unit 12: acid and bases
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Unit 12: Acid and Bases. Chapter 19. 4 th Hour. I am gone today so you will go through this Powerpoint to get the notes for objective 38 and the beginning of 39. Remember, there is an open-note quiz tomorrow and I expect a good report from the sub. - PowerPoint PPT PresentationTRANSCRIPT
Unit 12: Acid and BasesChapter 19
4th HourI am gone today so you will go
through this Powerpoint to get the notes for objective 38 and the beginning of 39.
Remember, there is an open-note quiz tomorrow and I expect a good report from the sub.
Before we get to the new material, I want to get an idea of how you are doing on the material from the previous few days.
Work on the following problems with the person sitting next to you.
Record your answers on a separate piece of paper (one per group).
Name the following:
1. HNO3
2. Mn(OH)3
3. HBr4. CuOH5. H3PO3
Write the complete dissociation reaction for the following:
6. HNO3 + H2O
7. H2CO3 + H2O
Calculate the pH for the following.8. [H3O-] = 0.0008 M
9. HCl + H2O H3O- + Cl-
[HCl] = 0.02 M
10.NaOH Na+ + OH-
[NaOH] = 0.036 M
Turn your answers.
At this point, we will begin objective 38.
This objective deals with weak acids and weak bases.
Writing Ka equations
The same principles that apply to equilibrium equations will also apply to acids and bases.
For example:
HBr(aq) + H2O(l) H3O+(aq) + Br- (aq)
Ka =
Using K equationsMost commonly, these K equations
are used to determine the pH of a substance.
This is done by calculating the concentration of the hydronium ion.
Ka =
Once that concentration is known, use the pH equation: -log[H3O+]
Example
HF(aq) + H2O(l) H3O+(aq) + F- (aq)
Given the following equation, calculate the pH with the following information.
◦[HF] = 0.5M◦Ka = 7.2 x 10-4
Example
HF(aq) + H2O(l) H3O+(aq) + F- (aq)
Given the following equation, calculate the pH with the following information.
◦[HF] = 0.5M◦Ka = 7.2 x 10-4
First, write the Ka equation.
Example
HF(aq) + H2O(l) H3O+(aq) + F- (aq)
Given the following equation, calculate the pH with the following information.
◦[HF] = 0.5M◦Ka = 7.2 x 10-4
Second, fill in the numbers. Since every HF breaks into H3O+ and F-, then we can assume that they both have the same concentration.
Example
HF(aq) + H2O(l) H3O+(aq) + F- (aq)
Given the following equation, calculate the pH with the following information.
◦[HF] = 0.5M◦Ka = 7.2 x 10-4 X2 = 3.6 x 10-4
X = 0.019 M = [H3O+]
Third, solve for X.
Example
HF(aq) + H2O(l) H3O+(aq) + F- (aq)
Given the following equation, calculate the pH with the following information.
◦[HF] = 0.5M X = 0.019 M = [H3O+]
◦Ka = 7.2 x 10-4 -log [0.019] = pH
pH = 1.72
Finally, solve for pH.
Kb
It is simple to calculate the pH for acids using a Ka equation because the hydronium ion (H3O+) concentration can be solved for.
For bases though, it is the hydroxide ion that is solved for:
Fe(OH)2(aq) Fe+2(aq) + 2OH-
(aq)
Thus: Kb =
Kb
Once [OH-] is known, it is possible to calculate pOH
◦pOH = -log[OH-]
Since most bases are aqueous solutions, we can use the water dissociation to complete the problem.
◦pKw =pH + pOH
Kb Recap
Write the dissociation equation.Calculate [OH-] from the Kb
equation.Determine pOHUse the water dissociation to
determine pH:◦14 = pH + pOH
ExampleCa(OH)2(aq) Ca+2
(aq) + 2OH-(aq)
Calculate the pH given the following information:
◦[Ca(OH)2] = 0.05 M
◦[Ca+2] = 0.0136 M◦Kb = 3.7 x 10-3
ExampleCa(OH)2(aq) Ca+2
(aq) + 2OH-(aq)
Calculate the pH given the following information:
◦[Ca(OH)2] = 0.05 M
◦[Ca+2] = 0.0136 M◦Kb = 3.7 x 10-3
First, write the Kb equation.
ExampleCa(OH)2(aq) Ca+2
(aq) + 2OH-(aq)
Calculate the pH given the following information:
◦[Ca(OH)2] = 0.05 M
◦[Ca+2] = 0.0136 M◦Kb = 3.7 x 10-3
Second, fill in the numbers.
ExampleCa(OH)2(aq) Ca+2
(aq) + 2OH-(aq)
Calculate the pH given the following information:
◦ [Ca(OH)2] = 0.05 M X2 = 1.36 x 10-2
◦ [Ca+2] = 0.0136 M◦Kb = 3.7 x 10-3 x = 0.12 M = [OH]
Third, solve for X.
ExampleCa(OH)2(aq) Ca+2
(aq) + 2OH-(aq)
Calculate the pH given the following information:
◦[Ca(OH)2] = 0.05 M x = 0.12 M = [OH]
◦[Ca+2] = 0.0136 M -log [0.12] = pOH
◦Kb = 3.7 x 10-3 0.93 = pOH
Fourth, solve for pOH.
ExampleCa(OH)2(aq) Ca+2
(aq) + 2OH-(aq)
Calculate the pH given the following information:
◦ [Ca(OH)2] = 0.05 M 0.93 = pOH
◦ [Ca+2] = 0.0136 M 14 = pH + pOH
◦Kb = 3.7 x 10-3 14 = pH + 0.93
pH = 13.07
Fifth, solve for pH.
39 NeutralizationWhen acids and bases react, they will
create a neutralization reaction. ◦This is because the pH begins to return to 7
(neutral)Neutralization reactions are essentially
double replacement reactions in which the products are always a salt and water.◦A salt does not refer to NaCl but rather the
product of an acid/base reaction.
Acid + Base Salt + Water
Two Examples
HCl + NaOH H2O + NaCl
2HBr + Ca(OH)2 2H2O + CaBr2
Acid Base Water Salt
Remember to balance the equations
The rest of class is designated to work on your homework packet.
Remember, we have an open-note quiz tomorrow.