Upcoming SlideShare
×

# Lect w8 152 - ka and kb calculations_abbrev_alg

1,413 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,413
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
23
0
Likes
0
Embeds 0
No embeds

No notes for slide
• Update for Tro.
• Kb = 6.9e-4
• ### Lect w8 152 - ka and kb calculations_abbrev_alg

1. 1. General Chemistry II CHEM 152 Unit 2 Week 8
2. 2. Week 8 Reading Assignment Chapter 15 – Sections 15.6 (finding pH of acids), 15.7 (bases), 15.8 (salts)
3. 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. 4. Equilibria Involving Weak Acids <ul><li>Consider the equilibrium in water for the weak acid, acetic acid CH 3 COOH: </li></ul><ul><li>CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq) </li></ul><ul><li>Acid Conj. base </li></ul>(K is designated K a for ACID ) Because [H 3 O + ] and [OAc - ] are SMALL, K a << 1
5. 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. 6. Ionization Constants for Acids Increase strength
7. 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. 8. Ionization Constants for Bases Increase strength
9. 9. pH for a Weak Acid <ul><li>Imagine that you have 1.00 M CH 3 COOH . Calculate the equilibrium concentrations of CH 3 COOH , H 3 O + , CH 3 COO - , and the pH of the solution (K a = 1.8 x 10 -5 ). How do you do it? </li></ul>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
10. 10. pH for a Weak Acid <ul><li>Step 2. Write K a expression </li></ul>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.
11. 11. pH for a Weak Acid <ul><li>Step 4. Solve K a approximate expression </li></ul>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
12. 12. pH for a Weak Acid <ul><li>Consider the approximate expression in the last problem: </li></ul>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
13. 13. Calculate the pH of a 0.0010 M solution of formic acid, HCOOH . K a = 1.8 x 10 -4
14. 14. Your Turn <ul><li>Calculate the pH of a 0.0010 M solution of formic acid, HCOOH . K a = 1.8 x 10 -4 </li></ul>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
15. 15. pH for a Weak Base <ul><li>Imagine that you have 0.010 M NH 3 . Calculate the pH. </li></ul><ul><li>NH 3 + H 2 O  NH 4 + + OH - </li></ul><ul><li>K b = 1.8 x 10 -5 </li></ul>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
16. 16. pH for a Weak Base <ul><li>Step 2. Solve the equilibrium expression </li></ul>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
17. 17. Calculate the pH of a 0.15 M solution of (CH 3 ) 3 N. K b = 6.3 x 10 -5
18. 18. <ul><li>MX + H 2 O  acidic or basic solution? </li></ul>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?
19. 19. <ul><li>NH 4 Cl(aq)  NH 4 + (aq) + Cl - (aq) </li></ul><ul><li>(b) Reaction of NH 4 + with H 2 O </li></ul><ul><li>NH 4 + + H 2 O  NH 3 + H 3 O + </li></ul><ul><li>acid base base acid </li></ul>Acid-Base Properties of Salts NH 4 + ion is a moderate acid because its conjugate base is weak. Therefore, NH 4 +  acidic solution
20. 20. A solution of NaCl would be: <ul><li>acidic </li></ul><ul><li>basic </li></ul><ul><li>neutral </li></ul>
21. 21. A solution of K 3 PO 4 would be: <ul><li>acidic </li></ul><ul><li>basic </li></ul><ul><li>neutral </li></ul>
22. 22. A solution of NH 4 F would be: <ul><li>acidic </li></ul><ul><li>basic </li></ul><ul><li>neutral </li></ul>
23. 23. <ul><li>Calculate the pH of a 0.10 M solution of Na 2 CO 3 . Step 1: Decide which ion will affect pH. </li></ul><ul><li>Na + + H 2 O  ? </li></ul><ul><li>CO 3 2- + H 2 O  ? K b = 2.1 x 10 -4 </li></ul>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
24. 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. 25. Calculate the pH of a 0.15 M solution of KNO 2 . K a (HNO 2 ) = 4.6 x 10 -4
26. 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. 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. 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. 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. 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. 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?
32. 32. Why is CH 3 COOH an Acid? <ul><li>1 .The electronegativity of the O atoms causes the H attached to O to be highly positive. </li></ul><ul><li>2.The O—H bond is highly polar. </li></ul><ul><li>3.The H atom of O—H is readily attracted to polar H 2 O . </li></ul>Example of Carboxylic Acid ( -COOH group )  Weak Acid
33. 33. <ul><li>Trichloroacetic acid is much stronger acid owing to the high electronegativity of Cl, which withdraws electrons from the rest of the molecule. This makes the O—H bond highly polar. The H of O—H is very positive. </li></ul><ul><li> Inductive effect </li></ul>Acetic acid Trichloroacetic acid Which acid is stronger? Why?
34. 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 +
35. 35. Which of the following is a weak base? <ul><li>A </li></ul><ul><li>B </li></ul><ul><li>Both </li></ul><ul><li>Neither </li></ul>(A) (B)
36. 36. 1. What is the pH of a solution of 0.050 M (C 2 H 5 ) 2 NH?
37. 37. 2. What is the pH of a solution of 0.050 M HClO 4 ?
38. 38. Summary Activity 3. Label the following as acidic, basic, or neutral solutions: KCl NaNO 2 (CH 3 ) 2 NH 2 Cl
39. 39. Which of the following is the strongest acid? HClO 2 , HBrO 2 , HIO 2
40. 40. Which of the following is the strongest acid? H 3 N, H 2 O, HF
41. 41. What kind of substance is C 2 H 5 NH 2 ? <ul><li>Strong acid </li></ul><ul><li>Strong base </li></ul><ul><li>Weak Acid </li></ul><ul><li>Weak base </li></ul>