unit 12: acid and bases

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] = 0.5M◦Ka = 7.2 x 10-4

First, write the Ka equation.

Example



◦[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] = 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] = 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


(aq) + 2OH-(aq)


◦[Ca(OH)2] = 0.05 M

◦[Ca+2] = 0.0136 M◦Kb = 3.7 x 10-3

First, write the Kb equation.


(aq) + 2OH-(aq)


◦[Ca(OH)2] = 0.05 M

◦[Ca+2] = 0.0136 M◦Kb = 3.7 x 10-3

Second, fill in the numbers.


(aq) + 2OH-(aq)


◦ [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.


(aq) + 2OH-(aq)


◦[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.


(aq) + 2OH-(aq)


◦ [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.

unit 12: acid and bases

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