1 Chapter 10 Acids and Bases

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5  Arrhenius acids  Produce H + ions in water. H 2 O HCl H + (aq) + Cl – (aq)  Are electrolytes.  Have a sour taste.  Corrode metals.  React with bases to form salts and water Acids and Bases in Aqueous Solution

6  Arrhenius bases  Produce OH – ions in water.  Taste bitter or chalky.  Are electrolytes.  Feel soapy and slippery.  React with acids to form salts and water. Bases

7 Identify each as a characteristic of an A) acid or B) base ____1. Has a sour taste. ____2. Produces OH - in aqueous solutions. ____3. Has a chalky taste. ____4. Is an electrolyte. ____5. Produces H + in aqueous solutions. Learning Check

8 Identify each as a characteristic of an A) acid or B) base A 1. Has a sour taste. B 2. Produces OH – in aqueous solutions. B 3. Has a chalky taste. A, B 4. Is an electrolyte. A 5. Produces H + in aqueous solutions. Solution

9 Names of Acids Acids with H and one nonmetal are named with the prefix hydro- and end with -ic acid. HClhydrochloric acid Acids with H and a polyatomic ion are named by changing the end of an –ate ion to -ic acid and an –ite ion to -ous acid. HClO 3 chloric acid HClO 2 chlorous acid

10 Name each of the following as acids: A. HBr 1. bromic acid 2. bromous acid 3. hydrobromic acid B. H 2 CO 3 1. carbonic acid 2. hydrocarbonic acid 3. carbonous acid Learning Check

11 A. HBr3. hydrobromic acid The name of an acid with H and one nonmetal begins with the prefix hydro- and ends with -ic acid. B. H 2 CO 3 1. carbonic acid An acid with H and a polyatomic ion is named by changing the end of an –ate ion to -ic acid. Solution

12  Bases with OH - ions are named as the hydroxide of the metal in the formula. NaOHsodium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide Al(OH) 3 aluminum hydroxide Fe(OH) 3 iron (III) hydroxide Some Common Bases

13 Match the formulas with the names: A. ___ HNO 2 1) hydrochloric acid B. ___Ca(OH) 2 2) sulfuric acid C. ___H 2 SO 4 3) sodium hydroxide D. ___HCl4) nitrous acid E. ___NaOH5) calcium hydroxide Learning Check

14 Match the formulas with the names: A. 4 HNO 2 4) nitrous acid B. 5 Ca(OH) 2 5) calcium hydroxide C. 2 H 2 SO 4 2) sulfuric acid D. 1 HCl 1) hydrochloric acid E. 3 NaOH3) sodium hydroxide Solution

15 According to the Brønsted-Lowry theory, Acids are hydrogen ion (H + ) donors. Bases are hydrogen ion (H + ) acceptors. donor acceptor hydronium ion HCl + H 2 O H 3 O + + Cl Br Ø nsted-Lowry Acids and Bases

16 When NH 3 dissolves in water, a few NH 3 molecules react with water to form ammonium ion NH 4 + and a hydroxide ion. NH 3 + H 2 O NH 4 + (aq) + OH - (aq) acceptor donor NH 3, A Bronsted-Lowry Base

17 Conjugate Acids and Bases An acid that donates H + forms a conjugate base. A base that accepts a H + forms a conjugate acid. In an acid-base reaction, there are two conjugate acid-base pairs. acid 1 conjugate base base 2 conjugate acid 2 HF H2OH2O H3O+H3O+ F -

18 Conjugate Acid-Base Pairs A conjugate acid-base pair is two substances related by a loss or gain of H acid 1– conjugate base 1 base 2– conjugate acid 2 HF H 2 O H3O+H3O+ F - HF, F - H 2 O, H 3 O +

19 Learning Check A. Write the conjugate base of the following: 1. HBr 2. H 2 S 3. H 2 CO 3 B. Write the conjugate acid of the following: 1. NO 2 – 2. NH 3 3. OH –

20 Solution A. Write the conjugate base of the following: 1. HBrBr - 2. H 2 SHS – 3. H 2 CO 3 HCO 3 – B. Write the conjugate acid of the following: 1. NO 2 – HNO 2 2. NH 3 NH OH – H 2 O

21 Learning Check Identify the following that are acid-base conjugate pairs. 1. HNO 2, NO 2 – 2. H 2 CO 3, CO 3 2 – 3. HCl, ClO 4 – 4. HS –, H 2 S 5. NH 3, NH 4 +

22 Solution Identify the following that are acid-base conjugate pairs. 1. HNO 2, NO 2 – 4. HS –, H 2 S 5. NH 3, NH 4 +

23 Learning Check A. The conjugate base of HCO 3 – is 1. CO 3 2– 2. HCO 3 – 3. H 2 CO 3 B. The conjugate acid of HCO 3 – is 1. CO 3 2– 2. HCO 3 – 3. H 2 CO 3 C. The conjugate base of H 2 O is 1. OH – 2. H 2 O 3. H 3 O + D. The conjugate acid of H 2 O is 1. OH – 2. H 2 O 3. H 3 O +

24 Solution A. The conjugate base of HCO 3 – is 1. CO 3 2– B. The conjugate acid of HCO 3 – is 3. H 2 CO 3 C. The conjugate base of H 2 O is 1. OH – D. The conjugate acid of H 2 O is 3. H 3 O +

25 Strong acids completely ionize (100%) in aqueous solutions. HCl + H 2 O H 3 O + (aq) + Cl – (aq) (100 % ions)  Strong bases completely (100%) dissociate into ions in aqueous solutions. H 2 O NaOH Na + (aq) + OH – (aq) (100 % ions) 10.6 Acid and Base Strength

26 Strong and Weak Acids In an HCl solution, the strong acid HCl dissociates 100%. A solution of the weak acid CH 3 COOH contains mostly molecules and a few ions.

27 Only a few acids are strong acids. The conjugate bases of strong acids are weak bases. Strong Acids

28 Most acids are weak acids. The conjugate bases of weak acids are strong bases. Weak Acids

29 Most bases in Groups 1A and 2A are strong bases. They include LiOH, NaOH, KOH, and Mg(OH) 2, Ca(OH) 2 Most other bases are weak bases. Strong Bases

30 Learning Check Identify each of the following as a strong or weak acid or base. A. HBr B. HNO 2 C. NaOH D. H 2 SO 4 E. Cu(OH) 2

31 Solution Identify each of the following as a strong or weak acid or base. A. HBrstrong acid B. HNO 2 weak acid C. NaOHstrong base D. H 2 SO 4 strong acid E. Cu(OH) 2 weak base

32 Table 10.1 Relative strengths of acids and bases

33 Learning Check A. Identify the stronger acid in each pair. 1. HNO 2 or H 2 S 2. HCO 3 – or HBr 3. H 3 PO 4 or H 3 O + B. Identify the strong base in each pair. 1. NO 3 – or F - 2. CO 3 2– or NO 2 – 3. OH – or H 2 O

34 Solution A. Identify the stronger acid in each pair. 1. HNO 2 2. HBr 3. H 3 O + B. Identify the stronger base in each pair. 1. F - 2. CO 3 2– 3. OH –

35 Strong Acids In water, the dissolved molecules of a strong acid are essentially all separated into ions. The concentrations of H 3 O + and the anion (A – ) are large.

36 Weak Acids In weak acids, The equilibrium favors the undissociated (molecular) form of the acid. The concentrations of the H 3 O + and the anion (A – ) are small.

37  In water, H + is transferred from one H 2 O molecule to another.  One water molecule acts as an acid, while another acts as a base. H 2 O + H 2 O H 3 O + + OH – : O : H + : O : H H : O : H + + : O : H – H H H water molecules hydronium hydroxide ion (+) ion (–) 10.8 Dissociation of Water

38  In pure water, the ionization of molecules produces small, but equal quantities of H 3 O + and OH - ions. H 2 O + H 2 O H 3 O + + OH -  Molar concentrations are indicated as [H 3 O + ] and [OH - ]. [H 3 O + ] = 1.0 x M [OH - ] = 1.0 x M Pure Water is Neutral

39  Adding an acid to pure water increases the [H 3 O + ].  In acids, [H 3 O + ] exceeds 1.0 x 10 –7 M.  As [H 3 O + ] increases, [OH – ] decreases. Acidic Solutions

40  Adding a base to pure water increases the [OH – ].  In bases, [OH – ] exceeds 1.0 x 10 –7 M.  As [OH – ] increases, [H 3 O + ] decreases. Basic Solutions

41 Comparison of [H 3 O + ] and [OH – ]

42 The ion product constant, K w, for water is the product of the concentrations of the hydronium and hydroxide ions. K w = [H 3 O + ][OH – ] We can obtain the value of K w using the concentrations in pure water. K w = [1.0 x 10 –7 ][1.0 x 10 –7 ] = 1.0 x 10 –14 Ion Product of Water, K w

43 K w in Acids and Bases In neutral, acidic, and basic solutions, the K w is equal to 1.0 x 10 –14.

44 What is the [H 3 O + ] of a solution if [OH – ] is 1.0 x M? Rearrange the K w expression for [H 3 O + ]. K w = [H 3 O + ][OH – ] = 1.0 x [H 3 O + ] = 1.0 x [OH – ] [H 3 O + ] = 1.0 x = 1.0 x M 1.0 x Calculating [H 3 O + ]

45 The [H 3 O + ] of lemon juice is 1.0 x 10 –3 M. What is the [OH – ] of the solution? 1) 1.0 x 10 3 M 2) 1.0 x 10 –11 M 3) 1.0 x M Learning Check

46 The [H 3 O + ] of lemon juice is 1.0 x 10 –3 M. What is the [OH – ] of the solution? 2) 1.0 x 10 –11 M Rearrange the K w to solve for [OH – ] K w = [H 3 O + ][OH – ] = 1.0 x 10 –14 [OH - ] = 1.0 x = 1.0 x 10 –11 M 1.0 x Solution

Measuring Acidity in Aqueous Solution: pH The pH scale:  Is used to indicate the acidity of a solution.  Has values that usually range from 0 to 14.  Indicates an acidic solution when the values are less than 7.  Indicates a neutral solution with a pH of 7.  Indicates a basic solution when the values are greater than 7.

48 Fig 10.2 The pH scale and pH of some common substances Fig 10.3 relationship of pH to H + and OH - ion concentrations

Neutral [H 3 O + ]>[OH - ] [H 3 O + ] = [OH - ] [H 3 O + ]<[OH - ] Acidic Basic pH Range

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51 Identify each solution as 1. acidic 2. basic3. neutral A. ___ HCl with a pH = 1.5 B. ___ Pancreatic fluid [H 3 O + ] = 1 x 10 –8 M C. ___ Sprite soft drink pH = 3.0 D. ___ pH = 7.0 E. ___ [OH – ] = 3 x 10 –10 M F. ___ [H 3 O + ] = 5 x 10 –12 M Learning Check

52 Identify each solution as 1. acidic 2. basic3. neutral A. 1 HCl with a pH = 1.5 B. 2 Pancreatic fluid [H 3 O + ] = 1 x 10 –8 M C. 1 Sprite soft drink pH = 3.0 D. 3 pH = 7.0 E. 1 [OH – ] = 3 x 10 –10 M F. 2 [H 3 O + ] = 5 x 10 –12 M Solution

53 Testing the pH of Solutions The pH of solutions can be determined using a a) pH meter, b) pH paper, and c) indicators that have specific colors at different pHs.

54 pH is:  Defined as the negative log of the hydrogen ion concentration. pH = - log [H 3 O + ]  The value of the negative exponent of [H 3 O + ] for concentrations with a coefficient of 1. [H 3 O + ] =1 x pH = 4.0 [H 3 O + ] =1 x pH = 11.0 pH Scale

55 A calculator can be used to find the pH of a solution with a [H 3 O + ] of 1 x as follows. 1. Enter 1 x by pressing 1 (EE) 3 (+/–). The EE key gives an exponent of 10 and the +/– key changes the sign. 2. Press the log key to obtain log 1 x log (1 x ) = –3 3. Press the +/– key to multiply by –1. (-3) +/– = 3 Calculations of pH

56 Significant Figures in pH When expressing log values, the number of decimal places in the pH is equal to the number of significant figures in the coefficient of [H 3 O + ]. [H 3 O + ] = 1 x pH = 4.0 [H 3 O + ] = 1.0 x pH = 6.00 [H 3 O + ] = 2.4 x pH = 7.62

57 A. The [H 3 O + ] of tomato juice is 2 x M. What is the pH of the solution? 1) 4.02) 3.73) 10.3 B. The [OH - ] of a solution is 1.0 x M. What is the pH of the solution? 1) 3.002) ) –11.00 Learning Check

58 A. 2) 3.7 pH = – log [ 2 x ] = (EE) 4 (+/–) log (+/–) B. 2) Use the K w to obtain [H 3 O + ] = 1.0 x pH = – log [1.0 x ] = – (+/–) = Solution

59 Calculating [H 3 O + ] from pH The [H 3 O + ] can be expressed with the pH as the negative power of 10. [H 3 O + ] = 1 x 10 -pH If the pH is 3.0, the [H 3 O + ] = 1 x On a calculator 1. Enter the pH value Use (+/–) to change sign 3.0 (+/–) = – Use the inverse log key (or 10 x ) to obtain the [H ]. =1 x M

60 A. What is the [H 3 O + ] of a solution with a pH of 10.0? 1) 1 x M2) 1 x M 3) 1 x M B. What is the [H 3 O + ] of a solution with a pH of 5.7? 1) 2 x M2) 5 x 10 5 M 3) 1 x M Learning Check

61 A. What is the [H 3 O + ] of a solution with a pH of 10.0? 3) 1 x M1 x 10 -pH B. What is the [H 3 O + ] of a solution with a pH of 5.7? 1) 2 x M 5.7 (+/–) inv log (or 10 x ) = 2 x Solution

62 pOH The pOH of a solution:  Is related to the [OH - ]. Is similar to pH. pOH = – log [OH - ] Added to the pH value is equal to pH + pOH = 14.0

63 Learning Check What is the pH and the pOH of coffee if the [H 3 O + ] is 1 x M? 1) pH = 5.0 pOH =7.0 2) pH = 7.0 pOH = 7.0 3) pH = 5.0pOH = 9.0

64 Solution What is the pH and the pOH of coffee if the [H 3 O + ] is 1 x M? 3) pH = 5.0pOH = 9.0 pH = – log [1 x ] = –(– 5.0) = 5.0 pH + pOH = 14.0 pOH = 14.0 – pH = 14.0 – 5.0 = 9.0 or [OH - ] = 1 x pOH = – log [1 x ] = –(– 9.0) = 9.0

65 An acid such as HCl produces H 3 O + and a base such as NaOH provides OH -. HCl + H 2 OH 3 O + + Cl - NaOHNa + + OH - In a neutralization reaction, the H 3 O + and the OH - combine to form water. H 3 O + + OH - 2 H 2 O The positive ion (metal) and the negative ion (nonmetal) are the ions of a salt. Neutralization Reactions

66 In the equation for neutralization, an acid and a base produce a salt and water. NaOH + HCl NaCl + H 2 O base acid salt water Ca(OH) HCl CaCl 2 + 2H 2 O base acid salt water Neutralization Equations

67 Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. 1. Write the formulas of the acid and base. Mg(OH) 2 + HNO 3 2. Balance to give equal OH - and H +. Mg(OH) HNO 3 3. Write the products (a salt and water) and balance: Mg(OH) 2 + 2HNO 3 Mg(NO 3 ) 2 + 2H 2 O Balancing Neutralization Reactions

68 Write strong acids, bases, and salts as ions. H + + Cl - + Na + + OH - Na + + Cl - + H 2 O Cross out matched ions. H + + Cl - + Na + + OH - Na + + Cl - + H 2 O Write a net ionic reaction. H + + OH - H 2 O Net Ionic Equations for Neutralization NaOH + HCl NaCl + H 2 O

69 Select the correct group of coefficients for the following neutralization equations. A. __ HCl + __ Al(OH) 3 __AlCl 3 + __ H 2 O 1) 1, 3, 3, 1 2) 3, 1, 1, 1 3) 3, 1, 1, 3 B. __ Ba(OH) 2 + __H 3 PO 4 __Ba 3 (PO 4 ) 2 + __ H 2 O 1) 3, 2, 2, 2 2) 3, 2, 1, 6 3) 2, 3, 1, 1 Learning Check

70 A. 3) 3, 1, 1 3 3HCl + 1Al(OH) 3 1AlCl 3 + 3H 2 O B. 2) 3, 2, 1, 6 3Ba(OH) 2 + 2H 3 PO 4 1Ba 3 (PO 4 ) 2 + 6H 2 O Solution

71 Antacids neutralize stomach acid (HCl). Alka-Seltzer NaHCO 3, citric acid, aspirin Chooz, TumsCaCO 3 Di-gel CaCO 3 and Mg(OH) 2 Gelusil, Maalox,Al(OH) 3 and Mg(OH) 2 Mylanta Al(OH) 3 and Mg(OH) 2 Milk of MagnesiaMg(OH) 2 Antacids

72 Learning Check Write the neutralization reactions for stomach acid HCl and Mylanta. Mylanta Al(OH) 3 and Mg(OH) 2

73 Solution Write the neutralization reactions for stomach acid HCl and Mylanta. Mylanta: Al(OH) 3 and Mg(OH) 2 3HCl + Al(OH) 3 AlCl 3 + 3H 2 O 2HCl + Mg(OH) 2 MgCl 2 + 2H 2 O

74 Some Salt Solutions are Neutral A salt containing the ions of a strong acid and a strong base forms a neutral solution. The metal ion does not produce H + and the negative ion does not attract H + from water. For example, KNO 3 contains ions from a strong base (KOH) and a strong acid (HNO 3 ). A solution of KNO 3 would be neutral.

75 Some Salt Solutions are Basic A salt containing the ions of a weak acid and a strong base forms a basic solution. The ion of the weak acid attracts H + from water to form a basic solution. For example, KHCO 3 contains ions from a strong base (KOH) and a weak acid (H 2 CO 3 ). A solution of KHCO 3 is basic. HCO H 2 O H 2 CO 3 + OH -

76 Some Salt Solutions are Acidic A salt containing the ions of a strong acid and a weak base forms an acidic solution. The ion of the weak base produces H + in water to form an acidic solution. For example, NH 4 Cl contains ions from a weak base (NH 4 OH) and a strong acid (HCl). A solution of NH 4 Cl is acidic. NH H 2 O H 3 O + + NH 3

Acidity and basicity of Salt Solution Salt solutions can be neutral, acidic, or basic depending on ions present, because some ions react with water to produce H 3 O + and some other ions react with water to produce OH - ions. The general rule for predicting acidity or basicity of a salt solution is that the stronger partner from which the salt was formed dominates:

78 A salt from strong acid and weak base will be acidic. A salt from weak acid and strong base will be basic. A salt from strong acid and strong base will be neutral.

79 Learning Check Predict whether a solution of each salt will be 1) acidic2) basicor 3) neutral A. Li 2 CO 3 B. Mg(NO 3 ) 2 C. NH 4 Cl

80 Solution Predict whether a solution of each salt will be 1) acidic2) basicor 3) neutral A. Li 2 CO 3 2) basic; CO 3 -2 forms HCO 3 - and OH - B. Mg(NO 3 ) 2 3) neutral C. NH 4 Cl 1) acidic; NH 4 + produces H 3 O +

81 When an acid or base is added to water, the pH changes drastically. A buffer solution resists a change in pH when an acid or base is added Buffer Solutions

82 Buffers Absorb H 3 O + or OH - from foods and cellular processes to maintain pH. Are important in the proper functioning of cells and blood. In blood maintain a pH close to 7.4. A change in the pH of the blood affects the uptake of oxygen and cellular processes. Buffers Animation

83 A buffer solution Typically has equal concentrations of a weak acid and its salt. Or equal concentration of a weak base and its salt. Contains a combination of acid-base conjugate pairs. Contains a weak acid and a salt of the conjugate base of that acid. Components of a Buffer

84 The acetic acid/acetate buffer contains acetic acid (CH 3 COOH) and sodium acetate (CH 3 COONa). The salt produces sodium and acetate ions. CH 3 COONa CH 3 COO - + Na + The salt provides a higher concentration of the conjugate base CH 3 COO - than the weak acid. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Large amount Large amount Buffer Action

85 The function of the weak acid is to neutralize a base. The acetate ion in the product adds to the available acetate. CH 3 COOH + OH – CH 3 COO – + H 2 O Function of the Weak Acid

86 The function of the acetate ion CH 3 COO - (conjugate base) is to neutralize H 3 O + from acids. The weak acid product adds to the weak acid available. CH 3 COO - + H 3 O + CH 3 COOH + H 2 O Function of the Conjugate Base

87 The weak acid in a buffer neutralizes base. The conjugate base in the buffer neutralizes acid. The pH of the solution is maintained. Summary of Buffer Action

88 Learning Check Does each combination make a buffer solution? 1) yes2) noExplain. A. HCl and KCl B. H 2 CO 3 and NaHCO 3 C. H 3 PO 4 and NaCl D. CH 3 COOH and CH 3 COOK

89 Solution Does each combination make a buffer solution? 1) yes2) no Explain. A. HCl + KClno; HCl is a strong acid. B. H 2 CO 3 + NaHCO 3 yes; weak acid and its salt. C. H 3 PO 4 + NaCl no; NaCl does not contain a conjugate base of H 3 PO 4. D. CH 3 COOH + CH 3 COOK yes; weak acid and its salt.

90 pH of a Buffer The [H 3 O + ] in the K a expression is used to determine the pH of a buffer. Weak acid + H 2 O H 3 O + + Conjugate base K a = [H 3 O + ][conjugate base] [weak acid] For values of K a see Page 303 of your book. [H 3 O + ] = K a x [weak acid] [conjugate base] pH = –log [H 3 O + ]

91 Calculation of Buffer pH The weak acid H 2 PO 4 - in a blood buffer H 2 PO 4 - /HPO 4 2- has K a = 6.2 x What is the pH of the buffer if it is 0.20 M in both H 2 PO 4 - and HPO 4 2- ? [H 3 O + ] = K a x [H 2 PO 4 - ] [HPO 4 2- ] [H 3 O + ] = 6.2 x x [0.20 M] = 6.2 x [0.20 M] pH = –log [6.2 x ] = 7.21

92 Learning Check What is the pH of a H 2 CO 3 buffer that is 0.20 M H 2 CO 3 and 0.10 M HCO 3 - ? K a = 4.3 x 10 -7

93 Solution What is the pH of a H 2 CO 3 buffer that is 0.20 M H 2 CO 3 and 0.10 M HCO 3 - ? K a (H 2 CO 3 ) = 4.3 x [H 3 O + ] = K a x [H 2 CO 3 ] [HCO 3 - ] [H 3 O + ] = 4.3 x x [0.20 M] = 8.6 x [0.10 M] pH = –log [8.6 x ] = 6.07

Buffers in the Body The pH of body fluid is maintained by the following three major buffer systems: Carbonic acid – bicarbonate system Dihydrogen phosphate – hydrogen phosphate system Proteins acting as either proton acceptor or proton donors at different pH values. Relationships of the bicarbonate buffer system to the lungs and kidneys shown in the following figure 10.7.

95 Fig 10.7 Lungs and kidneys relation with the bicarbonate buffer system

TitrationTitration Titration is a laboratory procedure used to determine the molarity of an acid. In a titration, a base such as NaOH is added to a specific volume of an acid. Base (NaOH) Acid solution

97 Indicator A few drops of an indicator is added to the acid in the flask. The indicator changes color when the base (NaOH) has neutralized the acid.

98 End Point of Titration At the end point, the indicator has a permanent color. The volume of the base used to reach the end point is measured. The molarity of the acid is calculated using the neutralization equation for the reaction.

99 Fig 10.9 Flow diagram for acid-base titration

100 Calculating Molarity What is the molarity of an HCl solution, if 18.0 mL of a 0.25 M NaOH, are required to neutralize 10.0 mL of the acid? 1. Write the neutralization equation. HCl + NaOH NaCl + H 2 O 2. Calculate the moles of NaOH mL NaOH x 1 L x 0.25 mole NaOH 1000 mL 1 L = moles NaOH

101 Calculating Molarity (continued) 3. Calculate the moles of HCl. HCl + NaOH NaCl + H 2 O mole NaOH x 1 mole HCl 1 mole NaOH = moles HCl 4. Calculate the molarity of HCl mL HCl = L HCl mole HCl = 0.45 M HCl L HCl

102 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 + 2KOH K 2 SO 4 + 2H 2 O 1)12.5 mL 2) 50.0 mL 3) 200. mL Learning Check

103 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 + 2KOH K 2 SO 4 + 2H 2 O 1)12.5 mL L x 1.00 mole KOH x 1 mole H 2 SO 4 x 1 L 2 mole KOH 1 L x 1000 mL = 12.5 mL 2 mole H 2 SO 4 1 L Solution

104 Chapter Summary Arrhenius acid is a substance that gives hydrogen ions, H +, when dissolved in water. Arrhenius base is a substance that gives hydroxide ions, OH -, when dissolved in water. The neutralization reaction of an acid with a base yields water plus a salt. Bronsted-Lowry Acids: Any substance that is able to give a hydrogen ion, H +, to another ion or molecule. Bronsted-Lowry bases: Any substance that is able to accept a hydrogen ion, H +, from an acid.

105 Chapter Summary Contd. Conjugate acid-base pair: Two substances whose formula differ by only a hydrogen ion, H +. According to Bronsted-Lowry acid-base theory, water is both an acid and a base, property known as amphoteric. Strong acid: An acid that gives up H + easily and is essentially 100% dissociated (splits to produce H + and an anion) in water.

106 Chapter Summary Contd. Weak acid: An acid that gives up H + with difficulty and is less than 100% dissociated in water. Strong base: A base that has a high affinity for H + and holds it tightly. Weak base: A base that has only a slight affinity for H + and holds it weakly. The stronger the acid, the weaker its conjugate base; the weaker the acid, the stronger its conjugate base.

107 Chapter Summary Contd. Strong acids have k a value much greater than 1; Weak acids have k a value much less than 1. Ion product constant for water, k w : k w = k a [H 2 O] = [H 3 O + ][OH - ] = [1.00 x ][1.00 x ] = 1.00 x at 25 o C. Product of [H 3 O + ] and [OH - ] is a constant. Therefore, in an acidic solution where [H 3 O + ] is large and [OH-] must be small.

108 Chapter Summary Contd. Acidic solution: pH 1.00 x M Neutral solution: pH = 7 [H 3 O + ] = 1.00 x M Basic solution:pH > 7 [H 3 O + ] < 1.00 x M Buffer: A combination of substances that act together to prevent a drastic change in pH.

109 End of Chapter 10