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1 The Chemistry of Acids and Bases Chemistry I – Chapter 19 Chemistry I HD – Chapter 16 ICP – Chapter 23 SAVE PAPER AND INK!!! When you print out the notes.

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Presentation on theme: "1 The Chemistry of Acids and Bases Chemistry I – Chapter 19 Chemistry I HD – Chapter 16 ICP – Chapter 23 SAVE PAPER AND INK!!! When you print out the notes."— Presentation transcript:

1 1 The Chemistry of Acids and Bases Chemistry I – Chapter 19 Chemistry I HD – Chapter 16 ICP – Chapter 23 SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

2 2 Acid and Bases

3 3

4 4

5 5 Acids Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases

6 6 Some Properties of Acids þ Produce H + (as H 3 O + ) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) þ Taste sour þ Corrode metals þ Electrolytes þ React with bases to form a salt and water þ pH is less than 7 þ Turns blue litmus paper to red “Blue to Red A-CID”

7 7 Acid Nomenclature Review No Oxygen  w/Oxygen An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”

8 8 Some Properties of Bases  Produce OH - ions in water  Taste bitter, chalky  Are electrolytes  Feel soapy, slippery  React with acids to form salts and water  pH greater than 7  Turns red litmus paper to blue “Basic Blue”

9 9 Some Common Bases NaOHsodium hydroxidelye KOHpotassium hydroxideliquid soap Ba(OH) 2 barium hydroxidestabilizer for plastics Mg(OH) 2 magnesium hydroxide“MOM” Milk of magnesia Al(OH) 3 aluminum hydroxideMaalox (antacid) Al(OH) 3 aluminum hydroxideMaalox (antacid)

10 10 Acid/Base definitions Definition #1: Arrhenius (traditional) Acids – produce H + ions (or hydronium ions H 3 O + ) Bases – produce OH - ions (problem: some bases don’t have hydroxide ions!)

11 11 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water

12 12 Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron!

13 13 A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acid conjugate base base conjugate acid

14 14 ACID-BASE THEORIES The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID

15 15 Learning Check! Label the acid and base HONORS ONLY! HCl + OH -  Cl - + H 2 O H 2 O + H 2 SO 4  HSO H 3 O +

16 16 The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H + (or OH - ) ion. Under 7 = acid 7 = neutral Over 7 = base

17 17 pH of Common Substances

18 18 Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of HONORS ONLY!

19 19 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of HONORS ONLY!

20 20 Relation of K a, K b, [H 3 O + ] and pH HONORS ONLY!

21 21 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib x+x+x 1.00-xxx HONORS ONLY!

22 22 Equilibria Involving A Weak Acid Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. Solve using quadratic formula. or you can make an approximation if x is very small! (Rule of thumb: or smaller is ok) HONORS ONLY!

23 23 Equilibria Involving A Weak Acid Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Now we can more easily solve this approximate expression. HONORS ONLY!

24 24 Approximating If K is really small, the equilibrium concentrations will be nearly the same as the initial concentrations. Example: 0.20 – x is just about 0.20 if x is really small. If the K is or smaller (10 -6, 10 -7, etc.), you should approximate. Otherwise, you have to use the quadratic.

25 25 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = 4.2 x M pH = - log [ H 3 O + ] = -log (4.2 x ) = 2.37 HONORS ONLY!

26 26 Equilibria Involving A Weak Acid Calculate the pH of a M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O  HCO H 3 O + HCO 2 H + H 2 O  HCO H 3 O + K a = 1.8 x Approximate solution [H 3 O + ] = 4.2 x M, pH = 3.37 [H 3 O + ] = 4.2 x M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x M [H 3 O + ] = [HCO 2 - ] = 3.4 x M [HCO 2 H] = x = M [HCO 2 H] = x = M pH = 3.47 pH = 3.47 HONORS ONLY!

27 27 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O  NH OH - NH 3 + H 2 O  NH OH - K b = 1.8 x Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib x+x+x xx x HONORS ONLY!

28 28 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O  NH OH - NH 3 + H 2 O  NH OH - K b = 1.8 x Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib x+x+x xx x HONORS ONLY!

29 29 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O  NH OH - NH 3 + H 2 O  NH OH - K b = 1.8 x Step 2. Solve the equilibrium expression Assume x is small, so x = [OH - ] = [NH 4 + ] = 4.2 x M x = [OH - ] = [NH 4 + ] = 4.2 x M and [NH 3 ] = x ≈ M The approximation is valid! HONORS ONLY!

30 30 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O  NH OH - NH 3 + H 2 O  NH OH - K b = 1.8 x Step 3. Calculate pH [OH - ] = 4.2 x M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = HONORS ONLY!

31 31 Types of Acid/Base Reactions: Summary HONORS ONLY!

32 32 pH testing There are several ways to test pHThere are several ways to test pH –Blue litmus paper (red = acid) –Red litmus paper (blue = basic) –pH paper (multi-colored) –pH meter (7 is neutral, 7 base) –Universal indicator (multi-colored) –Indicators like phenolphthalein –Natural indicators like red cabbage, radishes

33 33 Paper testing Paper tests like litmus paper and pH paperPaper tests like litmus paper and pH paper –Put a stirring rod into the solution and stir. –Take the stirring rod out, and place a drop of the solution from the end of the stirring rod onto a piece of the paper –Read and record the color change. Note what the color indicates. –You should only use a small portion of the paper. You can use one piece of paper for several tests.

34 34 pH paper

35 35 pH meter Tests the voltage of the electrolyteTests the voltage of the electrolyte Converts the voltage to pHConverts the voltage to pH Very cheap, accurateVery cheap, accurate Must be calibrated with a buffer solutionMust be calibrated with a buffer solution

36 36 pH indicators Indicators are dyes that can be added that will change color in the presence of an acid or base. Some indicators only work in a specific range of pH Once the drops are added, the sample is ruined Some dyes are natural, like radish skin or red cabbage

37 37 ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

38 38 Setup for titrating an acid with a base

39 39 TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3.Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION. This is called NEUTRALIZATION.

40 mL of NaOH is neutralized with 25.2 mL of M HCl by titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

41 41 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!

42 42 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add?

43 43 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution

44 44 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M V = M V = (3.0 mol/L)(0.050 L) = 0.5 M NaOH X V Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH) / (0.50 M) = 0.30 L or 300 mL

45 45 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

46 46 A shortcut A shortcut M 1 V 1 = M 2 V 2 Preparing Solutions by Dilution

47 47 You try this dilution problem You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400. mL of 0.10 M HCl. How much of the acid and how much water will you need?


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