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Packet 17:Principles of Reactivity:Chemistry of Acids and Bases: Packet 17: Principles of Reactivity: Chemistry of Acids and Bases

Concept Area I: Terminology : Concept Area I: Terminology strong/weak acids & bases monoprotic acid diprotic acid triprotic acid polyprotic acid aprotic amphoteric conjugate acid-base pair autoionization complex ion ionization constant, Ka or Kb water ionization constant, Kw pH pOH Arrhenius acid Arrhenius base Brønsted-Lowry acid Brønsted-Lowry base Lewis acid Lewis base

Concept Area II: Acids and Bases Revisited: Concept Area II: Acids and Bases Revisited You should still know the strong acids and bases learned in CHE 105. You should know the difference between Arrhenius, Brønsted-Lowry and Lewis definitions of acids and bases. You should be able to recognize acids/bases and be able to tell which definition(s) they meet. You should be able to write an acid-base reaction and describe the result. You should be able to label conjugate acid-base pairs in an acid-base reaction. You should be able to tell whether an acid is monoprotic, diprotic and so on. You should understand what an amphoteric compound is and be able to recognize one.

Recognizing Strong Acids & Bases: Recognizing Strong Acids & Bases Strong Acids HCl – hydrochloric acid HBr – hydrobromic acid HI – hydroiodic acid HNO3 – nitric acid HClO4 – perchloric acid H2SO4 – sulfuric acid Strong Bases hydroxides of the alkali and alkaline earth metals Memorize – if you don’t remember!

Slide5: Kotz, Treichel & Weaver page 187 And the rest of the alkali & alkaline-earth metal hydroxides. From Chapter/Packet 5

Recognizing Weak Acids & Bases: Recognizing Weak Acids & Bases Is it an acid or base? Is it on the strong list? That was easy! Right?

What determines if an acid/base is strong or weak?: What determines if an acid/base is strong or weak?

Definition time!: Definition time! Arrhenius acid: Arrhenius base: Which of these acids/bases meet the definition? H3PO4 AlCl3 H2SO4 NH3 Mg(OH)2 CH3Cl HCl NaOH

An example of an Arrhenius acid:: An example of an Arrhenius acid: Kotz, Treichel & Weaver page 799

Definition time!: Definition time! Brønsted-Lowry acid: Brønsted-Lowry base: Which of these acids/bases meet the definition? H3PO4 AlCl3 H2SO4 NH3 Mg(OH)2 CH3Cl HCl NaOH

An example of a Brønsted-Lowry base: An example of a Brønsted-Lowry base Kotz, Treichel & Weaver page 800

Definition time!: Definition time! Lewis acid: Lewis base: Which of these acids/bases meet the definition? H3PO4 AlCl3 H2SO4 NH3 Mg(OH)2 CH3Cl HCl NaOH

Review: writing acid-base reactions:: Review: writing acid-base reactions: Remember we saw in 105: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ) ACID + BASE YIELDS SALT & “WATER”! And now, CH3CO2H + NH3 NH4+ + CH3CO2– CH3Cl + AlCl3 [CH3]+[Cl–AlCl3]– Why does the top reaction have a one-way arrow and the bottom two don’t?

Conjugate Acid-Base Pairs: Conjugate Acid-Base Pairs Let’s use the Brønsted-Lowry definitions…. Once an acid has donated its proton, like acetic acid, CH3CO2H it becomes a base, or a proton acceptor, CH3CO2–! So, we call the acetate ion, the conjugate base of acetic acid.

Slide15: Kotz & Treichel 5th edition page 696 similar image Kotz, Treichel & Weaver 6th edition page 802 Conjugate Acid-Base Pairs + –

Conjugate Acid-Base Pairs: Conjugate Acid-Base Pairs Kotz, Treichel & Weaver page 803

Now you try!Label the conjugate acid-base pairs in the reactions below.: Now you try! Label the conjugate acid-base pairs in the reactions below. HCl + H2O  H3O+ + Cl– NH3 + H2O NH4+ + OH– H2SO4 + H2O  HSO4– + H3O+ HCO3– + H2O CO32– + H3O+

More definitions:: More definitions: protic acid: an acid capable of donating a proton monoprotic acid: diprotic acid: triprotic acid: . polyprotic acid: aprotic acid: Just be careful! Was that, “a protic acid” or “aprotic acid”?

What is an amphoteric compound?: What is an amphoteric compound? What are some examples? Note, the book also talks about amphiprotic molecules. These are compounds that can act as either a Brønsted-Lowry acid or base. We won’t worry about this one!

Amphoteric Metal Hydroxides: Amphoteric Metal Hydroxides Complete the following two reactions Al(OH)3(s) + H+(aq)  Al(OH)3(s) + OH–(aq)  So, in the top reaction aluminum hydroxide acted as a(n) _____ by what definition(s)? In the bottom reaction it aluminum hydroxide acted as a(n) _____ to form a complex ion. By what definition(s) did it act this way?

Some other reactions:: Some other reactions: Is copper(II) hydroxide amphoteric? Does it react with an acid? Does it react with a base? Well, let’s continue and find out! Kotz, Treichel & Weaver page 831

Closer look at metals that can form complex ions:: Closer look at metals that can form complex ions: Some metals form soluble complex ions (also called coordination complexes) in aqueous solutions. They usually form these complex ions with H2O, OH–, or NH3. If you’d like to read more about them than what we’re going to discuss, see chapter 22 section 3. Kotz, Treichel & Weaver page 830

Copper and Ammonia: Copper and Ammonia Cu2+(aq) + NH3(aq)  ? First, what does ammonia do in water? So, what would copper do with that? Now, what if we add more ammonia? So, is copper(II) hydroxide amphoteric? Kotz, Treichel & Weaver page 830 Now, let’s look at a symbolic picture of the copper ammonia complex:

Copper and Ammonia: Copper and Ammonia Cu(OH)2(s) + 4 NH3(aq) [Cu(NH3)4]2+(aq) + 2 OH–(aq) Kotz, Treichel & Weaver page 830

Concept Area III: pH and pOH: Concept Area III: pH and pOH You should remember (from CHE 105) how to use and understand the pH scale (this includes calculating pH from H+ and H+ from pH). You should be able to use and understand the pOH scale (this includes calculating pOH from OH– and OH– from pOH). You should be able to convert between the pH and the pOH scales.

pH: a special concentration scale for acids and bases: pH: a special concentration scale for acids and bases pH is defined as the negative log of the hydrogen ion concentration or: pH = –log [H+] = log (1/ [H+]) What if we know the pH and we want to know the hydrogen ion concentration? Simple, solve for [H+]: Remember, logs are base 10 so: [H+] = 10–pH At neutral pH [H+] =[OH–]= 1.00×10–7 M at 25 °C See Screen 5.17 (on CD or Web-site) for a tutorial From Chapter/Packet 5

Why pH? Why not just use M?: Why pH? Why not just use M? Because we like the simple numbers pH gives us from the Molarity! You can use these values to make sure you can use your calculator. From Chapter/Packet 5

The pH Scale: The pH Scale Low pH: high [H+] High pH: low [H+] Kotz, Treichel & Weaver page 212 From Chapter/Packet 5

Slide30: From Chapter/Packet 5

How can I get a pH of a solution with [H+]=2.35×10–5 M using my calculator?: How can I get a pH of a solution with [H+]=2.35×10–5 M using my calculator? If you have a non-graphing calculator: 1st way: 2.35 EE 5 +/– log +/– 2nd way: 1 ÷ 2.35 EE 5 +/– = log If you have a graphing calculator: 1st way: – log 2.35 EE – 5 = 2nd way: log ( 1 ÷ 2.35 EE – 5 ) = Answer is pH = 4.628932138 (or the more reasonable 4.62 ) Learn how to perform these operations on your specific calculator; your key names may differ from those given here. From Chapter/Packet 5

[H+] to pH: [H+] to pH If the [H+] of an compound is 1.6×10–3 M, what is the pH? Because pH = – log [H+] then: pH= – log (1.6×10–3) pH = – (–2.80) pH = 2.80 So, is this an acid or a base? From Chapter/Packet 5

pH to [H+]: pH to [H+] If the pH of Coke is 3.12, it is . What is the hydrogen ion concentration? Remember that pH = – log [H+] and, log [H+] = – pH Take antilog of both sides to get: [H+] = 10–pH [H+] = 10–3.12 [H+] = 7.6×10–4 M H+ Kotz, Treichel & Weaver page 213 From Chapter/Packet 5

pOH: pOH pOH is similar to pH in that just like pH is the –log[H+], pOH is the –log[OH–]. So, we solve pOH problems the same way we solved pH problems! In addition, the pH + pOH = 14 always for a solution must! Make sense? If not, wait until we cover pKw.

Slide35: Kotz, Treichel & Weaver page 806

Concept Area IV: Kw: Concept Area IV: Kw You should remember the concept of water autoionization from CHE 105. You should know and be able to use the water ionization constant, Kw.

2 H2O H3O+ + HO–: 2 H2O H3O+ + HO– Water is in constant equilibrium between hydronium ions and hydroxide ions. A proton, H+, never exists alone in an aqueous solution, it is always a hydronium ion, H3O+. However, chemists will write simply H+ in an aqueous solution as a shorthand. You may also use this shorthand when presence of the hydronium ion itself is not important. Just don’t forget that when you write H+, it’s really H3O+! From Chapter/Packet 5

Water autoionization!: Water autoionization! Kotz, Treichel & Weaver page 803 Water is in constant equilibrium between hydronium ions and hydroxide ions. This is called water autoionization!

Water Autoionization: Water Autoionization Because of water autoionization, even pure water conducts a very amount of electricity. Let’s write the equilibrium constant expression for: 2 H2O(ℓ) H3O+(aq) + HO–(aq). small

Writing the Kw: Writing the Kw Let’s write the Kc with all the terms (including the liquid). So we’d write: Let’s get these constants together since the concentration of water is considered a constant 55.5 M! This is an awfully awkward constant. Let’s simplify: Instead of calling it Kc′ we just call it a Kw.

More about the Kw: More about the Kw So, we have: Kw = [H3O+][OH–] At 25ºC, Kw = 1.0×10–14 Does this number sound familiar? Well, what is the normal range of pH’s? Of course, pure water has neutral pH. Thus, [H3O+] = [OH–]. So, they must both equal 1.0×10–7. So in an acidic solution [H3O+] [OH–] So in an basic solution [H3O+] [OH–]

Using the Kw in simple calculations:: Using the Kw in simple calculations: To find the pH or pOH of a acid or a base, use the Kw! You can’t use the Kw for a acid or base, why? strong weak strong

Using the Kw in simple calculations:: Using the Kw in simple calculations: What are the concentrations of hydronium and hydroxide ions, and what is the pH and pOH for a 4.0×10–2 M solution of HCl? HCl(aq) + H2O(ℓ)  H3O+(aq) + Cl–(aq)

Concept Area V: Ka and Kb : Concept Area V: Ka and Kb You should be able to write equilibrium constant expressions for acids and bases. You should understand the relationship between Ka for a weak acid to its Kb for its conjugate base and vice versa. You should be able to solve acid-base equilibrium problems.

Writing Equilibrium Expressions for Weak Acids and Bases : Writing Equilibrium Expressions for Weak Acids and Bases You should know how by now! What are they for the following? HA(aq) + H2O(ℓ) H3O+(aq) + A–(aq) B(aq) + H2O(ℓ) BH+(aq) + OH–(aq) Ka Kb

Relationship between Ka and Kb: Relationship between Ka and Kb Now, would a acid have a large or small Ka? Now since Ka× Kb = Kw, do you suppose the Kb of a strong acid’s conjugate base is large or small? Make sense? strong Kotz & Treichel 5th edition page 703 similar image Kotz, Treichel & Weaver 6th edition page 807

Slide47: Kotz, Treichel & Weaver page 808

Do you get it?: Do you get it? It should make sense to you that a strong acid would have a weak conjugate base, and a strong base would have a weak conjugate acid. If both were strong, how could they both react 100%?!? Kotz, Treichel & Weaver page 809

A very useful rule:: A very useful rule: Whenever you are doing an acid-base reaction, to determine whether or not products or reactants are favored: How do you tell which is the strongest acid?

Slide50: Kotz & Treichel 5th edition page 709 similar image Kotz, Treichel & Weaver 6th edition pages 814-815 Which side is favored in these reactions?

Enough theory! Let’s do calculations!: Enough theory! Let’s do calculations! A 0.100 M aqueous solution of lactic acid has a pH of 2.43. What is the value of Ka for lactic acid? CH3CHOHCO2H(aq) + H2O(ℓ) CH3CHOHCO2–(aq) + H3O+(aq) Solution shown on notes page.

Slide52: The pH of a 0.100 M acetic acid solution is 2.88. Calculate Ka for acetic acid. solution shown on notes page

Slide53: Benzoic acid, C6H5CO2H, occurs in nature. For example, most berries contain up to 0.05% by mass of this acid. Calculate the pH of a 0.020 M solution of benzoic acid if Ka is 6.3×10–5.

Slide54: Calculate the pH of a 0.10 M solution of hypochlorous acid, HClO. Its Ka is 3.5×10–8. solution shown on notes page

Slide55: Calculate the pH of household ammonia, a 1.44 M solution. The Kb for ammonia is 1.8×10–5.

Slide56: What is the pH of a 0.010 M aqueous solution of pyridine, C5H5N? The Kb for pyridine is 1.8×10–9. solution shown on notes page

Slide57: What is the pH of a 0.10 M aqueous solution of sulfuric acid? The Ka for sulfuric acid is large since strong; the Ka for hydrogen sulfate ion is 1.2×10–2.

What about weak polyprotic acids?: What about weak polyprotic acids? Do weak acids completely dissociate? Do you think the second proton would come off easier or harder than the first once the first one comes off? So, if less than 100% of the first H comes off, and less than 100% of the second H comes off (for those where the first one came off), do you think we need to consider anything except the first H?

Concept Area VI: pKa and pKb : Concept Area VI: pKa and pKb You should be able to calculate the pKa or pKb from the Ka or Kb and vice versa. You should understand how pKa and pKb is correlated with acid and base strength. You should be able to describe the acid-base properties of the salts of acids and bases. You should be able to calculate the pH or pOH of a solution of a salt of a weak acid or a weak base. You should be able to calculate the pH or pOH after any acid-base reaction.

pKa and pKb: pKa and pKb pKa and pKb are similar to pH and pOH in that just like pH is the –log[H+] and pOH is the –log[OH–]: pKa = –log Ka pKb = –log Kb We can even take the –log of the Kw. Thus: pKw = pH + pOH = 14.00!

Ka and pKa Kb and pKb: Ka and pKa Kb and pKb Now, would a acid have a large or small Ka? So, would a acid would have a large or small pKa? And vice versa for a base. strong strong pKa=4.89 pKa=4.74 pKa=3.74 pKa increases, acid strength decreases Kotz, Treichel & Weaver page 809

What happens when…: What happens when… We add sodium chloride to water? NaCl(aq) + H2O(ℓ) Sodium is the conjugate _____ of _______, a _____________. Chloride is the conjugate _____ of _______, a _____________. Since they are the conjugates of strong acids and bases, essentially no HCl or NaOH forms. Thus, our solution remains neutral.

What happens when…: What happens when… We add sodium acetate to water? CH3CO2Na(aq) + H2O(ℓ) Sodium is the conjugate _____ of _______, a _____________. Acetate is the conjugate _____ of _________, a _____________. So, we still get a neutral solution from the sodium. The acetate ion is another story. Let’s see what happens to acetate when we add it to water:

Acetate Ion plus Water: Acetate Ion plus Water CH3CO2–(s) + H2O(ℓ) CH3CO2H(aq) + OH–(aq) Which way will this equilibrium lie? Thus, to destroy the stronger acid which side is favored? Okay, so the ____ side is favored. Does that mean we’ll make no acetic acid? Well, is acetic acid a strong or a weak acid? Therefore, will our pH be basic or acidic?

Slide65: Kotz, Treichel & Weaver page 811

You decide!: You decide! Will the following give rise to acidic, basic or neutral solutions in water? NaNO3 K3PO4 FeCl2 NaHCO3 NH4F See example 17.2 in text for further explanation.

Determination of pH for Solutions of Salts: Determination of pH for Solutions of Salts Most of these types of problems are very straightforward if you have followed how to work equilibrium problems thus far. So, do the worksheet and ask if you have difficulties!

What is the pH of a solution after an acid-base reaction?: What is the pH of a solution after an acid-base reaction? Calculate the pH of the solution that results when 22.0 mL of 0.15 M acetic acid is mixed with 22.0 mL of 0.15 M sodium hydroxide.

What is the pH of a solution after an acid-base reaction?: Since the moles of acid and base are equal – the acid and base are both neutralized. However, the conjugate base of the weak acid will reform some of the acid. The question is how much? Initially, there is no more acetic acid and we know that we have 0.0033 moles of acetate ion… CH3CO2H(aq) + OH–(aq) CH3CO2–(aq) + H2O(ℓ) i: 22 mL of 0.15 M 22 mL of 0.15 M 0 M 0 M after r’xn: 0 M 0 M 0.0033 mol /44 mL What is the pH of a solution after an acid-base reaction?

What is the pH of a solution after an acid-base reaction?: Initially, there is no more acetic acid and we know that we have 0.0033 moles of acetate ion… Let’s write a new equation showing what the acetate ion will do: CH3CO2–(aq) + H2O(ℓ) CH3CO2H (aq) + OH–(aq) What is the pH of a solution after an acid-base reaction?

What is the pH of a solution after an acid-base reaction?: What if we had started with 30.0 mL of NaOH instead? How would this change things? Well, we’d still have 0.0033 mol of acetic acid. But this time we’d have: 0.0045 mol of base. So, how much base will be leftover? What is the pH of a solution after an acid-base reaction? CH3CO2H(aq) + OH–(aq) CH3CO2–(aq) + H2O(ℓ) i: 22.0 mL of 0.15 M 30.0 mL of 0.15 M 0 M 0 M

What is the pH of a solution after an acid-base reaction?: Once again there is no more acetic acid, we still know that we have 0.0033 moles of acetate ion and now we also have 0.0012 mol OH–… Let’s write that new equation : CH3CO2–(aq) + H2O(ℓ) CH3CO2H (aq) + OH–(aq) What is the pH of a solution after an acid-base reaction?

What is the pH of a solution after an acid-base reaction?: What if we had started with 30.0 mL of acetic acid instead? How would this change things? Well, we’d still have 0.0033 mol of NaOH. But this time we’d have: 0.0045 mol of acid. So, how much acid will be leftover? What is the pH of a solution after an acid-base reaction? CH3CO2H(aq) + OH–(aq) CH3CO2–(aq) + H2O(l) i: 22.0 mL of 0.15 M 30.0 mL of 0.15 M 0 M 0 M

What is the pH of a solution after an acid-base reaction?: This time we have left over acetic acid and acetate ion. So, let’s write a chemical reaction to use our Ka since all the base is gone. CH3CO2H (aq) + H2O(ℓ) CH3CO2–(aq) + H3O+(aq) What is the pH of a solution after an acid-base reaction?

The End of Packet 17!: The End of Packet 17! Questions? Make sure you do lots and lots and lots of practice problems, or you won’t get it!