Chapter 14 Acid and Base Equilibria pH of Weak Acids.

Chapter 14 Acid and Base Equilibria pH of Weak Acids

Chapter 14 Table of Contents Copyright © Cengage Learning. All rights reserved 2 14.5 Calculating the pH of Weak Acid Solutions 14.6Bases 14.7Polyprotic Acids 14.8Acid–Base Properties of Salts 14.9The Effect of Structure on Acid–Base Properties 14.10Acid–Base Properties of Oxides 14.11The Lewis Acid–Base Model 14.12Strategy for Solving Acid–Base Problems: A Summary

Section 14.2 Atomic MassesAcid Strength Return to TOC 1.To review how to calculate pH and pOH for STRONG acids 2.To understand how equilibrium relates to acid and base theory 3.To rank the strength of bases related to the strength of their conjugate acids Objectives

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 4 Exercise Calculate the pH for each of the following solutions. a) 1.0 × 10 –4 M H + pH = 4.00 b) 0.040 M OH – pH = 12.60

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 5 Exercise Calculate the pOH for each of the following solutions. a) 1.0 × 10 –4 M H + pOH = 10.00 b) 0.040 M OH – pOH = 1.40

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 6 Exercise The pH of a solution is 5.85. What is the [OH – ] for this solution? [OH – ] = 7.1 × 10 –9 M

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 7 Strong acid:  Ionization equilibrium lies far to the right.  Value of K a ?  Yields a weak conjugate base  (low ability to attract H+)  HCl, H 2 SO 4, HNO 3 - recognize as strong  HA + H 2 O  H 3 O + + A -  CB (A - ) is weak, won’t hold H +, [H + ] = [HA]

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 8 Weak acid:  Ionization equilibrium lies far to the left.  Value of K a ?  Weaker the acid, stronger its conjugate base.  High ability to attract H+ and hold on to it!  HCN (K a = 6.2 x 10 –10 ) K a means weak acid  HA + H 2 O  H 3 O + + A -  HA weak, CB (A - ) strong, will hold H +  Can’t assume [H + ] = [HA], ICE table for pH

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 11 Water as an Acid and a Base Water is amphoteric:  Behaves either as an acid or as a base. At 25°C: K w = [H + ][OH – ] = 1.0 × 10 –14 No matter what the solution contains, the product of [H + ] and [OH – ] must always equal 1.0 × 10 –14 at 25°C.

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 12 Three Possible Situations [H + ] = [OH – ]; neutral solution [H + ] > [OH – ]; acidic solution [H + ] < [OH – ]; basic solution

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 14 Concept Check HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid base conjugate conjugate acid base What is the equilibrium constant expression for an acid acting in water?

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 15 Concept Check If the equilibrium lies to the right, the value for K a is __________. large (or >1) If the equilibrium lies to the left, the value for K a is ___________. small (or <1) How do you calculate pH for each condition?

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 16 Concept Check HA(aq) + H 2 O(l) H 3 O + (aq) + A – (aq) If water is a better base than A –, do products or reactants dominate at equilibrium? Does this mean HA is a strong or weak acid? Is the value for K a greater or less than 1?

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 18 Concept Check Consider a 1.0 M solution of HCl. Order the following from strongest to weakest base and explain: A – (aq) (from weak acid HA) Cl – (aq) H 2 O…………………..next slide

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 19 Let’s Think About It… A – (aq) (from weak acid HA) Cl – (aq) H2OH2O Is A – (aq) a good base? (From weak acid HA) How good is Cl – (aq) as a base? HCl – strong or weak?..............VERY WEAK! From Very Strong Acid. Water? Weak Base, but stronger than Cl- The bases from strongest to weakest are: A –, H 2 O, Cl –

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 20 Concept Check Acetic acid (HC 2 H 3 O 2 ) and HCN are both weak acids. Acetic acid is a stronger acid than HCN. Arrange these bases from weakest to strongest and explain your answer: H 2 O Cl – CN – C 2 H 3 O 2 –

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 21 Let’s Think About It… Cl – CN – C 2 H 3 O 2 – The bases from weakest to strongest are: Cl –, H 2 O, C 2 H 3 O 2 –, CN –

Section 14.3 The MoleThe pH Scale Return to TOC 1.To review how to calculate pH and pOH for STRONG acids 2.To understand how equilibrium relates to acid and base theory 3.To rank the strength of bases related to the strength of their conjugate acids 4.Work Session: page 674 # 33, 35 Objectives Review

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 23 Objectives To realize what we were missing when we calculated pH of strong acids To use the ICE table and equilibrium to calculate pH of weak acids solutions

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check Calculate the pH of a 1.5 x 10 –2 M solution of HNO 3. Name HNO 3 Is it a strong or weak acid? pH = 1.8

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 25 Calculate the pH of a 1.5 x 10 –2 M solution of HNO 3. When HNO 3 is added to water, write the reaction that takes place immediately (dissociation of the acid): HNO 3 + H 2 O  H 3 O + + NO 3 – Why is this reaction not likely? NO 3 – (aq) + H 2 O(l) HNO 3 (aq) + OH – (aq)

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 26 What else is happening in aq? H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) In aqueous solutions, this reaction is always taking place. But is water the major contributor of H + (H 3 O + )? Which reaction controls the pH? HNO 3 + H 2 O  H 3 O + + NO 3 – or H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) We assumed that [H+] = M of sol’n for strong acids We assumed that the autoionization of water was a small effect of the overall H+ (aq) for strong acids.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 27 Solving Weak Acid Equilibrium Problems 1.List the major species in the solution. 2.Choose the species that can produce H +, and write balanced equations for the reactions producing H +. 3.Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H +. 4.Write the equilibrium expression for the dominant equilibrium.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 28 Solving Weak Acid Equilibrium Problems 5.List the initial concentrations of the species participating in the dominant equilibrium. 6.Define the change needed to achieve equilibrium; that is, define x. 7.Write the equilibrium concentrations in terms of x. 8.Substitute the equilibrium concentrations into the equilibrium expression.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 29 Solving Weak Acid Equilibrium Problems 9.Solve for x the “easy” way, that is, by assuming that [HA] 0 – x about equals [HA] 0. 10.Use the 5% rule to verify whether the approximation is valid. 11.Calculate [H + ] and pH.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 30 Solving Weak Acid Equilibrium Problems 1.List the major species in the solution that can produce H +, and write balanced equations for the reactions producing H +. 2.Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H +. (bigger K) 3.Write the equilibrium expression for the dominant equilibrium and use the ICE table to calculate pH of weak acids

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 31 Concept Check Calculate the pH of a 0.80 M aqueous solution of the acid HCN (K a = 6.2 x 10 –10 ). Is this a strong or weak acid? Write Keq = Ka = What are the major species in solution? HCN, H 2 O

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 32 Let’s Think About It… Why aren’t H + or CN – major species? Write the balanced equations for the reactions producing H+.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 33 Consider This HCN(aq) + H 2 O(l) H 3 O + (aq) + CN – (aq) K a = 6.2 x 10 -10 H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) K w = ? K w = 1.0 x 10 -14 Which reaction controls the pH? Explain.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 34 HCN(aq) + H 2 O(l) H 3 O + (aq) + CN – (aq) K a = 6.2 x 10 -10 Original Question? Calculate the pH of a 0.80 M aqueous solution of the acid HCN (K a = 6.2 x 10 –10 ). I C ------------------------------------------------------------------ E [H+] = 2.2 X10 -5 pH = 4.7

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 35 HCN(aq) + H 2 O(l) H 3 O + (aq) + CN – (aq) K a = 6.2 x 10 -10 Original Question? Calculate the pH of a 0.80 M aqueous solution of the acid HCN (K a = 6.2 x 10 –10 ). I C ------------------------------------------------------------------ E [H+] = 2.2 X10 -5 pH = 4.7

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 36 Exercise Calculate the pH of a 0.50 M aqueous solution of the weak acid HF. (K a = 7.2 x 10 –4 )

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 37 Let’s Think About It… What are the major species in solution? HF, H 2 O Why aren’t H + and F – major species?

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 38 Let’s Think About It… What are the possibilities for the dominant reaction? HF(aq) + H 2 O(l) H 3 O + (aq) + F – (aq) K a =7.2 x 10 -4 H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) K w =1.0 x 10 -14 Which reaction controls the pH? Why?

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 39 Steps Toward Solving for pH K a = 7.2 x 10 –4 pH = 1.72 HF(aq) +H2OH2OH 3 O + (aq) + F – (aq) Initial0.50 M~ 0 Change–x+x Equilibrium0.50–xxx

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 40 Objectives To realize what we were missing when we calculated pH of strong acids To use the ICE table and equilibrium to calculate pH of weak acids solutions Work Session: Page 674 #56, #55 (CH 3 COOH = Acetic Acid), 58, 59, 60, 61, 69, 70 pH = -log[H+] must find [H+] first through ICE

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 42 Percent Dissociation (Ionization) For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 43 Exercise A solution of 8.00 M formic acid (HCHO 2 ) is 0.47% ionized in water. Calculate the K a value for formic acid. ICE Table K a = 1.8 x 10 –4

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 44 Exercise Calculate the pH of an 8.00 M solution of formic acid. Use the data from the previous slide to help you solve this problem. pH = 1.42

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 45 Concept Check The value of K a for a 4.00 M formic acid solution should be: higher than lower thanthe same as the value of K a of an 8.00 M formic acid solution. Explain.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 46 Concept Check The percent ionization of a 4.00 M formic acid solution should be: higher than lower thanthe same as the percent ionization of an 8.00 M formic acid solution. Write the ionization reaction of formic acid and explain in terms of equilibrium shift.

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 47 Concept Check The pH of a 4.00 M formic acid solution should be: higher than lower thanthe same as the pH of an 8.00 M formic acid solution. Explain. Calculate…

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 48 Exercise Calculate the percent ionization of a 4.00 M formic acid solution in water. Same Ka… % Ionization = 0.67%

Section 14.5 Calculating the pH of Weak Acid Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 50 Objectives Review To calculate percent dissociation of weak acids Work Session: 63, 64, 65, 66, 67, 69, 70

Section 14.6 Bases Return to TOC Copyright © Cengage Learning. All rights reserved 52 Arrhenius: bases produce OH – ions. Brønsted–Lowry: bases are proton acceptors. In a basic solution at 25°C, pH > 7. Ionic compounds containing OH  are generally considered strong bases.  LiOH, NaOH, KOH, Ca(OH) 2 pOH = –log[OH – ] pH = 14.00 – pOH

Section 14.6 Bases Return to TOC Copyright © Cengage Learning. All rights reserved 56 pH calculations for solutions of weak bases are very similar to those for weak acids. K w = [H + ][OH – ] = 1.0 × 10 –14 pOH = –log[OH – ] pH = 14.00 – pOH

Section 14.6 Bases Return to TOC Copyright © Cengage Learning. All rights reserved 57 Concept Check Calculate the pH of a 2.0 M solution of the weak base ammonia (NH 3 ). (K b = 1.8 × 10 –5 ) pH = 11.78 How about a 4.0 M solution? 8.0 M solution? 12.0 M solution? 11.93, 12.08, 12.17

Section 14.6 Bases Return to TOC Copyright © Cengage Learning. All rights reserved 58 Objectives Review To calculate pH of strong and weak bases To name a few bases that aren’t hydroxides (OH-) Work Session: 71, 73, 75, 77, 79, 81, 83, 85

Section 14.7 Polyprotic Acids Return to TOC Copyright © Cengage Learning. All rights reserved 60 Acids that can furnish more than one proton. Always dissociates in a stepwise manner, one proton at a time. The conjugate base of the first dissociation equilibrium becomes the acid in the second step. For a typical weak polyprotic acid like H 3 PO 4 : K a1 > K a2 > K a3 For a typical polyprotic acid in water, only the first dissociation step is important to pH.

Section 14.7 Polyprotic Acids Return to TOC Copyright © Cengage Learning. All rights reserved 61 Exercise Calculate the pH of a 1.00 M solution of H 3 PO 4. K a1 = 7.5 x 10 -3 K a2 = 6.2 x 10 -8 K a3 = 4.8 x 10 -13 pH = 1.08

Section 14.7 Polyprotic Acids Return to TOC Copyright © Cengage Learning. All rights reserved 62 Concept Check Calculate the equilibrium concentration of PO 4 3- in a 1.00 M solution of H 3 PO 4. K a1 = 7.5 x 10 -3 K a2 = 6.2 x 10 -8 K a3 = 4.8 x 10 -13 [PO 4 3- ] = 3.6 x 10 -19 M

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 64 Salts Ionic compounds. When dissolved in water, break up into its ions (which can behave as acids or bases).

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 66 Salts A basic solution is formed if the anion of the salt is the conjugate base of a weak acid.  NaF, KC 2 H 3 O 2  K w = K a × K b  Use K b when starting with base.

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 67 Salts An acidic solution is formed if the cation of the salt is the conjugate acid of a weak base.  NH 4 Cl  K w = K a × K b  Use K a when starting with acid.

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 70 Exercise HC 2 H 3 O 2 K a = 1.8 x 10 -5 HCN K a = 6.2 x 10 -10 Calculate the K b values for: C 2 H 3 O 2 − and CN − K b (C 2 H 3 O 2 - ) = 5.6 x 10 -10 K b (CN - ) = 1.6 x 10 -5

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 71 Concept Check Arrange the following 1.0 M solutions from lowest to highest pH. HBr NaOH NH 4 Cl NaCN NH 3 HCN NaCl HF Justify your answer. HBr, HF, HCN, NH 4 Cl, NaCl, NaCN, NH 3, NaOH

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 72 Concept Check Consider a 0.30 M solution of NaF. The K a for HF is 7.2 x 10 -4. What are the major species? Na +, F -, H 2 O

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 73 Let’s Think About It… Why isn’t NaF considered a major species? What are the possibilities for the dominant reactions?

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 74 Let’s Think About It… The possibilities for the dominant reactions are: 1.F – (aq) + H 2 O(l) HF(aq) + OH – (aq) 2.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 3.Na + (aq) + H 2 O(l) NaOH + H + (aq) 4.Na + (aq) + F – (aq) NaF

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 75 Let’s Think About It… How do we decide which reaction controls the pH? F – (aq) + H 2 O(l) HF(aq) + OH – (aq) H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) Determine the equilibrium constant for each reaction.

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 76 Exercise Calculate the pH of a 0.75 M aqueous solution of NaCN. K a for HCN is 6.2 x 10 –10.

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 77 Let’s Think About It… What are the major species in solution? Na +, CN –, H 2 O Why isn’t NaCN considered a major species?

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 78 Let’s Think About It… What are all possibilities for the dominant reaction? The possibilities for the dominant reaction are: 1.CN – (aq) + H 2 O(l) HCN(aq) + OH – (aq) 2.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 3.Na + (aq) + H 2 O(l) NaOH + H + (aq) 4.Na + (aq) + CN – (aq) NaCN Which of these reactions really occur?

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 79 Let’s Think About It… How do we decide which reaction controls the pH? CN – (aq) + H 2 O(l) HCN(aq) + OH – (aq) H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq)

Section 14.8 Acid–Base Properties of Salts Return to TOC Copyright © Cengage Learning. All rights reserved 80 Steps Toward Solving for pH K b = 1.6 x 10 –5 pH = 11.54 CN – (aq) +H2OH2OHCN(aq) +OH – (aq) Initial0.75 M0~ 0 Change–x+x Equilibrium0.75–xxx

Section 14.9 The Effect of Structure on Acid–Base Properties Return to TOC Copyright © Cengage Learning. All rights reserved 81 Models of Acids and Bases Two factors for acidity in binary compounds:  Bond Polarity (high is good)  Bond Strength (low is good)

Section 14.9 The Effect of Structure on Acid–Base Properties Return to TOC Copyright © Cengage Learning. All rights reserved 83 Oxyacids Contains the group H–O–X. For a given series the acid strength increases with an increase in the number of oxygen atoms attached to the central atom. The greater the ability of X to draw electrons toward itself, the greater the acidity of the molecule.

Section 14.10 Acid–Base Properties of Oxides Return to TOC Copyright © Cengage Learning. All rights reserved 86 Oxides Acidic Oxides (Acid Anhydrides):  OX bond is strong and covalent. SO 2, NO 2, CO 2 When HOX grouping is dissolved in water, the OX bond will remain intact. It will be the polar and relatively weak HO bond that will tend to break, releasing a proton.

Section 14.10 Acid–Base Properties of Oxides Return to TOC Copyright © Cengage Learning. All rights reserved 87 Oxides Basic Oxides (Basic Anhydrides):  OX bond is ionic. K 2 O, CaO If X has a very low electronegativity, the OX bond will be ionic and subject to being broken in polar water, producing a basic solution.

Section 14.11 The Lewis Acid–Base Model Return to TOC Copyright © Cengage Learning. All rights reserved 88 Lewis Acids and Bases Lewis acid: electron pair acceptor Lewis base: electron pair donor Lewis acid Lewis base

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 90 Concept Check SKIP>>>>1,2,3 Consider a solution of NaA where A – is the anion from weak acid HA: A – (aq) + H 2 O(l) HA(aq) + OH – (aq) base acid conjugate conjugate acid base a)Which way will equilibrium lie? Write Ka… left

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 91 Concept Check SKIP>>>> Consider a solution of NaA where A – is the anion from weak acid HA: A – (aq) + H 2 O(l) HA(aq) + OH – (aq) base acid conjugate conjugate acid base b) Is the value for K b greater than or less than 1? SKIP>>>>> less than 1

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 92 Concept Check SKIP>>>>>>> Consider a solution of NaA where A – is the anion from weak acid HA: A – (aq) + H 2 O(l) HA(aq) + OH – (aq) base acid conjugate conjugate acid base c) Does this mean A – is a strong or weak base? weak base

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 93 Concept Check Discuss whether the value of K for the reaction: HCN(aq) + F – (aq) CN – (aq) + HF(aq) is  1  1 =1 (K a for HCN is 6.2×10 –10 ; K a for HF is 7.2×10 –4.) Explain your answer.

Section 14.2 Atomic MassesAcid Strength Return to TOC Copyright © Cengage Learning. All rights reserved 94 Concept Check Calculate the value for K for the reaction: HCN(aq) + F – (aq) CN – (aq) + HF(aq) (K a for HCN is 6.2×10 –10 ; K a for HF is 7.2×10 –4.) K = 8.6 x 10 –7

Chapter 14 Acid and Base Equilibria pH of Weak Acids.

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Chapter 14 Acid and Base Equilibria pH of Weak Acids.

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