15 Acid-Base Equilibria The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion already involved in the equilibrium

More information

Chem Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation

Chem Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation Chem 106 3--011 Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation 3//011 1 The net ionic equation for the reaction of KOH(aq) and

More information

CHEM1902/ N-4 November 2004

CHEM1902/ N-4 November 2004 CHEM190/4 004-N-4 November 004 Teeth are made from hydroxyapatite, Ca 5 (PO 4 ) 3 OH. Why does an acidic medium promote tooth decay and how can the decay be stopped using fluoridation of drinking water?

More information

ph + poh = 14 G = G (products) G (reactants) G = H T S (T in Kelvin) 1. Which of the following combinations would provide buffer solutions?

ph + poh = 14 G = G (products) G (reactants) G = H T S (T in Kelvin) 1. Which of the following combinations would provide buffer solutions? JASPERSE CHEM 210 PRACTICE TEST 3 VERSION 3 Ch. 17: Additional Aqueous Equilibria Ch. 18: Thermodynamics: Directionality of Chemical Reactions Key Equations: For weak acids alone in water: [H + ] = K a

More information

1) Write the Brønsted-Lowry reaction for weak acid HCN reacting with H 2 O.

1) Write the Brønsted-Lowry reaction for weak acid HCN reacting with H 2 O. 1) Write the Brønsted-Lowry reaction for weak acid HCN reacting with H O. HCN + H O º H O + + CN ) Write the Brønsted-Lowry reaction for weak base NH reacting with H O NH + H O º OH + NH + ) Using the

More information

Judith Herzfeld 1996,1998. These exercises are provided here for classroom and study use only. All other uses are copyright protected.

Judith Herzfeld 1996,1998. These exercises are provided here for classroom and study use only. All other uses are copyright protected. Judith Herzfeld 1996,1998 These exercises are provided here for classroom and study use only. All other uses are copyright protected. 3.3-010 According to Bronsted-Lowry Theory, which of the following

More information

Chemistry 112 Spring 2007 Prof. Metz Exam 3 Each question is worth 5 points, unless otherwise indicated.

Chemistry 112 Spring 2007 Prof. Metz Exam 3 Each question is worth 5 points, unless otherwise indicated. Chemistry 112 Spring 2007 Prof. Metz Exam 3 Each question is worth 5 points, unless otherwise indicated. 1. The ph of a 0.150 M solution of formic acid, HCOOH is (K a (formic acid) = 1.8 x 10-4 ). (A)

More information

Chemistry 222. Start mol End mol

Chemistry 222. Start mol End mol Chemistry Spring 019 Exam 3: Chapters 8-10 Name 80 Points Complete problem 1 and four of problems 6. CLEARLY mark the problem you do not want graded. You must show your work to receive credit for problems

More information

Equilibri acido-base ed equilibri di solubilità. Capitolo 16

Equilibri acido-base ed equilibri di solubilità. Capitolo 16 Equilibri acido-base ed equilibri di solubilità Capitolo 16 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

More information

Chemistry 102 Discussion #8, Chapter 14_key Student name TA name Section

Chemistry 102 Discussion #8, Chapter 14_key Student name TA name Section Chemistry 102 Discussion #8, Chapter 14_key Student name TA name Section 1. If 1.0 liter solution has 5.6mol HCl, 5.mol NaOH and 0.0mol NaA is added together what is the ph when the resulting solution

More information

Chapter 17. Additional Aspects of Equilibrium

Chapter 17. Additional Aspects of Equilibrium Chapter 17. Additional Aspects of Equilibrium Sample Exercise 17.1 (p. 726) What is the ph of a 0.30 M solution of acetic acid? Be sure to use a RICE table, even though you may not need it. (2.63) What

More information

ph + poh = 14 G = G (products) G (reactants) G = H T S (T in Kelvin)

ph + poh = 14 G = G (products) G (reactants) G = H T S (T in Kelvin) JASPERSE CHEM 210 PRACTICE TEST 3 VERSION 2 Ch. 17: Additional Aqueous Equilibria Ch. 18: Thermodynamics: Directionality of Chemical Reactions Key Equations: For weak acids alone in water: [H + ] = K a

More information

Chem 222 #28 Review Dec 2, 2004

Chem 222 #28 Review Dec 2, 2004 Chem 222 #28 Review Dec 2, 2004 Announcement Please receive your quiz from the TAs. If you have any corrections about your quiz, labs, and notebook, please ask for corrections by 5pm of Dec 3. Final exam

More information

Volume NaOH Delivered (ml)

Volume NaOH Delivered (ml) Chemistry Spring 011 Exam 3: Chapters 8-10 Name 80 Points Complete five (5) of the following problems. Each problem is worth 16 points. CLEARLY mark the problems you do not want graded. You must show your

More information

BCIT Winter Chem Exam #2

BCIT Winter Chem Exam #2 BCIT Winter 2015 Chem 0012 Exam #2 Name: Attempt all questions in this exam. Read each question carefully and give a complete answer in the space provided. Part marks given for wrong answers with partially

More information

Acid-Base Equilibria and Solubility Equilibria Chapter 17

Acid-Base Equilibria and Solubility Equilibria Chapter 17 PowerPoint Lecture Presentation by J. David Robertson University of Missouri Acid-Base Equilibria and Solubility Equilibria Chapter 17 The common ion effect is the shift in equilibrium caused by the addition

More information

AP Chapter 15 & 16: Acid-Base Equilibria Name

AP Chapter 15 & 16: Acid-Base Equilibria Name AP Chapter 15 & 16: Acid-Base Equilibria Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. AP Chapter 15 & 16: Acid-Base Equilibria 2 Warm-Ups (Show

More information

Chapter 17 Homework Problem Solutions

Chapter 17 Homework Problem Solutions Chapter 17 Homework Problem Solutions 17.40 D 2 O D + + OD, K w = [D + ] [OD ] = 8.9 10 16 Since [D + ] = [OD ], we can rewrite the above expression to give: 8.9 10 16 = ([D + ]) 2, [D + ] = 3.0 10 8 M

More information

Chem 116 POGIL Worksheet - Week 11 - Solutions Titration. millimol NaOH added = millimol HCl initially present

Chem 116 POGIL Worksheet - Week 11 - Solutions Titration. millimol NaOH added = millimol HCl initially present Chem 116 POGIL Worksheet - Week 11 - Solutions Titration Key Questions 1. A 25.0-mL sample of 0.100 M HCl(aq) is titrated with 0.125 M NaOH(aq). How many milliliters of the titrant will be need to reach

More information

AP Chemistry - Packet #1 - Equilibrium. 1) a) 5 points K p = (P NH3 ) (P H2S )

AP Chemistry - Packet #1 - Equilibrium. 1) a) 5 points K p = (P NH3 ) (P H2S ) 1) a) 5 points K p = (P NH3 ) (P H2S ) P NH3 = P H2S = (0.659 atm / 2) = 0.330 atm K p = (0.330 atm) (0.330 atm) = 0.109 atm 2 b) 5 points P NH3 = 2 P H2S (2x) (x) = 0.109 x = 0.233 atm = P H2S 2x = 0.466

More information

Chapter 17: Additional Aspects of Aqueous equilibria. Common-ion effect

Chapter 17: Additional Aspects of Aqueous equilibria. Common-ion effect Chapter 17: Additional Aspects of Aqueous equilibria Learning goals and key skills: Describe the common ion effect. Explain how a buffer functions. Calculate the ph of a buffer solution. Calculate the

More information

Acid-Base Titration Solution Key

Acid-Base Titration Solution Key Key CH3NH2(aq) H2O(l) CH3NH3 (aq) OH - (aq) Kb = 4.38 x 10-4 In aqueous solution of methylamine at 25 C, the hydroxide ion concentration is 1.50 x 10-3 M. In answering the following, assume that temperature

More information

Chapter 15 - Applications of Aqueous Equilibria

Chapter 15 - Applications of Aqueous Equilibria Neutralization: Strong Acid-Strong Base Chapter 15 - Applications of Aqueous Equilibria Molecular: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) SA-SB rxn goes to completion (one-way ) Write ionic and net ionic

More information

Aqueous Equilibria Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry

Aqueous Equilibria Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry 2012 Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry The Common-Ion Effect Consider a solution of acetic acid: HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) If

More information

Consider a normal weak acid equilibrium: Which direction will the reaction shift if more A is added? What happens to the % ionization of HA?

Consider a normal weak acid equilibrium: Which direction will the reaction shift if more A is added? What happens to the % ionization of HA? ch16blank Page 1 Chapter 16: Aqueous ionic equilibrium Topics in this chapter: 1. Buffers 2. Titrations and ph curves 3. Solubility equilibria Buffersresist changes to the ph of a solution. Consider a

More information

Chem 112, Fall 05 Exam 3A

Chem 112, Fall 05 Exam 3A Before you begin, make sure that your exam has all 10 pages. There are 32 required problems (3 points each, unless noted otherwise) and two extra credit problems (3 points each). Stay focused on your exam.

More information

Chapter 17. Additional Aspects of Equilibrium

Chapter 17. Additional Aspects of Equilibrium Chapter 17. Additional Aspects of Equilibrium 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak

More information

Announcements. Exam 1 is on Thursday, September 23 from 7-8:10pm; Conflict exam is from 5:35-6:45pm

Announcements. Exam 1 is on Thursday, September 23 from 7-8:10pm; Conflict exam is from 5:35-6:45pm Announcements Print worksheet #6 prior to your Tuesday discussion section LON-CAPA assignment 4 is now available Don t forget to bring your Clicker to class EVERY day The points from last week have been

More information

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.3 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): 0.40 0.00

More information

Exam 2 Sections Covered: 14.6, 14.8, 14.9, 14.10, 14.11, Useful Info to be provided on exam: K K [A ] [HA] [A ] [B] [BH ] [H ]=

Exam 2 Sections Covered: 14.6, 14.8, 14.9, 14.10, 14.11, Useful Info to be provided on exam: K K [A ] [HA] [A ] [B] [BH ] [H ]= Chem 101B Study Questions Name: Chapters 14,15,16 Review Tuesday 3/21/2017 Due on Exam Thursday 3/23/2017 (Exam 3 Date) This is a homework assignment. Please show your work for full credit. If you do work

More information

Chem 222 #18 Ch 10, Review Mar 10, 2005

Chem 222 #18 Ch 10, Review Mar 10, 2005 Chem 222 #18 Ch 10, Review Mar 10, 2005 Announcement Next Tuesday Review Class (Lecture note will be probably uploaded on this Saturday) Midterm Exam on the next Thursday this room in the same time (50

More information

ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor

ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Acid Base Conjugate acid Conjugate

More information

Acids, Bases and Buffers

Acids, Bases and Buffers 1 Acids, Bases and Buffers Strong vs weak acids and bases Equilibrium as it relates to acids and bases ph scale: [H+(aq)] to ph, poh, etc ph of weak acids ph of strong acids Conceptual about oxides (for

More information

Chapter 16 Aqueous Ionic Equilibrium

Chapter 16 Aqueous Ionic Equilibrium Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall The Danger of Antifreeze

More information

CHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA

CHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA CHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA The introduction refers to the formation of stalactites and stalagmites as an example of the material in this chapter: CO 2

More information

Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria

Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria Solutions of Acids or Bases Containing a Common Ion A common ion often refers to an ion that is added by two or more species. For

More information

For problems 1-4, circle the letter of the answer that best satisfies the question.

For problems 1-4, circle the letter of the answer that best satisfies the question. CHM 106 Exam II For problems 1-4, circle the letter of the answer that best satisfies the question. 1. Which of the following statements is true? I. A weak base has a strong conjugate acid II. The strength

More information

Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College

Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College The Common Ion Effect Consider a solution of acetic acid: HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq)

More information

CHAPTER 9 SUPPLEMENTARY SOLUTIONS 22 MONOPROTIC ACID-BASE EQUILIBRIA ] + [ + H 3 NCH 2 CO 2 H] = 0.05 M

CHAPTER 9 SUPPLEMENTARY SOLUTIONS 22 MONOPROTIC ACID-BASE EQUILIBRIA ] + [ + H 3 NCH 2 CO 2 H] = 0.05 M CHAPTER 9 SUPPLEMENTARY SOLUTIONS MONOPROTIC ACIDBASE EQUILIBRIA S91. [H ] [ H 3 NCH CO H] [OH ] [H NCH CO ] S9. [H ] 3[Al 3 ] [AlOH ] [Al(OH) ] [K ] [OH ] [Al(OH ) 4 ] S93. [H NCH CO ] [ H 3 NCH CO ]

More information

More About Chemical Equilibria

More About Chemical Equilibria 1 More About Chemical Equilibria Acid-Base & Precipitation Reactions Chapter 15 & 16 1 Objectives Chapter 15 Define the Common Ion Effect (15.1) Define buffer and show how a buffer controls ph of a solution

More information

X212F Which of the following is a weak base in aqueous solution? A) H 2 CO 3 B) B(OH) 3 C) N 2 H 4 D) LiOH E) Ba(OH) 2

X212F Which of the following is a weak base in aqueous solution? A) H 2 CO 3 B) B(OH) 3 C) N 2 H 4 D) LiOH E) Ba(OH) 2 PX212SP14 Practice Exam II / Spring 2014 1. Which of the following statements are characteristic of acids? 1. They are proton donors. 2. They react with bases to produce a salt and water. 3. They taste

More information

5 Acid Base Reactions

5 Acid Base Reactions Aubrey High School AP Chemistry 5 Acid Base Reactions 1. Consider the formic acid, HCOOH. K a of formic acid = 1.8 10 4 a. Calculate the ph of a 0.20 M solution of formic acid. Name Period Date / / 5.2

More information

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.5 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H (aq) CH 3 COO (aq) Initial (): 0.40 0.00 0.00

More information

Part One: Pure Solutions of Weak Acids, Bases (water plus a single electrolyte solute)

Part One: Pure Solutions of Weak Acids, Bases (water plus a single electrolyte solute) CHAPTER 16: ACID-BASE EQUILIBRIA Part One: Pure Solutions of Weak Acids, Bases (water plus a single electrolyte solute) A. Weak Monoprotic Acids. (Section 16.1) 1. Solution of Acetic Acid: 2. See Table

More information

Chapter 17: Additional Aspects of Aqueous Equilibria

Chapter 17: Additional Aspects of Aqueous Equilibria Chapter 17: Additional Aspects of Aqueous Equilibria -Buffer Problems -Titrations -Precipitation Rx Common Ion Effect- The effect that a common ion (from two different sources) has on an equilibrium. (LeChatelier's

More information

KEY. Practice Problems: Applications of Aqueous Equilibria

KEY. Practice Problems: Applications of Aqueous Equilibria Practice Problems: Applications of Aqueous Equilibria KEY CHEM 1B 1. Ammonia (NH3) is a weak base with a Kb = 1.8 x 1 5. a) Write the balanced chemical equation for the reaction of ammonia with water.

More information

Completion of acid/base/buffer chemistry. Hanson Activity Clicker quiz 3/11/2013. Chs 7 8 of Zumdahl

Completion of acid/base/buffer chemistry. Hanson Activity Clicker quiz 3/11/2013. Chs 7 8 of Zumdahl Completion of acid/base/buffer chemistry Chs 7 8 of Zumdahl Hanson Activity 16 3 Discuss Key Questions 1 of Activity 16 3, page 301, with your partner for three minutes. The clicker quiz will commence

More information

AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS

AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS THE COMMON ION EFFECT The common ion effect occurs when the addition of an ion already present in the system causes the equilibrium to shift away

More information

Operational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA

Operational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA APPLICATIONS OF AQUEOUS EQUILIBRIA Operational Skills Calculating the common-ion effect on acid ionization Calculating the ph of a buffer from given volumes of solution Calculating the ph of a solution

More information

ANSWER KEY CHEMISTRY F14O4 FIRST EXAM 2/16/00 PROFESSOR J. MORROW EACH QUESTION IS WORTH 1O POINTS O. 16.

ANSWER KEY CHEMISTRY F14O4 FIRST EXAM 2/16/00 PROFESSOR J. MORROW EACH QUESTION IS WORTH 1O POINTS O. 16. discard 1 2 ANSWER KEY CHEMISTRY F14O4 FIRST EXAM 2/16/00 PROFESSOR J. MORROW PRINT NAME, LAST: FIRST: I.D.# : EACH QUESTION IS WORTH 1O POINTS 1. 7. 13. 2. 8. 14. 3. 9. 15. 4. 1O. 16. 5. 11. 17. 6. 12.

More information

Chapter 17. Additional Aspects of Equilibrium

Chapter 17. Additional Aspects of Equilibrium Chapter 17. Additional Aspects of Equilibrium 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak

More information

Dougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria

Dougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria Dougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria This is a PRACTICE TEST. Complete ALL questions. Answers will be provided so that you may check your work. I strongly

More information

CHEM1109 Answers to Problem Sheet Isotonic solutions have the same osmotic pressure. The osmotic pressure, Π, is given by:

CHEM1109 Answers to Problem Sheet Isotonic solutions have the same osmotic pressure. The osmotic pressure, Π, is given by: CHEM1109 Answers to Problem Sheet 5 1. Isotonic solutions have the same osmotic pressure. The osmotic pressure, Π, is given by: Π = MRT where M is the molarity of the solution. Hence, M = Π 5 (8.3 10 atm)

More information

Lecture #11-Buffers and Titrations The Common Ion Effect

Lecture #11-Buffers and Titrations The Common Ion Effect Lecture #11-Buffers and Titrations The Common Ion Effect The Common Ion Effect Shift in position of an equilibrium caused by the addition of an ion taking part in the reaction HA(aq) + H2O(l) A - (aq)

More information

Chapter 8: Applications of Aqueous Equilibria

Chapter 8: Applications of Aqueous Equilibria Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations

More information

Chem 116 POGIL Worksheet - Week 11 Titration

Chem 116 POGIL Worksheet - Week 11 Titration Chem 116 POGIL Worksheet - Week 11 Titration Why? Titration is the addition of a standard solution of precisely known concentration (the titrant) to a precisely measured volume of a solution with unknown

More information

Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2

Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2 Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2 Creative Commons License Images and tables in this file have been used from the following sources: OpenStax: Creative Commons Attribution

More information

5. What is the percent ionization of a 1.4 M HC 2 H 3 O 2 solution (K a = ) at 25 C? A) 0.50% B) 0.36% C) 0.30% D) 0.18% E) 2.

5. What is the percent ionization of a 1.4 M HC 2 H 3 O 2 solution (K a = ) at 25 C? A) 0.50% B) 0.36% C) 0.30% D) 0.18% E) 2. Name: Date: 1. For which of the following equilibria does K c correspond to an acid-ionization constant, K a? A) NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O(l) B) NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H

More information

Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq)

Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq) Dealing with chemical stoichiometry Steward Fall 08 of Not including volumetric stoichiometry of Chapter 6.0x10 A 6.0x10 Mol/mol ratio from balanced equation B 6.0x10 6.0x10 s, Equations, and Moles: II

More information

Acid-Base Equilibria (Chapter 10.) Problems: 2,3,6,13,16,18,21,30,31,33

Acid-Base Equilibria (Chapter 10.) Problems: 2,3,6,13,16,18,21,30,31,33 Acid-Base Equilibria (Chapter 10.) Problems: 2,3,6,13,16,18,21,30,31,33 Review acid-base theory and titrations. For all titrations, at the equivalence point, the two reactants have completely reacted with

More information

Chapter 17 Answers. Practice Examples [H3O ] 0.018M, 1a. HF = M. 1b. 30 drops. 2a.

Chapter 17 Answers. Practice Examples [H3O ] 0.018M, 1a. HF = M. 1b. 30 drops. 2a. Chapter 17 Answers Practice Examples 1a. + [HO ] 0.018M, 1b. 0 drops [HF] = 0.8 M. [H O + ] = 0.10 M, HF = 0.97 M. a. + HO 1.10 M, CHO = 0.150 M. b. 15g NaCHO a. The hydronium ion and the acetate ion react

More information

Chapter 17 Additional Aspects of

Chapter 17 Additional Aspects of Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 17 Additional Aspects of John D. Bookstaver St. Charles Community College Cottleville,

More information

CH 223 SPRING EXAM 1 April 21, 2011 DO NOT OPEN THIS EXAM UNTIL TOLD TO DO SO

CH 223 SPRING EXAM 1 April 21, 2011 DO NOT OPEN THIS EXAM UNTIL TOLD TO DO SO CH 223 SPRING 2011 - EXAM 1 April 21, 2011 This exam has 22 questions worth 5 pts each, plus one question worth 2 pts. The total possible score is 112 pts. DO NOT OPEN THIS EXAM UNTIL TOLD TO DO SO Before

More information

The Common Ion Effect

The Common Ion Effect Chapter 17 ACID BASE EQUILIBRIA (Part I) Dr. Al Saadi 1 17.1 The Common Ion Effect A phenomenon known as the common ion effect states that: When a compound containing an ion in common with an already dissolved

More information

Chapter 9. Aqueous Solutions and Chemical Equilibria. Classifying Solutions of Electrolytes

Chapter 9. Aqueous Solutions and Chemical Equilibria. Classifying Solutions of Electrolytes Chapter 9 Aqueous Solutions and Chemical Equilibria Classifying Solutions of Electrolytes Electrolytes solutes form ions when dissolved in water (or certain other solvents, e.g. acetonitrile) Strong (weak,

More information

Chapter 10. Acids, Bases, and Salts

Chapter 10. Acids, Bases, and Salts Chapter 10 Acids, Bases, and Salts Topics we ll be looking at in this chapter Arrhenius theory of acids and bases Bronsted-Lowry acid-base theory Mono-, di- and tri-protic acids Strengths of acids and

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Still having trouble understanding the material? Check

More information

Acid-Base Equilibria. And the beat goes on Buffer solutions Titrations

Acid-Base Equilibria. And the beat goes on Buffer solutions Titrations Acid-Base Equilibria And the beat goes on Buffer solutions Titrations 1 Common Ion Effect The shift in equilibrium due to addition of a compound having an ion in common with the dissolved substance. 2

More information

Chapter 17. Additional Aspects of Aqueous Equilibria 蘇正寬 Pearson Education, Inc.

Chapter 17. Additional Aspects of Aqueous Equilibria 蘇正寬 Pearson Education, Inc. Chapter 17 Additional Aspects of Aqueous Equilibria 蘇正寬 chengkuan@mail.ntou.edu.tw Additional Aspects of Aqueous Equilibria 17.1 The Common-Ion Effect 17.2 Buffers 17.3 Acid Base Titrations 17.4 Solubility

More information

Chemistry Lab Equilibrium Practice Test

Chemistry Lab Equilibrium Practice Test Chemistry Lab Equilibrium Practice Test Basic Concepts of Equilibrium and Le Chatelier s Principle 1. Which statement is correct about a system at equilibrium? (A) The forward and reverse reactions occur

More information

CHEMISTRY 102 Fall 2010 Hour Exam III Page My answers for this Chemistry 102 exam should be graded with the answer sheet associated with:

CHEMISTRY 102 Fall 2010 Hour Exam III Page My answers for this Chemistry 102 exam should be graded with the answer sheet associated with: Hour Exam III Page 1 1. My answers for this Chemistry 102 exam should be graded with the answer sheet associated with: a) Form A b) Form B c) Form C d) Form D e) Form E Consider the titration of 30.0 ml

More information

2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect

2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect Chemistry: Atoms First Julia Burdge & Jason Overby 17 Acid-Base Equilibria and Solubility Equilibria Chapter 15 Acid-Base Equilibria and Solubility Equilibria Kent L. McCorkle Cosumnes River College Sacramento,

More information

Homework #7 Chapter 8 Applications of Aqueous Equilibrium

Homework #7 Chapter 8 Applications of Aqueous Equilibrium Homework #7 Chapter 8 Applications of Aqueous Equilibrium 15. solution: A solution that resists change in ph when a small amount of acid or base is added. solutions contain a weak acid and its conjugate

More information

Course Notes Chapter 9, 11, 12. Charge balance All solutions must be electrically neutral!!!!!!!! Which means they carry a no net charge.

Course Notes Chapter 9, 11, 12. Charge balance All solutions must be electrically neutral!!!!!!!! Which means they carry a no net charge. Course Notes Chapter 9, 11, 12 Chapter 9 Mass balance Let s look at a triprotic system, such as H 3 PO 4 There are four different species within this system H 3 PO 4 H 2 PO 4 - HPO 4 2- PO 4 3- For a 0.0500

More information

Titration a solution of known concentration, called a standard solution

Titration a solution of known concentration, called a standard solution Acid-Base Titrations Titration is a form of analysis in which we measure the volume of material of known concentration sufficient to react with the substance being analyzed. Titration a solution of known

More information

Buffer Effectiveness, Titrations & ph curves. Section

Buffer Effectiveness, Titrations & ph curves. Section Buffer Effectiveness, Titrations & ph curves Section 16.3-16.4 Buffer effectiveness Buffer effectiveness refers to the ability of a buffer to resist ph change Effective buffers only neutralize small to

More information

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria and Solubility Equilibria ACIDS-BASES COMMON ION EFFECT SOLUBILITY OF SALTS Acid-Base Equilibria and Solubility Equilibria Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2 The common

More information

CHEMISTRY 102 Fall 2010 Hour Exam III. 1. My answers for this Chemistry 102 exam should be graded with the answer sheet associated with:

CHEMISTRY 102 Fall 2010 Hour Exam III. 1. My answers for this Chemistry 102 exam should be graded with the answer sheet associated with: 1. My answers for this Chemistry 10 exam should be graded with the answer sheet associated with: a) Form A b) Form B c) Form C d) Form D e) Form E Consider the titration of 30.0 ml of 0.30 M HCN by 0.10

More information

1. (3) The pressure on an equilibrium mixture of the three gases N 2, H 2 and NH 3

1. (3) The pressure on an equilibrium mixture of the three gases N 2, H 2 and NH 3 1. (3) The pressure on an equilibrium mixture of the three gases N 2, H 2 and NH 3 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) is suddenly decreased by doubling the volume of the container at constant temperature.

More information

CHEMISTRY 31 FINAL EXAM May 18, points total Lab Section Number

CHEMISTRY 31 FINAL EXAM May 18, points total Lab Section Number NAME CHEMISTRY 31 FINAL EXAM May 18, 2015-150 points total Lab Section Number Constants: K w (autoprotolysis of H 2 O) = 1.0 x 10-14 h = Planck's constant = 6.63 x 10-34 J s c = speed of light in a vacuum

More information

Try this one Calculate the ph of a solution containing M nitrous acid (Ka = 4.5 E -4) and 0.10 M potassium nitrite.

Try this one Calculate the ph of a solution containing M nitrous acid (Ka = 4.5 E -4) and 0.10 M potassium nitrite. Chapter 17 Applying equilibrium 17.1 The Common Ion Effect When the salt with the anion of a is added to that acid, it reverses the dissociation of the acid. Lowers the of the acid. The same principle

More information

Advanced Placement Chemistry Chapters Syllabus

Advanced Placement Chemistry Chapters Syllabus As you work through the chapter, you should be able to: Advanced Placement Chemistry Chapters 14 16 Syllabus Chapter 14 Acids and Bases 1. Describe acid and bases using the Bronsted-Lowry, Arrhenius, and

More information

CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA

CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA Advanced Chemistry Name Hour Advanced Chemistry Approximate Timeline Students are expected to keep up with class work when absent. CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA Day Plans for the day Assignment(s)

More information

Chapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO

Chapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation Chapter 17 Additional Aspects of John D. Bookstaver St. Charles Community College Cottleville, MO The Common-Ion Effect Consider a solution of acetic acid: CH 3 COOH(aq) + H 2 O(l)

More information

Chemistry 265 December Exam 2011 Smith-Palmer

Chemistry 265 December Exam 2011 Smith-Palmer 1 Chemistry 265 December Exam 2011 Smith-Palmer NAME: [1] 1. Define an anode [1] Define a cathode [2] What is the E o for the following reaction: Ag + + Cu Ag (s) + Cu 2+ Ag + + e - Ag (s) E o = 0.799

More information

CHEM 212 Practice Exam 2 1

CHEM 212 Practice Exam 2 1 CHEM 212 Practice Exam 2 1 1. In the following reaction NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) a. NH 4 + is an acid and NH 3 is its b. H 2 O is an acid and H 3 O + is its c. NH 4 + is an acid

More information

Solutions to: Acid-Base and Solubility Homework Problem Set S.E. Van Bramer 1/3/17

Solutions to: Acid-Base and Solubility Homework Problem Set S.E. Van Bramer 1/3/17 Solutions to: Acid-Base and Solubility Homework Problem Set S.E. Van Bramer /3/7.Calculate the H 3 O + concentration, OH - concentration, ph, and poh of the following solutions. First solve assuming that

More information

Buffer Effectiveness 19

Buffer Effectiveness 19 Buffer Effectiveness 19 Buffer Effectiveness What makes a buffer effective? A buffer should be able to neutralize small to moderate amounts of added acid or base Too much added acid or base will destroy

More information

Solutions, mixtures, and media

Solutions, mixtures, and media Chapter2 Solutions, mixtures, and media n Introduction Whether it is an organism or an enzyme, most biological activities function optimally only within a narrow range of environmental conditions. From

More information

ACID-BASE EQUILIBRIA. Chapter 14 Big Idea Six

ACID-BASE EQUILIBRIA. Chapter 14 Big Idea Six ACID-BASE EQUILIBRIA Chapter 14 Big Idea Six Acid-Base Equilibria Common Ion Effect in Acids and Bases Buffer SoluDons for Controlling ph Buffer Capacity ph-titradon Curves Acid-Base TitraDon Indicators

More information

A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base.

A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base. 1 A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base. after addition of H 3 O + equal concentrations of weak

More information