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Chapter 19 - Neutralization 1. Section 19.1 Neutralization Reactions  OBJECTIVES:  Explain how acid-base titration is used to calculate the concentration.

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Presentation on theme: "Chapter 19 - Neutralization 1. Section 19.1 Neutralization Reactions  OBJECTIVES:  Explain how acid-base titration is used to calculate the concentration."— Presentation transcript:

1 Chapter 19 - Neutralization 1

2 Section 19.1 Neutralization Reactions  OBJECTIVES:  Explain how acid-base titration is used to calculate the concentration of an acid or a base 2

3 Section 19.1 Neutralization Reactions  OBJECTIVES:  Explain the concept of equivalence in neutralization reactions. 3

4 Acid-Base Reactions  Acid + Base Water + Salt  Properties related to every day:  antacids depend on neutralization  farmers use it to control soil pH  formation of cave stalactites  human body kidney stones 4

5 Acid-Base Reactions  Neutralization Reaction - a reaction in which an acid and a base react in an aqueous solution to produce a salt and water: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) H 2 SO 4(aq) + 2KOH (aq) K 2 SO 4(aq) + 2 H 2 O (l)  Table 19.1, page 458 lists some salts 5

6 Titration  Titration is the process of adding a known amount of solution of known concentration to determine the concentration of another solution  Remember? - a balanced equation is a mole ratio  Sample Problem 19-1, page 460 6

7 Titration  The concentration of acid (or base) in solution can be determined by performing a neutralization reaction  An indicator is used to show when neutralization has occurred  Often use phenolphthalein- colorless in neutral and acid; turns pink in base 7 SimulationSimulation Worksheet

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9 Steps - Neutralization reaction 1. A measured volume of acid of unknown concentration is added to a flask 2. Several drops of indicator added 3. A base of known concentration is slowly added, until the indicator changes color-measure the volume  Figure 19.4, page 461 9

10 Neutralization  The solution of known concentration is called the standard solution  added by using a buret  Continue adding until the indicator changes color  called the “end point” of the titration  Sample Problem 19-2, page

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13 Equivalents  One mole of hydrogen ions reacts with one mole of hydroxide ions  does not mean that 1 mol of any acid will neutralize 1 mol of any base  because some acids and bases can produce more than 1 mole of hydrogen or hydroxide ions  example: H 2 SO 4(aq) 2H + + SO

14 Equivalents  Made simpler by the existence of a unit called an equivalent  One equivalent (equiv) is the amount of acid (or base) that will give 1 mol of hydrogen (or hydroxide) ions  1 mol HCl = 1 equiv HCl  1 mol H 2 SO 4 = 2 equiv H 2 SO 4 14

15 Equivalents  In any neutralization reaction, the equivalents of acid must equal the equivalents of base  How many equivalents of base are in 2 mol Ca(OH) 2 ?  The mass of one equivalent is it’s gram equivalent mass (will be less than or equal to the formula mass): HCl = 36.5 g/mol; H 2 SO 4 = 49.0 g/mol 15

16 Equivalents  Sample Problem 19-3, page 462  Sample Problem 19-4, page

17 Normality (N)  Useful to know the Molarity of acids and bases  Often more useful to know how many equivalents of acid or base a solution contains  Normality (N) of a solution is the concentration expressed as number of equivalents per Liter 17

18 Normality (N)  Normality (N) = equiv/L  equiv = Volume(L) x N;  and also know: N=M x eq;  M = N / eq  Sample Problem 21-5, page 621  Diluting solutions of known Normality: N 1 x V 1 = N 2 x V 2  N 1 and V 1 are initial solutions  N 2 and V 2 are final solutions 18

19 Normality (N)  Titration calculations often done more easily using normality instead of molarity  In a titration, the point of neutralization is called the equivalence point  the number of equivalents of acid and base are equal 19

20 Normality (N)  Doing titrations with normality use: N A x V A = N B x V B  Sample Problem 19-6, page 464  Sample Problem 19-7, page 464  Sample Problem 19-8, page

21 Section 19.2 Salts in Solution  OBJECTIVES:  Demonstrate with equations how buffers resist changes in pH 21

22 Section 19.2 Salts in Solution  OBJECTIVES:  Calculate the solubility product constant (K sp ) of a slightly soluble salt 22

23 Salt Hydrolysis  A salt:  comes from the anion of an acid (Cl -)  comes from the cation of a base (Na +)  formed from a neutralization reaction  some neutral; others acidic or basic  Salt hydrolysis- salt reacts with water to produce acid or base solution 23

24 Salt Hydrolysis  Hydrolyzing salts usually made from:  strong acid + weak base, or  weak acid + strong base  Strong refers to the degree of ionization (100%) What pH will result from the above combinations? 24

25 Salt Hydrolysis  To see if the resulting salt is acidic or basic, check the “parent” acid and base that formed it: NaCl HCl + NaOH NH 4 OH H 2 SO 4 + NH 4 OH CH 3 COOK CH 3 COOH + KOH 25

26 26 Strong Acids HCl HClO 4 H 2 SO 4 HI HNO 3 HBr Strong Bases Mg(OH) 2 NaOH Ca(OH) 2 KOH Na Cl OHH To determine if a salt is made From a combination which is: acid/base weak/weak weak/strong strong/strong strong/weak

27 27 Lab 42 : Salt Hydrolysis Universal Indicator Colors

28 Buffers  Buffers are solutions in which the pH remains relatively constant when small amounts of acid or base are added  made from a pair of chemicals  a weak acid and one of it’s salts;  HA / A -  or a weak base and one of it’s salts  NH 3 / NH Simulation

29 Buffers  A buffer system is better able to resist changes in pH than pure water  Since it is a pair of chemicals:  one chemical neutralizes any acid added, while the other chemical would neutralize any additional base  they make each other in the process! 29

30 30 pH Add strong acid pH Add strong acid Unbuffered reaction between and acid an base Buffered solution and reaction of an acid with a base HA / A - HCl+A-A- HA+Cl - HCl + NaOH  NaCl + HOH

31 Buffers  Example: Ethanoic (acetic) acid and sodium ethanoate (also called sodium acetate)  HC 2 H / NaC 2 H becomes  HC 2 H / C 2 H  Weak acid weak base  The buffer capacity is the amount of acid or base that can be added before a significant change in pH 31

32 Buffers  Buffers that are crucial to maintain the pH of human blood: 1.carbonic acid - hydrogen carbonate H 2 CO 3 / HCO dihydrogen phosphate - monohydrogen phoshate H 2 PO 4 - / HPO 4 2-  Table 19.2, page 469 has some important buffer systems  Sample Problem 19-9, page

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34 34 Calculating k sp or Solubility Product Constant Copy Example 10/11 pg 470/471 into your notes What does a high Ksp mean? What does a low Ksp mean?

35 Solubility Product Constant  Salts differ in their solubilities  Table 19.3, page 470  Most “insoluble” salts will actually dissolve to some extent in water  said to be slightly, or sparingly, soluble in water 35

36 Solubility Product Constant  Consider: AgCl (s)  The “equilibrium expression” is: 36 K eq = Ag + (aq) + Cl - (aq) [ Ag + ] x [ Cl - ] [ AgCl ]

37 Solubility Product Constant  But, the [ AgCl ] is constant as long as some undissolved solid is present  Thus, a new constant is developed, and is called the “solubility product constant” (K sp ): K eq x [ AgCl ] = [ Ag + ] x [ Cl - ] = K sp 37

38 Solubility Product Constant  Values of solubility product constants are given for some sparingly soluble salts in Table 19.4, page 471  Although most compounds of Ba are toxic, BaSO 4 is so insoluble that it is used in gastrointestinal examinations by doctors! - p

39 Solubility Product Constant  To solve problems:  a) write equation,  b) write expression, and  c) fill in values using x for unknowns  Sample Problem 21-10, page 634  Sample Problem 21-11, page

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41 Common Ion Effect  A “common ion” is an ion that is common to both salts in solution  example: You have a solution of lead (II) chromate. You now add some lead (II) nitrate to the solution.  The lead (II) ion is the common ion 41

42 42 PbCrO 4  Add Pb(NO 3 ) 2 This causes a shift in equilibrium (due to Le Chatelier’s principle), and is called the common ion effect PbCrO 4  Pb 2+ + CrO 4 2-  shift Pb 2+ + CrO 4 2-

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44 Common Ion Effect  Sample Problem The Ksp of silver iodide is 8.3x What is the iodide concentration of a 1.00L saturated solution of AgI to which mol of AgNO 3 is added?  1. Write the equilibrium equation. AgI (s) Ag 1+ + I Write the Ksp expression Ksp = [Ag 1+ ] 1 [I 1- ] 1 = 8.3x Ksp = (x) 1 (x) 1 = 8.3x Ksp = (x ) 1 (x) 1 = 8.3x Ksp = (0.020) (x) = 8.3x x = 4.2x The [ ] of iodide ion is 4.2x M Note: X is So small that it can be ignored

45  The solubility product constant (K sp ) can be used to predict whether a precipitate will form or not in a reaction  if the calculated ion-product concentration is greater than the known K sp, a precipitate will form 45

46 46 ksp Sample Problem: A student prepares a solution by combining mol CaCl 2 with mol Pb(NO 3 ) 2 in a 1 L container. Will a precipitate form? WHAT WE KNOW!!! CaCl 2 +Pb(NO 3 ) 2 Ca(NO 3 ) 2 (aq) + PbCl 2 (s) mol/L0.015 mol/L X ppt PbCl 2 Pb Cl 1- ksp = [Pb 2+ ] 1 [Cl 1- ] 2 The values we need to solve the problem!

47 47 1)Calculate the concentration of the ion’s used to make the precipitate. CaCl 2 Ca Cl mol CaCl 2 1L 2 mol Cl - 1 mol CaCl mol Cl - L mol/L Pb(NO 3 ) 2 Pb 2+ +2NO mol/L 1L1 mol Pb(NO 3 ) mol Pb(NO 3 ) 2 1 mol Pb mol Pb 2+ 1L

48 48 2) Calculate the Ksp for the precipitate PbCl 2  Pb Cl - Ksp = [Pb 2+ ] 1 [Cl - ] 2 Ksp = ( M) 1 ( M) 2 Ksp for PbCl 2 = 3.75 x Calculated Ksp 3.75 x is > Known Ksp 1.7 x for PbCl 2 Calc > Known = PPT Known> Calc = No PPT Known = Calc = No PPT See pg 471 Table 19-4 for known values

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