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General Chemistry II CHEM 152  Unit 2 Week 8
Week 8 Reading Assignment Chapter 15 – Sections 15.6 (finding pH of  acids), 15.7 (bases), 15.8 (salts)
Methods for  Calculating  the pH of an Aqueous Solution With a strong acid or base – we assume 100% dissociation and directly get the concentration of  H 3 O +  or  OH ¯  to get the pH. But what happens when we are dealing with weak acids or bases? To answer this question we need to apply the ideas we have learned about  chemical equilibrium.
Equilibria Involving  Weak  Acids ,[object Object],[object Object],[object Object],(K is designated K a  for  ACID ) Because [H 3 O + ] and [OAc - ] are SMALL,  K a  << 1
Equilibrium Constants  for  Weak  Acids Weak acid has K a  < 1  Leads to  more [H 3 O + ] than in pure water  and a  pH of 2 - 7   Acid Ionization Constant In general:
Ionization Constants for Acids Increase strength
Equilibrium Constants  for Weak Bases Weak base has K b  < 1  Leads to  more [OH ¯ ] than in pure water  and a  pH of 7-12   Base Ionization Constant Write the ionization equation and the ionization constant for :  H 2 PO 4 -  (acid) and C 6 H 5 NH 2  (base) .
Ionization Constants for Bases  Increase strength
pH for a  Weak Acid ,[object Object],Step 1.  Define equilibrium concentrations: [CH 3 COOH] + H 2 O    [H 3 O + ] + [CH 3 COO - ] I  1.00     ~0   0 C -x     +x   +x E  1.00-x    x   x Note that we neglect [H 3 O + ] from H 2 O
pH for a  Weak Acid ,[object Object],Step 3.  Solve K a  expression to find x  We can assume x is very small because K a  is so small. Now we can more easily solve this approximate expression.
pH for a  Weak Acid ,[object Object],x =  [ H 3 O + ]  =  [ CH 3 COO - ]  = [K a  • 1.00] 1/2   x  =  [ H 3 O + ]  =  [ CH 3 COO - ]  =  4.2 x 10 -3  M pH = -log [ H 3 O + ]= -log (4.2 x 10 -3 )=  2.37
pH for a  Weak Acid ,[object Object],For many weak acids [H 3 O + ]= [conjugate base]= [K a • C o ] 1/2 where C 0  = initial concentration of acid In general: If  100•K a  <  C o , then  [H 3 O + ]  =  [K a •C o ] 1/2
Calculate the pH of a 0.0010 M solution of formic acid,  HCOOH .  K a  = 1.8 x 10 -4
Your Turn ,[object Object],Exact Solution [H 3 O + ]= [ HCO 2 - ] =  3.4 x 10 -4  M [ HCO 2 H ]= 0.0010-3.4 x 10 -4  = 0.0007 M  pH = 3.46  Approximate solution [H 3 O + ] = [K a  • C o ] 1/2  = 4.2 x 10 -4  M,  pH = 3.37
pH for a  Weak Base ,[object Object],[object Object],[object Object],Step 1.  Define equilibrium concentrations: [NH 3 ] + H 2 O    [NH 4 + ] + [OH - ] Initial 0.010     0   0 Change  -x     +x  +x Equilib  0.010 - x    +x  +x
pH for a  Weak Base ,[object Object],Assume x is small (100•K b  < C o ), so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4  M and  [NH 3 ] = 0.010 - 4.2 x 10 -4  ≈  0.010 M Step 3.  Calculate pH [OH - ] =  4.2 x 10 -4  M so pOH = - log [OH - ]  =  3.37 Because pH + pOH = 14,  pH = 10.63
Calculate the pH of a 0.15 M solution of (CH 3 ) 3 N.  K b  = 6.3 x 10 -5
[object Object],Acid-Base Properties of  Salts Cl -  ion is a VERY weak base because its conjugate acid is strong  Therefore,  Cl -     neutral solution Consider  NH 4 Cl NH 4 Cl(aq)     NH 4 + (aq)  +  Cl - (aq) (a) Reaction of Cl -  with H 2 O Cl -   +  H 2 O   HCl  +  OH - base   acid acid base What happens with the pH if we dissolve a salt in water?
[object Object],[object Object],[object Object],[object Object],Acid-Base Properties of  Salts NH 4 +  ion is a moderate acid because its conjugate base is weak.  Therefore,  NH 4 +     acidic solution
A solution of  NaCl  would be: ,[object Object],[object Object],[object Object]
A solution of  K 3 PO 4  would be: ,[object Object],[object Object],[object Object]
A solution of  NH 4 F  would be: ,[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],Acid-Base Properties of Salts Step 2 . Set up ICE table [CO 3 2- ] + H 2 O    [HCO 3 - ] + [OH - ]  initial 0.10 0 0 change -x +x +x equilib 0.10-x  x  x
Acid-Base Properties of  Salts Assume 0.10 - x  ≈  0.10, because 100•K b  < C o x =[HCO 3 - ] =[OH - ] = 0.0046 M  Step 2 . Solve the equilibrium expression Step 3. Calculate the pH [OH - ]= 0.0046 M, pOH= - log [OH - ]  =  2.34 pH  +  pOH = 14,  so  pH  =  11.66
Calculate the pH of a 0.15 M solution of KNO 2 .  K a (HNO 2 ) = 4.6 x 10 -4
Polyprotic Acids Some acids donate more than one H +  per molecule:  H 2 SO 4 , H 2 CO 3 , H 3 PO 4 These acids donate their protons in an stepwise manner: H 3 PO 4 (aq) + H 2 O(l)    H 3 O + (aq) + H 2 PO 4 - (aq)   K a =7.5 x 10 -3 H 2 PO 4 -(aq) + H 2 O(l)    H 3 O + (aq) + HPO 4 2- (aq)   K a =6.2 x 10 -8 HPO 4 2- (aq) + H 2 O(l)    H 3 O + (aq) + PO 4 3-  (aq)   K a =3.6 x 10 -13 The resulting acids are weaker
Molecular structure  and acid/base strength There are four main effects that influence the relative strength of an acid: Size of the anion Electronegativity of the H-bearing atom Inductive effect Resonance stabilization of the anion The first two factors influence  binary  acids The latter two factors influence  non-binary  acids
Binary Acid Strength Factors affecting  binary acid strength The polarity of the H-A bond   (most important factor when comparing atoms in the same row in the periodic table)  electronegativity The bond energy of the H-A bond  (dependent on the length of the bond; important when comparing atoms in the same column in the periodic table) size of the anion HA type of acid
Binary Acid Strength Which of these is a stronger acid: HCl or HBr? Why? Which of these is a stronger acid: H 3 P or H 2 S? Why?
Strength of Non-binary Acids For nonbinary acids –  Atoms with higher electronegativity can draw the electrons away from the bond with the acidic hydrogen – making that bond more polar.  This is known as the  inductive effect . Or if the anion of the acid has several equal resonance structures – the anion would be rather stable – and the acid more likely to lose its acidic hydrogen.  This is known as  resonance stablization .
Strength of Oxyacids Acids in which the acidic hydrogen is bonded directly to oxygen in an  H-O-Z  bond are called oxyacids. For each of the following pairs, which acid is stronger – and why? Strength?
Why is CH 3 COOH an Acid? ,[object Object],[object Object],[object Object],Example of Carboxylic Acid  ( -COOH  group )    Weak Acid
[object Object],[object Object],Acetic acid Trichloroacetic acid Which acid is stronger?  Why?
Weak Bases: Amines Amines are compounds that, like ammonia, have a nitrogen atom with three of its valence electrons in covalent bonds and an unshared electron pair on the nitrogen atom.  The lone pair of electrons can accept an H + . H 2 O H +
Which of the following is a weak base? ,[object Object],[object Object],[object Object],[object Object],(A) (B)
1. What is the pH of a solution of 0.050 M (C 2 H 5 ) 2 NH?
2. What is the pH of a solution of 0.050 M HClO 4 ?
Summary Activity 3. Label the following as acidic, basic, or neutral solutions: KCl NaNO 2 (CH 3 ) 2 NH 2 Cl
Which of the following is the strongest acid? HClO 2 , HBrO 2 , HIO 2
Which of the following is the strongest acid? H 3 N, H 2 O, HF
What kind of substance is C 2 H 5 NH 2 ? ,[object Object],[object Object],[object Object],[object Object]

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Lect w8 152 - ka and kb calculations_abbrev_alg

  • 1. General Chemistry II CHEM 152 Unit 2 Week 8
  • 2. Week 8 Reading Assignment Chapter 15 – Sections 15.6 (finding pH of acids), 15.7 (bases), 15.8 (salts)
  • 3. Methods for Calculating the pH of an Aqueous Solution With a strong acid or base – we assume 100% dissociation and directly get the concentration of H 3 O + or OH ¯ to get the pH. But what happens when we are dealing with weak acids or bases? To answer this question we need to apply the ideas we have learned about chemical equilibrium.
  • 4.
  • 5. Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to more [H 3 O + ] than in pure water and a pH of 2 - 7 Acid Ionization Constant In general:
  • 6. Ionization Constants for Acids Increase strength
  • 7. Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to more [OH ¯ ] than in pure water and a pH of 7-12 Base Ionization Constant Write the ionization equation and the ionization constant for : H 2 PO 4 - (acid) and C 6 H 5 NH 2 (base) .
  • 8. Ionization Constants for Bases Increase strength
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  • 13. Calculate the pH of a 0.0010 M solution of formic acid, HCOOH . K a = 1.8 x 10 -4
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  • 17. Calculate the pH of a 0.15 M solution of (CH 3 ) 3 N. K b = 6.3 x 10 -5
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  • 24. Acid-Base Properties of Salts Assume 0.10 - x ≈ 0.10, because 100•K b < C o x =[HCO 3 - ] =[OH - ] = 0.0046 M Step 2 . Solve the equilibrium expression Step 3. Calculate the pH [OH - ]= 0.0046 M, pOH= - log [OH - ] = 2.34 pH + pOH = 14, so pH = 11.66
  • 25. Calculate the pH of a 0.15 M solution of KNO 2 . K a (HNO 2 ) = 4.6 x 10 -4
  • 26. Polyprotic Acids Some acids donate more than one H + per molecule: H 2 SO 4 , H 2 CO 3 , H 3 PO 4 These acids donate their protons in an stepwise manner: H 3 PO 4 (aq) + H 2 O(l)  H 3 O + (aq) + H 2 PO 4 - (aq) K a =7.5 x 10 -3 H 2 PO 4 -(aq) + H 2 O(l)  H 3 O + (aq) + HPO 4 2- (aq) K a =6.2 x 10 -8 HPO 4 2- (aq) + H 2 O(l)  H 3 O + (aq) + PO 4 3- (aq) K a =3.6 x 10 -13 The resulting acids are weaker
  • 27. Molecular structure and acid/base strength There are four main effects that influence the relative strength of an acid: Size of the anion Electronegativity of the H-bearing atom Inductive effect Resonance stabilization of the anion The first two factors influence binary acids The latter two factors influence non-binary acids
  • 28. Binary Acid Strength Factors affecting binary acid strength The polarity of the H-A bond (most important factor when comparing atoms in the same row in the periodic table) electronegativity The bond energy of the H-A bond (dependent on the length of the bond; important when comparing atoms in the same column in the periodic table) size of the anion HA type of acid
  • 29. Binary Acid Strength Which of these is a stronger acid: HCl or HBr? Why? Which of these is a stronger acid: H 3 P or H 2 S? Why?
  • 30. Strength of Non-binary Acids For nonbinary acids – Atoms with higher electronegativity can draw the electrons away from the bond with the acidic hydrogen – making that bond more polar. This is known as the inductive effect . Or if the anion of the acid has several equal resonance structures – the anion would be rather stable – and the acid more likely to lose its acidic hydrogen. This is known as resonance stablization .
  • 31. Strength of Oxyacids Acids in which the acidic hydrogen is bonded directly to oxygen in an H-O-Z bond are called oxyacids. For each of the following pairs, which acid is stronger – and why? Strength?
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  • 34. Weak Bases: Amines Amines are compounds that, like ammonia, have a nitrogen atom with three of its valence electrons in covalent bonds and an unshared electron pair on the nitrogen atom. The lone pair of electrons can accept an H + . H 2 O H +
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  • 36. 1. What is the pH of a solution of 0.050 M (C 2 H 5 ) 2 NH?
  • 37. 2. What is the pH of a solution of 0.050 M HClO 4 ?
  • 38. Summary Activity 3. Label the following as acidic, basic, or neutral solutions: KCl NaNO 2 (CH 3 ) 2 NH 2 Cl
  • 39. Which of the following is the strongest acid? HClO 2 , HBrO 2 , HIO 2
  • 40. Which of the following is the strongest acid? H 3 N, H 2 O, HF
  • 41.

Notas do Editor

  1. Update for Tro.
  2. Kb = 6.9e-4