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Section 14.1

Section 14.1. Nature of Acids and Bases. Arrhenius Definition. Acids produce hydrogen ions in aqueous solution. HCl (aq)  H + (aq) + Cl - (aq) Bases produce hydroxide ions when dissolved in water. NaOH (aq)  Na + (aq) + OH - (aq)

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Section 14.1

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  1. Section 14.1 Nature of Acids and Bases

  2. Arrhenius Definition • Acids produce hydrogen ions in aqueous solution. • HCl (aq)  H+(aq) + Cl-(aq) • Bases produce hydroxide ions when dissolved in water. • NaOH (aq)  Na+(aq) + OH-(aq) • Limits what can be considered bases (as we’ll see with other definitions).

  3. Lewis Definition e- pair acceptor e- pair donor • An acid is an electron pair acceptor. • A base is an electron pair donor. • Easy to see if Lewis structures are drawn:

  4. Bronsted-Lowry Definition • This is the one we’ll focus on! • An acid is a proton (H+) donor and a base is a proton acceptor. • HCl is an acid. • When it dissolves in water it gives its proton to water. • HCl(g) + H2O(l) H3O+ + Cl- • Water is a base since it accepts the H+. • In the Arrhenius definition water would not be considered a base!

  5. Conjugate Acid-Base Pairs • General equation • HA(aq) + H2O(l) H3O+(aq) + A-(aq) acid+ base conjugate acid + conjugate base • This is an equilibrium. • Equilibrium favors the side with the weaker acid and base. • Refer to the handout for acid and base strength. • Note: conjugate bases of strong acids are weak, and conjugate bases of weak acids are strong. In other words, the stronger the acid/base, the weaker the conjugate base/acid and vice-versa.

  6. Acid dissociation constant Ka • Recall strong vs. weak acids/bases! • Strong = essentially completely dissociate in water (no equilibrium). • Weak = partially dissociate in water. Equilibrium is therefore present! • For a weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq) • Water is often left out: HA (aq) H+(aq) + A-(aq)

  7. Acid dissociation constant Ka • Since weak acids result in equilibrium, an equilibrium expression can be written: HA(aq) + H2O(l) H3O+(aq) + A-(aq) • Ka = [H3O+][A-] [HA] • Acid dissociation constant = Ka. • Note: technically all acids have a Ka value. Size of Ka indicates strength!

  8. Acid dissociation constant Ka • If water is left out: • HA(aq) H+(aq) + A-(aq) • Ka = [H+][A-] [HA] Shown without H3O+

  9. Section 14.2 Acid Strength

  10. Types of Acids • Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic). • Ex: H2SO4, H3PO4 • Oxyacids - Proton is attached to the oxygen of an ion. • Ex: H2SO4, HNO3 • Organic acids contain the carboxyl group -COOH (acidic H+ attached to O). • Ex: CH3COOH (acetic acid) • Generally very weak.

  11. Ka of Polyprotic Acids The first Ka value is much larger than the second, third, etc. The second Ka value is much larger than the third Ka value, etc. Since the first Ka value is by far the largest, second, third, etc. Ka values can be ‘ignored’ and the first Ka value can usually be used for the acid dissociation constant of the acid.

  12. Amphoteric • Can behave as either an acid or a base. • Water undergoes autonionization (it autoionizes). • 2H2O(l) H3O+(aq) + OH-(aq) • KW= [H3O+][OH-]=[H+][OH-] • At 25ºC KW = 1.0 x10-14 • Occurs in EVERY aqueous solution. • Neutral solution [H+] = [OH-]= 1.0 x10-7 • Acidic solution [H+] > [OH-] • Basic solution [H+] < [OH-] Notice it’s very small!

  13. Amphoteric Continued Note: although water dissociation occurs in all solutions and contributes to [H+], it is small in comparison and can be ignored when calculating [H+]. Adding other species to water can also allow water to act as an acid or a base.

  14. Strong Acids • HBr, HI, HCl, HNO3, H2SO4, HClO4 • Completely dissociated • [H+] = [HA] • Ex: What is [H+] in a 0.10M HNO3 solution? HNO3 H+ + NO3- *Mole ratio is 1:1 for HNO3:H+ *Thus [H+] = 0.10M

  15. Weak Acids • Ka will be small. • It will be an equilibrium problem to find [H+]. • Determine whether most of the H+ will come from the acid or the water. • Compare Ka and Kw. • Whichever is larger is the one that will donate more H+ (this is usually the acid). • Rest is just like last chapter.

  16. Weak Acids • Because you’re dealing with weak acids, it can be assumed that the acid won’t dissociate much. • Lets you make assumptions and simplify terms when solving for x. (If you forget and don’t simplify that’s OK- you will just need to use the quadratic formula). • Then calculate the [H+].

  17. Weak Acids • For weak acids, the concentrations of H+ and A- may be found if both the initial concentration of the acid and Ka are known. • Ex: What is the [H+] in 0.300M acetic acid solution? Ka = 1.8 x 10-5 • Remember- this is an equilibrium problem! Set up your ICE table, then solve for x. • Answer: x = [H+] = 2.3 x 10-3 M

  18. Practice Problem #1 Pg. 675 #55: A 0.0560g sample of acetic acid is added to enough water to make 50.00mL of solution. Calculate the [H+], [CH3COO-], and [CH3COOH]. The Ka for acetic acid = 1.8 x 10-5.

  19. Summary: Strong vs. Weak Remember- strong acids: [HA] = [H+] and weak acids = equilibrium problem to find [H+]. Otherwise, once the [H+] is known, pH problems are the same.

  20. Section 1 Homework • Pg. 672 # 19(a&c), 21, 24, 28, 29, 32

  21. pH Scale • pH= -log[H+] or pH = -log [H3O+] • pH of pure water = -log(1.0 x 10-7) = 7.00 • In other words, this is neutral. • pH < 7.00 is acidic; pH > 7.00 is basic • Notice that as pH decreases, [H+] increases • Sig figs: only the digits after the decimal place of a pH are significant • [H+] = 1.0 x 10-8 pH= 8.00 2 sig figs

  22. pH Scale Continued • The pH of a solution can be estimated if the [H+] is known. • Look at the exponent to estimate the pH • Ex: [H+] = 1 x 10-5; pH = 5 • pOH can also be calculated: • pOH = -log[OH-]

  23. [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 pH 0 1 3 5 7 9 11 13 14 14 13 11 9 7 5 3 1 0 pOH 10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100 [OH-] Acidic Neutral Basic

  24. Important Relationships • KW = 1.0 x 10-14 = [H+][OH-] • pKW = 14.00 = pH + pOH • pH = -log[H+] • pOH = -log[OH-] • [H+],[OH-],pH and pOH Given any one of these we can find the other three! • There are also often multiple ways to correctly solve these problems!

  25. Example #1 Note: if [H+] is given, no need to worry if acid is strong or weak. Answer: Either way you get pOH = 9.000 The [H+] = 1.00 x 10-5 M. What is the pOH for this solution? Step 1: Find pH. Step 2: Then use 14 = pH + pOH to solve for pOH. Can you think of another approach? Find [OH-] using: 1.0x10-14 = [H+][OH-]. Then use: pOH = -log[OH-].

  26. Example #2 [H+] is not given, so now we need to decide if the acid is strong or weak, and then solve for [H+]. Calculate the pH of a 1.00M solution of HF, whose Ka = 7.2x10-4. (1) Small Ka, so it’s a weak acid. (2) Major species: HF and H2O. (3) Which provides the most H+ ions? Ka = 7.2 x 10-4 and Kw = 1.0 x 10-14 So HF provides more H+ ions. (4) Equilibrium we’re investigating: HF (aq) H+ (aq) + F- (aq) So, Ka = 7.2 x 10-4 = [H+][F-]/[HF]

  27. Example #2 Simplification: since Ka is small, can assume the initial [HF] won’t change noticeably, so at equilibrium [HF] ≈ 1.00. (5) Set up ICE table: HF (aq) H+ (aq) + F- (aq) I 1.00 0 0 C -x +x +x E 1.00-x x x 7.2 x 10-4 = (x)(x) (1.00-x) So: 7.2 x 10-4 = x2 Solve for x: x = 0.027 1.00

  28. Example #2 (5) Verify assumption was OK- x divided by the initial concentration of HF must be less than or equal to 5%: 0.027 x 100% = 2.7% 1.00 Since 2.7% < 5%, the assumption that x was small enough to be neglected was valid. (6) Calculate [H+] and then the pH: [H+] = x = 0.027M pH = -log(0.027) = 1.57

  29. Practice Problem #1 Calculate the pH of a 0.100M solution of hypochlorous acid (HOCl). The Ka value = 3.5 x 10-8. [H+] = 5.9 x 10-5M, so pH = 4.23

  30. A mixture of Weak Acids • The process is the same. • Determine the major species. • The stronger will predominate (whichever has the largest Ka value). • Doubt you’ll see this on the AP exam, but just in case!

  31. Example Calculate the pH of a solution that contains 1.00MHCN whose Ka = 6.2x10-10 and 5.00MHNO2 whose Ka = 4.0x10-4. Approach: same as before, now just need to consider THREE K values (the two above and water). Since Ka for HNO2 is much larger than Kw and the Ka for HCN, this is the only one that needs to be used for finding [H+]. [H+] = 4.5 x 10-2, so pH = 1.35

  32. AP Practice Question A 0.1M solution of acetic acid (CH3COOH) has a pH of about: 1 3 7 10 You can answer this question without doing any math!

  33. AP Practice Question What is the ionization constant, Ka, for a weak monoprotic acid if a 0.30M solution has a pH of 4.0? a) 3.3 x 10-8 c) 1.7 x 10-6 b) 4.7 x 10-2 d) 3.0 x 10-4 Solve without using a calculator: -Estimate [H+]: pH = 4.0, so [H+] = 1 x 10-4 -Ka = (1 x 10-4)2/0.30  ÷1 = 1 x 10-8 so Ka should be a little bigger

  34. When Kw Matters…

  35. Percent Dissociation [H+] at equilibrium • = amount dissociated (M) x 100 initial concentration (M) • Example: Calculate the % dissociation of 1.00 M acetic acid, Ka = 1.8 x 10-5. • Approach is the same as weak acid problems! Solve like an equilibrium problem for the necessary concentrations, then calculate % dissociation. • % dissociation = 0.42%

  36. Sections 2-3 Homework Problems Pg. 673 #17, 22, 40, 43, 47, 51, 57

  37. Section 14.7 Polyprotic Acids

  38. Polyprotic acids • Always dissociate stepwise. • The first H+ comes off much easier than the second. • Ka for the first step is much bigger than Ka for the second, the second is bigger than the third, etc. • More difficult to lose the next H+ because the negative charge increases. • Denoted Ka1, Ka2, Ka3.

  39. Polyprotic acid • H2CO3 H+ + HCO3- Ka1= 4.3 x 10-7 • HCO3- H+ + CO3-2 Ka2= 4.3 x 10-10 • Conjugate base in first step is the acid in second. • In calculations we can normally ignore the second, third, etc. dissociation.

  40. Sulfuric Acid is Special • In the first step it is a strong acid. • No Ka value given- complete dissociation. • Second step is a weak acid. • Ka2 = 1.2 x 10-2 • Small, but not always small enough to ignore. • If the initial concentration of H2SO4 is low enough, the second H+ impacts pH!

  41. Sulfuric Acid Example • Calculate the pH of a 0.0100M H2SO4 solution. Ka2 = 0.012. • The first acidic proton (H+) dissociates completely (H2SO4 is strong at first): H2SO4 H+ + HSO4- Thus: [H2SO4] = [H+] = [HSO4-] = 0.0100M • Now, we need to consider the second acidic proton.

  42. Sulfuric Acid Example The second acidic proton comes from HSO4-, which is a weak acid. So this is treated like an equilibrium problem: HSO4- H+ + SO4-2 I 0.0100 0.0100 0 C -x +x +x E 0.0100-x 0.0100+x x Plug into Ka2 expression: 0.012 = (x)(0.0100+x)/(0.0100-x)

  43. Sulfuric Acid Example We can try to use the simplification that x is negligible with respect to HSO4- and H+: 0.012 = (x)(0.0100)/(0.0100), however the value for x = 0.012, which does not make sense. Thus the simplification is not valid! Use quadratic formula to solve! x = 0.0045 [H+] = 0.0100+0.0045 = 0.0145M pH = 1.84

  44. Notice that most Ka1 values are significantly larger than the Ka2 values, and thus typically do not impact the [H+] or the pH. Typically, exceptions are sulfuric acid and oxalic acid.

  45. Section 7 Homework Pg. 676 #93, 96, 97, 98

  46. Lesson Essential Question: How do calculations with acids differ from calculations with bases?

  47. Section 14.6 Bases

  48. Bases • The OH-is a strong base. • Hydroxides of the alkali metals are strong bases because they dissociate completely when dissolved. • The hydroxides of alkaline earth metals (Ca(OH)2, etc.) are also strong, but they don’t dissolve well in water. • Used as antacids because [OH-] can’t build up.

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