1. Chapter 19 - Neutralization

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 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

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) + 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
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

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
Simulation
Simulation Worksheet

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

10. 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

13. 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-

14. 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

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; H2SO4 = 49.0 g/mol

16. 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

17. 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

18. 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

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

20. 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

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 reaction
some neutral; others acidic or basic
Salt hydrolysis- salt reacts with water to produce acid or base solution

24. Hydrolyzing salts usually made from:
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?

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
NH4OH H2SO4 + NH4OH
CH3COOK CH3COOH + KOH

26. 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

27. Universal Indicator Colors
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
NH3 / 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 base
they make each other in the process!

30. 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

31. 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

32. 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

34. 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?

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

36. Solubility Product Constant
Consider: AgCl(s)
The “equilibrium expression” is:
Ag+(aq) + Cl-(aq)
[ Ag+ ] x [ Cl- ]
Keq =
[ 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” (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 471
Although 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, and
c) fill in values using x for unknowns
Sample Problem 21-10, page 634
Sample Problem 21-11, page 634

41. 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

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 effect
shift
PbCrO4  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-] = 8.3x10-17
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

45. 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

46. 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!

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.
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

48. 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