1. Ch. 18: Acids & Bases
Sec. 18.4: Neutralization

2. Objectives Write chemical equations for neutralization reactions.
Explain how neutralization reactions are used in acid-base titrations.
Compare the properties of buffered & unbuffered solutions.

3. Neutralization
A neutralization reaction is a reaction between an acid and a base that produces a salt and water.
A salt is an ionic compound made up of a cation from a base and an anion from an acid.
Example: HCl + NaOH --> NaCl + H2O

4. Neutralization Neutralization is a double-replacement reaction
replaces
2HCl + Mg(OH)2 --> MgCl2 + 2H2O
Note: in order to write the correct formula for the salt, you must “criss-cross” the ions, Mg+2 and Cl-. Then you must BALANCE the equation to conserve matter.

5. Practice Problems I
Write balanced chemical equations for the following neutralization reactions:
HNO3 + CsOH -->
HBr + Ca(OH)2 -->
sulfuric acid + potassium hydroxide -->
HC2H3O2 + NH4OH -->
H3PO4 + Ba(OH)2 

6. Practice Problems II
Complete the following neutralization reactions & then balance the equations.
___ + ___ --> Mg(NO3)2 + H2O
___ + ___ --> Ca3(PO4)2 + H2O
___ + ___ --> lithium sulfate + water
___ + ___ --> Na2CO3 + water
___ + ____  Al2(SO4)3 + water

7. Net Ionic Equations
It is important to realize that if the acid, base, and salt completely dissociate in aqueous solution, you can also write a complete and net ionic equation for neutralization reactions.
HCl + NaOH --> NaCl + H2O
H+ + Cl- + Na+ + OH- --> Na+ + Cl- + H2O
H+ + OH- --> H2O
Practice: Write complete and net ionic equations for #1 on previous slide.

8. Titration
The stoichiometry of acid-base reactions provides the basis for a procedure known as titration.
Titration is a method for determining the concentration of a solution by reacting a known volume of the solution with a solution of known volume AND concentration.
To find the concentration of an acid, you would titrate it with a base of known concentration (and vice versa).

9. The Procedure for Titration
- Place a measured volume (or KNOWN volume) of a acid or base of UNKNOWN concentration in a beaker. Add an indicator. (See pg. 662)
Note: Any indicator can be used as long as it has a color change at, or close to, a pH of 7. The color change will indicate to you
when enough of the
solution of KNOWN
concentration has been added
to neutralize the solution of
UNKNOWN concentration.

10. The Procedure for Titration
Phenolphthalein is generally used because there is a clear indication of an acid state (colorless) and a basic state (pink). In titration, a very pale pink is the color sought.

11. The Procedure for Titration
- Fill a buret with a KNOWN volume of the solution of KNOWN concentration. This solution is called the standard solution or titrant.

12. The Procedure for Titration
- Add measured volumes of the standard solution to a KNOWN volume of the solution of “unknown” concentration, mixing thoroughly. Continue UNTIL the indicator undergoes a color change. This indicates that the standard solution has neutralized the “unknown” solution.

13. Titration
The end point is the point at which the indicator in the titration changes color.
The endpoint should closely approximate the equivalence point.
The equivalence point is the stoichiometric point at which the moles of H+ ions from the acid equal the moles of OH- from the base.

14. Titration Curves
A graph like this shows how the pH changes during a titration.
In this titration, the “unknown” is a base. The pH slowly decreases when the standard acid is added.

15. Titration Curves
When nearly all the the OH- ions have been neutralized, a dramatic decrease in pH is seen with the addition of very small volumes of the standard acid.
This abrupt change in pH occurs at the equivalence point.
Additional amounts of acids result in lowering the pH even further.

16. Equivalence Points
Not all titrations have equivalence points at a pH of 7.
The pH at the equivalence point depends on the relative strengths of the reacting acid and base.

17. Calculating the Unknown Molarity
A chemist titrated 25 mL of an unknown concentration of hydrochloric acid with 14.5 mL of 0.50 M NaOH. What is the concentration of the acid?
Write the balanced equation for the reaction.
HCl + NaOH --> NaCl + H2O
Calculate the # of moles of the solution of KNOWN molarity in the volume given. In this reaction, 14.5 mL of 0.50 M NaOH was used in the neutralization.
Remember: M = moles solute/liters of solution,
0.50 M = x moles/ L
x = mol NaOH

18. Calculating the Unknown Molarity
Use the moles of the KNOWN solution and a mole ratio from the balanced equation to calculate the moles of the UNKNOWN solution needed for neutralization.
mol NaOH x 1 mol HCl =
1 mol NaOH mol HCl
Determine the molarity of the UNKNOWN by using the moles (just found) & the volume of the UNKNOWN solution used in the titration.
M = mol solute = = 0.29 M
liters sol’n

19. Practice Problems
What is the molarity of a Ca(OH)2 solution if 30.0 mL of the solution is neutralized by 26.4 mL of M HBr?
What is the molarity of H2SO4 solution if 43.3 mL of M KOH is needed to neutralize 20.0 mL of the unknown?
What is the concentration of household ammonia (NH3) if 49.9 mL of 0.59 M HCl is required to neutralize 25.9 mL of ammonia?

20. Salt Hydrolysis Recall the definition of a salt . . . .
When a salt is added to water, will any reaction occur?
Why then does an indicator added to a salt and water solution show that some salts are neutral, some are basic, and some are acidic?

21. Bromthymol blue and salts
The ammonium chloride solution turns yellow, indicating it is acidic.
Sodium nitrate (blue) is neutral.
Potassium fluoride (green) is basic.

22. Salt hydrolysis Many salts react with water in salt hydrolysis.
When dissolved, the ions of a salt separate.
The anions of the salt may accept H+ from water molecules OR the cations may donate H+ to water.
It all depends on the acid & base used to produce the salt.

23. An Example Potassium fluoride (KF) was produced from ?
_____ + _____  KF + H2O
Note that KF results from the neutralization of a weak acid (HF) by a strong base (KOH).
When the KF produced is then dissolved, the salt dissociates: KF  K+ + F-

24. Example Continued
Once dissolved, this what happens next: the “weak” ion of the salt reacts with water. In this example, the anion of the salt (F-) accepts H+ from water molecules.
F- + H2O  HF + OH-
Fluoride ions act as a Bronsted base and OH- is produced! The solution is basic.

25. Another example
NH4Cl is formed from a weak base (NH3) and a strong acid (HCl)
In solution, NH4Cl dissociates and the weak NH4+ reacts with water. When the cation donates H+ to water, an acidic solution is formed.
NH4Cl  NH Cl-
NH H2O  NH3 + H3O+
NH4Cl produces an acidic solution.

26. Conclusions Have you noticed?
The salt of a strong base produces a basic solution.
The salt of a strong acid produces a acidic solution.
Little or no hydrolysis occurs when a salt of a strong acid and a strong base dissociates in water. These salt solutions are neutral.

27. Predictions
Therefore, in order to predict whether a salt will produce a acidic, basic, or neutral solution, we must determine the acid and base that produced the salt.
In addition, we must know which acids are considered strong acids and which bases are strong bases.

28. Strong Acids (pg. 603) Strong acid ionize completely in water.
Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Perchloric (HClO4)
Nitric (HNO3)
Sulfuric (H2SO4)

29. Strong bases (pg. 606) Strong bases dissociate entirely in water.
Sodium hydroxide (NaOH)
Potassium hydroxide (KOH)
Rubidium hydroxide (RbOH)
Cesium hydroxide (CsOH)
Calcium hydroxide (Ca(OH)2)
Barium hydroxide (Ba(OH)2)

30. Practice Problems
First, write the equation for the neutralization reaction producing the following salts. Then, write equations for their salt hydrolysis reactions and classify their solutions as acidic, basic, or neutral.
Magnesium sulfate
Calcium carbonate
Rubidium acetate

31. Buffered Solutions
pH is an important abiotic factor for all living things.
Control of pH is as important in an aquarium as it is in your body.

32. Buffered Solutions
A buffer is a solution that resists a change in pH when small amounts of an acid or base are added.
Adding small amounts of HCl to pure water, for example, may lower its pH from 7 to 2. But adding the same amount of HCl to a buffered solution, the pH may decrease only to 6.

33. How do buffers work?
A buffer is a mixture of a weak acid and its conjugate base OR a weak base and its conjugate acid.
This solution reacts with any H+ or OH- added to it.

34. Example
Suppose a buffer solution contains HF (the weak acid) and F- (its conjugate base).
Since it is a weak acid, an equilibrium is established when the acid dissociates:
HF H+ + F-
Any small addition of H+ shifts the equilibrium to the left, forming more HF.
The pH will still remain fairly constant because the [H+] does not change that much.

35. HF H+ + F-
If a base (OH-) is added to the buffering solution, the H+ ions in the solution react with the OH- to form water.
The equilibrium will shift to the right but pH will still remain fairly constant. The [H+] does not change that much.

36. Buffered Solutions
Great amounts of acid or base added to a buffered solution will change the pH.
The amount of acid or base a buffer can absorb without a large change in pH is called the buffer capacity.
The greater the concentration of substances in the buffering solution, the greater the buffer capacity.

37. Buffering Systems in Blood
The pH of human blood must be kept within a narrow range – 7.1 to 7.7.
Outside of this range, among other things, proteins will denature (and not function properly).
There are a number of buffering systems in your blood that maintain pH.

38. This buffer is the most important.
Blood Buffers
CO2 + H2O H2CO H+ + HCO3-
This buffer is the most important.
If H+ ions enter the blood, the equilibriums shift left. The kidneys remove more water from your blood and the lungs expel greater amounts of CO2. Your rate of breathing increases.

39. CO2 + H2O H2CO H+ + HCO3-
If OH- ions enter the blood, H+ ions react with them and the equilibriums shift to the right. The kidneys remove more HCO3- ions and the lungs expel less CO2. Your rate of breathing slows.

40. Practice Problems
In the following situations, predict whether the blood pH will rise or fall, and which way the H2CO3/HCO3- equilibrium will shift.
A person becomes overexcited and hyperventilates.
A person takes an overdose of the antacid NaHCO3.
A person has the flu and vomits a lot.