Warmup (5 min)

1. Solve: log(3.9 x 10-4) =2. Name the compound HI and show

how it dissociates in solution.3. How do acids and bases differ?

Write down WHATEVER you remember.

Acids, Bases, and pH

Acid or Base? Aqueous solution Have a high pH (above 7-14) Have a low pH (below 0-7) Feel slippery Taste sour Conduct electricity Changes the color of pH paper Corrosive: can damage skin Neutralize before disposal

Both Base Acid Base Acid Both Both Both Both

Acids: produce H+ ions in water, forming H3O+ (Arrhenius)

or donate H+ ions to bases (BL)

HNO3

HCl

H2SO4

H3C6H5O7

H3PO4

HClO

Bases: produce OH- in solution (Arrhenius)

OR accept H+ from compounds (BL)

Mg(OH)2

KOH

CaCO3

NaHCO3

NH3

Any anion

Acidic solution: mostly H2O, [H3O+] > [OH-]

Neutral solution: mostly H2O, [H3O+] = [OH-]

Basic solution: mostly H2O, [H3O+] < [OH-]

Water molecules undergo autoionization and split into ions spontaneously:

HOH + HOH H3O+ + OH-

Count each particle and identify each solution as neutral, basic, or acidic

H2O OH- H2O H2O H3O+ H2O H2O OH- H2O H3O+

H2O H2O H3O+ H2O OH- H2O OH- H2O H2O H3O+

H2O H2O OH- H2O H3O+ H2O H2O H3O+ H2O OH-

H2O OH- H2O H2O H3O+ H2O H2O H3O+ H2O H3O+ H2O H2O H3O+ H2O H2O OH- H2O H2O H3O+ H2O H2O H3O+ H2O H2O H2O H2O H3O+ H2O

H2O OH- H2O H2O H3O+

H2O H2O H3O+ H2O OH-

H2O H2O OH- H2O H2O OH- H2O H2O H3O+ H2O H2O OH- H2O H2O H2O H2O H3O+ H2O

Strong Acids & Bases: (far away from pH 7)

NaOH → Na+ + OH- HCl → H+ + Cl-

- completely dissociate in waterWeak Acids & Bases: (closer to pH 7)

- partially dissociate in water

The pH scale: based on [H+] or [H3O+] Water (pH = 7) is neutral

Naming and Dissociation Practice:Fill in the blanks!

Formula

HI

HClO2

HC2H3O2

H2Se

HClO3

Dissociation H+ + I-

H+ + ClO2-

H+ + C2H3O2–

H+ + HSe-

H+ + ClO3-

Namehydroiodic acid

chlorous acidacetic acidhydroselenic

acidchloric acid

Write the complete ion equation and the net ionic equation for the neutralization of nitric acid (strong acid) by sodium hydroxide (strong base).

HNO3(aq) + NaOH(aq) H2O(l) + NaNO3(aq)

H++ NO3- + Na+ + OH- H2O(l) + Na+ + NO3

-

Get rid of everything that doesn’t change phase or compound from one side to another!

H+ (aq) + OH-(aq) H2O(l)

If the acid and base are both strong, the net ionic equation will be the same every time.

Neutralizationacid + base = salt + water

Write the molecular and net ionic equation when lithium hydroxide (strong base) is mixed with carbonic acid (weak acid).

LiOH(aq) + H2CO3(aq)

H2O(l) + Li2CO3 (aq)

Leave weak or insoluble things together. Separate strong or soluble things.

Li+ + OH- + H2CO3 H2O + Li+ + CO32-

OH-(aq) + H2CO3(aq) H2O(l) + CO32-(aq)

2

2

2 2

Write the molecular and net ionic equations when magnesium hydroxide(weak) is mixed with chlorous acid (weak).

Mg(OH)2(aq) + HClO2(aq)

H2O(l) + Mg(ClO2)2(aq)

*net ionic is the same!

Write the molecular and net ionic equations when aluminum hydroxide(weak) is mixed with sulfuric acid (strong).

Al(OH)3(aq) + H2SO4(aq)

Al2(SO4)3(aq) + H2O(l)

Al(OH)3(aq) + H+(aq) Al3+(aq) + H2O(l)

2

2 36

2

33

HYDROCATCH

The ball represents a H+

“Bronsted-Lowry (BL) Theory”

ConjugatesWrite the conjugate acid of F-

H+ + F- ↔ HF

Conjugate acid: formed after a reaction when a base gains a H+

Write the conjugate base of H2SO4

Conjugate base: formed after a reaction when an acid loses a H+

H2SO4 ↔ HSO4- + H+

HCl + H2O H3O+ + Cl-

gained H+

lost H+

Acid Conjugatebase

ConjugateacidBase

Let’s identify the acid and base, and their conjugates in the products

H2O and Cl- are electron pair DONORS in this reaction, so they are both “Lewis bases”

The H in HCl and H3O+ is an electron pair ACCEPTOR in this reaction, so they are both “Lewis acids”

NH3 + HOH NH4+ + OH-

gained H+

lost H+

Acid Conjugatebase

ConjugateacidBase

H2CO3 + H2O HCO3- + H3O+

gained H+

lost H+

Acid Conjugatebase

ConjugateacidBase

Calculating pH*let’s practice some logs…

Formulas :

pH = -log[H+]pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

-log 3.000 = -

0.4771

-log(3.9 x 10-

4) =3.4

10-5.6 = 2.5 x

10-6

1. A tap water sample is contaminated with acid! If the [H3O+] in the sample is 8.90 x 10-3 M, calculate the pH of the water.

Remember that [H3O+] isthe same thing as the [H+]pH = -log[H3O+]pH = -log(8.90 x 10-3)pH = 2.05 (pH has no units)

pH = -log[H+]pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

2. Most tap water samples are slightly basic. If the pH of a sample = 7.9, calculate the [H+] in the sample.

[H+] = 10-pH

[H+] = 10-7.9

[H+] = 1.3 x 10-8 MUnits for concentration are in M, “molar”

pH = -log[H+]pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

3. What is the [OH-] of a solution that has a pOH of 3.00?

[OH-] = 10-pOH

[OH-] = 10-3.00

[OH-] = 1.00 x 10-3 M

4. What is the pH of this solution?

pOH + pH = 14.00

3.00 + pH = 14.00

pH = 11.00

pH = -log[H+]pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

5. Calculate the [H3O+] AND [OH-] of human blood (pH =7.40)

[H3O+] = 10-pH

[H3O+] = 10-7.40

[H3O+] = 3.98 x 10-8 M

To find [OH-], use

pOH + pH = 14.00

pOH + 7.40 = 14.00

pOH = 6.60

pH = -log[H+]pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

then [OH-] = 10-pOH

[OH-] = 10-6.60

[OH-] = 2.51 x 10-7 M

pOH = -log[OH-]

pOH = -log(9.35 x 10-4)

pOH = 3.03

pOH + pH = 14.00

pH = 10.97

[H+] = 10-pH

[H+] = 10-10.97

[H+]= 1.07 x 10-11 M

6. What is the [H+] of a solution that has a [OH-] of 9.35 x 10-4 M?

pH = -log[H+]pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

Start on Problem Patent prelab. You may begin brainstorming with your group.