 Chapter 19 - Neutralization

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Chapter 19 - Neutralization

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

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

Properties related to every day:
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

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) + H2O(l) H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2 H2O(l) Table 19.1, page 458 lists some salts

Remember? - a balanced equation is a mole ratio
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

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 Simulation Simulation Worksheet

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

The solution of known concentration is called the standard solution
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 461

One mole of hydrogen ions reacts with one mole of hydroxide ions
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: H2SO4(aq) H+ + SO42-

Made simpler by the existence of a unit called an equivalent
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 H2SO4 = 2 equiv H2SO4

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; H2SO4 = 49.0 g/mol

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

Useful to know the Molarity of acids and bases
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

Diluting solutions of known Normality: N1 x V1 = N2 x V2
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: N1 x V1 = N2 x V2 N1 and V1 are initial solutions N2 and V2 are final solutions

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

Doing titrations with normality use: NA x VA = NB x VB
Normality (N) Doing titrations with normality use: NA x VA = NB x VB Sample Problem 19-6, page 464 Sample Problem 19-7, page 464 Sample Problem 19-8, page 464

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

Section 19.2 Salts in Solution
OBJECTIVES: Calculate the solubility product constant (Ksp) of a slightly soluble salt

A salt: Salt Hydrolysis 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

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?

Salt Hydrolysis To see if the resulting salt is acidic or basic, check the “parent” acid and base that formed it: NaCl HCl + NaOH NH4OH H2SO4 + NH4OH CH3COOK CH3COOH + KOH

Strong Acids HCl HClO4 H2SO4 HI HNO3 HBr Strong Bases Mg(OH)2 NaOH
Ca(OH)2 KOH To determine if a salt is made From a combination which is: acid/base weak/weak weak/strong strong/strong strong/weak Na Cl OH H

Universal Indicator Colors
Lab 42 : Salt Hydrolysis Universal Indicator Colors

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 NH3 / NH4+ Simulation

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!

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

Buffers Example: Ethanoic (acetic) acid and sodium ethanoate (also called sodium acetate) HC2H302 / NaC2H302 becomes HC2H302 / C2H302 1- Weak acid weak base The buffer capacity is the amount of acid or base that can be added before a significant change in pH

Buffers Buffers that are crucial to maintain the pH of human blood:
carbonic acid - hydrogen carbonate H2CO3 / HCO3 - 2. dihydrogen phosphate - monohydrogen phoshate H2PO4- / HPO4 2- Table 19.2, page 469 has some important buffer systems Sample Problem 19-9, page 468

Solubility Product Constant
Calculating ksp 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?

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

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

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” (Ksp): Keq x [ AgCl ] = [ Ag+ ] x [ Cl- ] = Ksp

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, BaSO4 is so insoluble that it is used in gastrointestinal examinations by doctors! - p.632

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 634

A “common ion” is an ion that is common to both salts in solution
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

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 effect shift PbCrO4  Pb CrO42-

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-] = 8.3x10-17 Ksp = (x) (x) = 8.3x10-17 Ksp = (x )1 (x) = 8.3x10-17 Ksp = (0.020) (x) = 8.3x10-17 x = 4.2x10-15 Note: X is So small that it can be ignored The [ ] of iodide ion is 4.2x10-15 M

The solubility product constant (Ksp) 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 Ksp, a precipitate will form

The values we need to solve the problem!
ksp Sample Problem: A student prepares a solution by combining mol CaCl2 with mol Pb(NO3)2 in a 1 L container. Will a precipitate form? WHAT WE KNOW!!! CaCl2 + Pb(NO3)2 Ca(NO3)2 (aq) + PbCl2 (s) 0.025 mol/L 0.015 mol/L X ppt PbCl Pb Cl1- ksp = [Pb2+]1 [Cl1-]2 The values we need to solve the problem!

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. CaCl2 Ca 2+ + 2Cl- 0.025 mol/L 0.050 mol Cl- L 0.025 mol CaCl2 2 mol Cl- 1L 1 mol CaCl2 Pb(NO3)2 Pb2+ + 2NO3- 0.015 mol/L 0.015 mol Pb2+ 1L 0.015 mol Pb(NO3)2 1 mol Pb2+ 1L 1 mol Pb(NO3)2

2) Calculate the Ksp for the precipitate
PbCl2  Pb Cl- Ksp = [Pb 2+]1 [Cl-]2 Ksp = ( M)1 ( M) 2 Ksp for PbCl2 = x 10 -5 Calculated Ksp 3.75 x is > Known Ksp 1.7 x for PbCl2 See pg 471 Table 19-4 for known values Calc > Known = PPT Known> Calc = No PPT Known = Calc = No PPT