2 Section 19.1 Neutralization Reactions OBJECTIVES:Explain how acid-base titration is used to calculate the concentration of an acid or a base
3 Section 19.1 Neutralization Reactions OBJECTIVES:Explain the concept of equivalence in neutralization reactions.
4 Properties related to every day: Acid-Base ReactionsAcid + Base Water + SaltProperties related to every day:antacids depend on neutralizationfarmers use it to control soil pHformation of cave stalactiteshuman body kidney stones
5 Acid-Base ReactionsNeutralization 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) + H2O(l)H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2 H2O(l)Table 19.1, page 458 lists some salts
6 Remember? - a balanced equation is a mole ratio TitrationTitration is the process of adding a known amount of solution of known concentration to determine the concentration of another solutionRemember? - a balanced equation is a mole ratioSample Problem 19-1, page 460
7 TitrationThe concentration of acid (or base) in solution can be determined by performing a neutralization reactionAn indicator is used to show when neutralization has occurredOften use phenolphthalein- colorless in neutral and acid; turns pink in baseSimulationSimulation Worksheet
9 Steps - Neutralization reaction 1. A measured volume of acid of unknown concentration is added to a flask2. Several drops of indicator added3. A base of known concentration is slowly added, until the indicator changes color-measure the volumeFigure 19.4, page 461
10 The solution of known concentration is called the standard solution NeutralizationThe solution of known concentration is called the standard solutionadded by using a buretContinue adding until the indicator changes colorcalled the “end point” of the titrationSample Problem 19-2, page 461
13 One mole of hydrogen ions reacts with one mole of hydroxide ions EquivalentsOne mole of hydrogen ions reacts with one mole of hydroxide ionsdoes not mean that 1 mol of any acid will neutralize 1 mol of any basebecause some acids and bases can produce more than 1 mole of hydrogen or hydroxide ionsexample: H2SO4(aq) H+ + SO42-
14 Made simpler by the existence of a unit called an equivalent EquivalentsMade simpler by the existence of a unit called an equivalentOne equivalent (equiv) is the amount of acid (or base) that will give 1 mol of hydrogen (or hydroxide) ions1 mol HCl = 1 equiv HCl1 mol H2SO4 = 2 equiv H2SO4
15 EquivalentsIn any neutralization reaction, the equivalents of acid must equal the equivalents of baseHow 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; H2SO4 = 49.0 g/mol
16 Sample Problem 19-3, page 462 Sample Problem 19-4, page 462 EquivalentsSample Problem 19-3, page 462Sample Problem 19-4, page 462
17 Useful to know the Molarity of acids and bases Normality (N)Useful to know the Molarity of acids and basesOften more useful to know how many equivalents of acid or base a solution containsNormality (N) of a solution is the concentration expressed as number of equivalents per Liter
18 Diluting solutions of known Normality: N1 x V1 = N2 x V2 Normality (N) = equiv/Lequiv = Volume(L) x N;and also know: N=M x eq;M = N / eqSample Problem 21-5, page 621Diluting solutions of known Normality: N1 x V1 = N2 x V2N1 and V1 are initial solutionsN2 and V2 are final solutions
19 Normality (N)Titration calculations often done more easily using normality instead of molarityIn a titration, the point of neutralization is called the equivalence pointthe number of equivalents of acid and base are equal
20 Doing titrations with normality use: NA x VA = NB x VB Normality (N)Doing titrations with normality use: NA x VA = NB x VBSample Problem 19-6, page 464Sample Problem 19-7, page 464Sample Problem 19-8, page 464
21 Section 19.2 Salts in Solution OBJECTIVES:Demonstrate with equations how buffers resist changes in pH
22 Section 19.2 Salts in Solution OBJECTIVES:Calculate the solubility product constant (Ksp) of a slightly soluble salt
23 A salt: Salt Hydrolysis comes from the anion of an acid (Cl-) comes from the cation of a base (Na+)formed from a neutralization reactionsome neutral; others acidic or basicSalt hydrolysis- salt reacts with water to produce acid or base solution
24 Hydrolyzing salts usually made from: Salt HydrolysisHydrolyzing salts usually made from:strong acid + weak base, orweak acid + strong baseStrong refers to the degree of ionization (100%)What pH will result from the abovecombinations?
25 Salt HydrolysisTo see if the resulting salt is acidic or basic, check the “parent” acid and base that formed it:NaCl HCl + NaOHNH4OH H2SO4 + NH4OHCH3COOK CH3COOH + KOH
26 Strong Acids HCl HClO4 H2SO4 HI HNO3 HBr Strong Bases Mg(OH)2 NaOH Ca(OH)2KOHTo determine if a salt is madeFrom a combination which is:acid/baseweak/weakweak/strongstrong/strongstrong/weakNa ClOHH
28 BuffersBuffers are solutions in which the pH remains relatively constant when small amounts of acid or base are addedmade from a pair of chemicalsa weak acid and one of it’s salts;HA / A-or a weak base and one of it’s saltsNH3 / NH4+Simulation
29 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 basethey make each other in the process!
30 HA / A- HCl + A- HA + Cl- pH pH Add strong acid Add strong acid Unbuffered reactionbetween and acid an baseBuffered solution and reactionof an acid with a baseHA / A-HCl + NaOH NaCl + HOHHCl+A-HA+Cl-pHpHAdd strong acidAdd strong acid
31 BuffersExample: Ethanoic (acetic) acid and sodium ethanoate (also called sodium acetate)HC2H302 / NaC2H302 becomesHC2H302 / C2H302 1-Weak acid weak baseThe buffer capacity is the amount of acid or base that can be added before a significant change in pH
32 Buffers Buffers that are crucial to maintain the pH of human blood: carbonic acid - hydrogen carbonateH2CO3 / HCO3 -2. dihydrogen phosphate - monohydrogen phoshateH2PO4- / HPO4 2-Table 19.2, page 469 has some important buffer systemsSample Problem 19-9, page 468
34 Solubility Product Constant Calculating ksporSolubility Product ConstantCopy Example 10/11 pg 470/471 into your notesWhat does a high Ksp mean?What does a low Ksp mean?
35 Solubility Product Constant Salts differ in their solubilitiesTable 19.3, page 470Most “insoluble” salts will actually dissolve to some extent in watersaid to be slightly, or sparingly, soluble in water
37 Solubility Product Constant But, the [ AgCl ] is constant as long as some undissolved solid is presentThus, a new constant is developed, and is called the “solubility product constant” (Ksp):Keq x [ AgCl ] = [ Ag+ ] x [ Cl- ] = Ksp
38 Solubility Product Constant Values of solubility product constants are given for some sparingly soluble salts in Table 19.4, page 471Although most compounds of Ba are toxic, BaSO4 is so insoluble that it is used in gastrointestinal examinations by doctors! - p.632
39 Solubility Product Constant To solve problems:a) write equation,b) write expression, andc) fill in values using x for unknownsSample Problem 21-10, page 634Sample Problem 21-11, page 634
41 A “common ion” is an ion that is common to both salts in solution Common Ion EffectA “common ion” is an ion that is common to both salts in solutionexample: 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
42 PbCrO4 Pb2+ + CrO42- Add Pb(NO3)2 PbCrO4 Pb2+ + CrO42- This causes a shift in equilibrium (due to Le Chatelier’s principle), and is called the common ion effectshiftPbCrO4 Pb CrO42-
44 Common Ion Effect Ksp = (x)1 (x)1 = 8.3x10-17 x = 4.2x10-15 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 AgNO3 is added?1. Write the equilibrium equation.AgI (s) Ag I1-2. Write the Ksp expression Ksp = [Ag1+] [I1-] = 8.3x10-17Ksp = (x) (x) = 8.3x10-17Ksp = (x )1 (x) = 8.3x10-17Ksp = (0.020) (x) = 8.3x10-17x = 4.2x10-15Note: X isSo small thatit can be ignoredThe [ ] of iodide ion is 4.2x10-15 M
45 The solubility product constant (Ksp) can be used to predict whether a precipitate will form or not in a reactionif the calculated ion-product concentration is greater than the known Ksp, a precipitate will form
46 The values we need to solve the problem! ksp Sample Problem:A student prepares a solution by combining mol CaCl2with mol Pb(NO3)2 in a 1 L container. Will a precipitate form?WHAT WE KNOW!!!CaCl2 +Pb(NO3)2Ca(NO3)2 (aq)+ PbCl2 (s)0.025 mol/L0.015 mol/LX pptPbCl Pb Cl1-ksp = [Pb2+]1 [Cl1-]2The values we need to solve the problem!
47 CaCl2 Ca 2+ + 2Cl- 0.025 mol/L 0.050 mol Cl- L 0.025 mol CaCl2 Calculate the concentration of the ion’s used to make the precipitate.CaCl2Ca 2++2Cl-0.025 mol/L0.050 mol Cl-L0.025 mol CaCl22 mol Cl-1L1 mol CaCl2Pb(NO3)2Pb2++2NO3-0.015 mol/L0.015 mol Pb2+1L0.015 mol Pb(NO3)21 mol Pb2+1L1 mol Pb(NO3)2
48 2) Calculate the Ksp for the precipitate PbCl2 Pb Cl-Ksp = [Pb 2+]1 [Cl-]2Ksp = ( M)1 ( M) 2Ksp for PbCl2 = x 10 -5Calculated Ksp 3.75 x is > Known Ksp 1.7 x for PbCl2See pg 471 Table 19-4 for known valuesCalc > Known = PPTKnown> Calc = No PPTKnown = Calc = No PPT