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1 Announcements & Agenda (02/23/07) You should be reading Ch 10 this weekend! Quiz Today! Open Review 3pm on Wed. Low attendance this week 

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Presentation on theme: "1 Announcements & Agenda (02/23/07) You should be reading Ch 10 this weekend! Quiz Today! Open Review 3pm on Wed. Low attendance this week "— Presentation transcript:

1 1 Announcements & Agenda (02/23/07) You should be reading Ch 10 this weekend! Quiz Today! Open Review Sessions @ 3pm on Wed. Low attendance this week  Low attendance this week Today Acid & base strength: Quantitative (8.4-8.5) Acid & base strength: Quantitative (8.4-8.5) the pH scale the pH scale Acid & base reactions Acid & base reactions

2 2 Last Time: Bronsted-Lowry Acids & Bases acids donate a proton (H + )acids donate a proton (H + ) bases accept a proton (H + )bases accept a proton (H + )

3 3 A strong acid/base completely ionizes (100%) in aqueous solutions.A strong acid/base completely ionizes (100%) in aqueous solutions. HCl(g) + H 2 O(l) H 3 O + (aq) + Cl − (aq) COMPARE TO STRONG ELECTROLYTES A weak acid/base dissociates only slightly in water to form a few ions in aqueous solutions.A weak acid/base dissociates only slightly in water to form a few ions in aqueous solutions. H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 − (aq) H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 − (aq) COMPARE TO WEAK ELECTROLYTES Last Time: Strengths of Acids/Bases - Ionization

4 4 make up six (just a few) of all the acids.make up six (just a few) of all the acids. have weak conjugate bases (the product formed after the proton is transferred).have weak conjugate bases (the product formed after the proton is transferred). Strong Acids (Know These)

5 5 Strong Bases are formed from metals of Groups 1A (1) and 2A (2).are formed from metals of Groups 1A (1) and 2A (2). include LiOH, NaOH, KOH, and Ca(OH) 2.include LiOH, NaOH, KOH, and Ca(OH) 2. dissociate completely in water.dissociate completely in water. KOH(s) K + (aq) + OH − (aq) KOH(s) K + (aq) + OH − (aq)

6 6 In water occasionally, H + is transferred from 1 H 2 O molecule to another.H + is transferred from 1 H 2 O molecule to another. one water acts an acid, the another acts as a base.one water acts an acid, the another acts as a base. H 2 O + H 2 O H 3 O + + OH −................ :O: H + H:O: H:O:H + + :O:H − :O: H + H:O: H:O:H + + :O:H −................ H H H H H H water water hydronium hydroxide ion (+) ion (-) water water hydronium hydroxide ion (+) ion (-) Ionization of Water: A Basis for Understanding pH (H+ concentrations)

7 7 Pure Water is Neutral (NOT ACIDIC OR BASIC) the ionization of water molecules produces small, but equal quantities of H 3 O + and OH − ions.the ionization of water molecules produces small, but equal quantities of H 3 O + and OH − ions. molar concentrations are indicated in brackets as [H 3 O + ] and [OH − ].molar concentrations are indicated in brackets as [H 3 O + ] and [OH − ]. [H 3 O + ] = 1.0 x 10 −7 M [H 3 O + ] = 1.0 x 10 −7 M [OH − ] = 1.0 x 10 −7 M [OH − ] = 1.0 x 10 −7 M

8 8 Acidic Solutions Adding an acid to pure water: increases the [H 3 O + ].increases the [H 3 O + ]. causes the [H 3 O + ] to exceed 1.0 x 10 -7 M.causes the [H 3 O + ] to exceed 1.0 x 10 -7 M. decreases the [OH − ].decreases the [OH − ].

9 9 Basic Solutions Adding a base to pure water: increases the [OH − ].increases the [OH − ]. causes the [OH − ] to exceed 1.0 x 10 − 7 M.causes the [OH − ] to exceed 1.0 x 10 − 7 M. decreases the [H 3 O + ].decreases the [H 3 O + ]. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

10 10 The ion product constant, K w, for water is the product of the concentrations of the hydronium and hydroxide ions.is the product of the concentrations of the hydronium and hydroxide ions. K w = [ H 3 O + ] [ OH − ] can be obtained from the concentrations in pure water.can be obtained from the concentrations in pure water. K w = [ H 3 O + ] [ OH − ] K w = [1.0 x 10 − 7 M] x [ 1.0 x 10 − 7 M] = 1.0 x 10 − 14 = 1.0 x 10 − 14 Ion Product of Water, K w

11 11 [H 3 O + ] and [OH − ] in Solutions IMPORTANT: K w is always 1.0 x 10 −14.

12 12 Calculating [H 3 O + ] What is the [H 3 O + ] of a solution if [OH − ] is 5.0 x 10 -8 M? STEP 1: Write the K w for water. K w = [H 3 O + ][OH − ] = 1.0 x 10 −14 STEP 2: Rearrange the K w expression. [H 3 O + ] = 1.0 x 10 -14 [OH − ] [OH − ] STEP 3: Substitute [OH − ]. [H 3 O + ] = 1.0 x 10 -14 = 2.0 x 10 -7 M [H 3 O + ] = 1.0 x 10 -14 = 2.0 x 10 -7 M 5.0 x 10 - 8 5.0 x 10 - 8

13 13 If lemon juice has [H 3 O + ] of 2 x 10 −3 M, what is the [OH − ] of the solution? 1) 2 x 10 −11 M 2) 5 x 10 −11 M 3) 5 x 10 −12 M 12345

14 14 3) 5 x 10 −12 M Rearrange the K w to solve for [OH - ] K w = [H 3 O + ][OH − ] = 1.0 x 10 −14 [OH − ] = 1.0 x 10 -14 = 5 x 10 −12 M 2 x 10 - 3 2 x 10 - 3 Solution

15 15 pH Scale The pH of a solution is used to indicate the acidity of a solution.is used to indicate the acidity of a solution. has values that usually range from 0 to 14.has values that usually range from 0 to 14. is acidic when the values are less than 7.is acidic when the values are less than 7. is neutral with a pH of 7.is neutral with a pH of 7. is basic when the values are > 7.is basic when the values are > 7. NOTE: pH is a logarithmic scale!!!

16 16 pH of Everyday Substances

17 17 Testing the pH of Solutions The pH of solutions can be determined using a) pH metera) pH meter b) pH paperb) pH paper c) indicators that have specific colors at different pH values.c) indicators that have specific colors at different pH values.

18 18 pH is the negative log of the hydronium ion concentration. pH = - log [H 3 O + ] Example: For a solution with [H 3 O + ] = 1 x 10 −4 pH =−log [1 x 10 −4 ] pH = - [-4.0] pH = 4.0 Note: The number of decimal places in the pH equals the significant figures in the coefficient of [H 3 O + ]. the significant figures in the coefficient of [H 3 O + ]. 4.0 1 SF in 1 x 10 -4 Calculating pH

19 19 A. The [H 3 O + ] of tomato juice is 2 x 10 −4 M. What is the pH of the solution? What is the pH of the solution? 1) 4.0 2) 3.73) 10.3 1) 4.0 2) 3.73) 10.3 B. The [OH − ] of a solution is 1.0 x 10 −3 M. What is the pH of the solution? What is the pH of the solution? 1) 3.00 2) 11.003) -11.00 1) 3.00 2) 11.003) -11.00 Learning Check

20 20 If an area received 1 inch of rain with a pH of 4, how much more neutral rain would be needed to have a final pH of 6? 1.Approximately 2 inches 2.Approximately 9 inches 3.Approximately 20 inches 4.Approximately 100 inches 12345

21 21 [H 3 O + ], [OH - ], and pH Values

22 22 Calculating [H 3 O + ] from pH The [H 3 O + ] can be expressed by using the pH as the negative power of 10. [H 3 O + ] = 1 x 10 -pH For pH = 3.0, the [H 3 O + ] = 1 x 10 -3 On a calculator 1. Enter the pH value 3.0 2. Change sign -3.0 3. Use the inverse log key (or 10 x ) to obtain the [H 3 0 + ]. = 1 x 10 -3 M the [H 3 0 + ]. = 1 x 10 -3 M

23 23 In a neutralization reaction: a base such as NaOH reacts with an acid such as HCl.a base such as NaOH reacts with an acid such as HCl. HCl + H 2 OH 3 O + + Cl − NaOHNa + + OH − the H 3 O + from the acid and the OH − from the base form water.the H 3 O + from the acid and the OH − from the base form water. H 3 O + + OH − 2 H 2 O H 3 O + + OH − 2 H 2 O Neutralization Rxns of Acids & Bases

24 24 Bases Used in Some Antacids Antacids are used to neutralize stomach acid (HCl). Antacids are used to neutralize stomach acid (HCl).

25 25 In the equation for neutralization, an acid and a base produce a salt and water. acid base salt water acid base salt water HCl + NaOH NaCl + H 2 O HCl + NaOH NaCl + H 2 O 2HCl + Ca(OH) 2 CaCl 2 + 2H 2 O 2HCl + Ca(OH) 2 CaCl 2 + 2H 2 O Neutralization Equations Balance these like any other reaction!

26 26 Solving Problems… What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l)Method: Get into moles with “known”: Given: 18.5 mL of 0.225 M NaOH Do a moles-to-moles conversion Get out of moles with “unknown”:

27 27 18.5 mL NaOH x 1 L NaOH x 0.225 mole NaOH 1000 mL NaOH 1 L NaOH 1000 mL NaOH 1 L NaOH x 1 mole HCl = 0.00416 mole HCl x 1 mole HCl = 0.00416 mole HCl 1 mole NaOH 1 mole NaOH M HCl = 0.00416 mole HCl = 0.416 M HCl 0.0100 L HCl 0.0100 L HCl

28 28 Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) 1) 12.5 mL 2) 50.0 mL 3) 200. mL 12345

29 29 Solution 1)12.5 mL 0.0500 L KOH x 1.00 mole KOH x 1 mole H 2 SO 4 x 1 L KOH 2 mole KOH 1 L KOH 2 mole KOH 1 L H 2 SO 4 x 1000 mL = 12.5 mL 1 L H 2 SO 4 x 1000 mL = 12.5 mL 2.00 mole H 2 SO 4 1 L H 2 SO 4 2.00 mole H 2 SO 4 1 L H 2 SO 4

30 30 Two More Acid/Base Reactions… 1. Acids react with metals such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn.such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn. to produce hydrogen gas and the salt of the metal.to produce hydrogen gas and the salt of the metal. Molecular equations: 2K(s) + 2HCl(aq) 2KCl(aq) + H 2 (g) Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g)

31 31 Acids and Carbonates Acids react with carbonates & hydrogen carbonateswith carbonates & hydrogen carbonates to produce carbon dioxide gas, a salt, & water.to produce carbon dioxide gas, a salt, & water. 2HCl(aq) + CaCO 3 (s) CO 2 (g) + CaCl 2 (aq) + H 2 O(l) HCl(aq) + NaHCO 3 (s) CO 2 (g) + NaCl (aq) + H 2 O(l)


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