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Solve: log(3.9 x 10 -4 ) = Name the compound HI and show how it dissociates in solution. How do acids and bases differ? Write down WHATEVER you remember. Warmup (5 min). Acids, Bases, and pH. Acid or Base?. Both Base Acid Base Acid Both Both Both Both. Aqueous solution

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Warmup (5 min)


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warmup 5 min

Solve: log(3.9 x 10-4) =

  • Name the compound HI and show how it dissociates in solution.
  • How do acids and bases differ? Write down WHATEVER you remember.

Warmup (5 min)

acid or base
Acid or Base?
  • Both
  • Base
  • Acid
  • Base
  • Acid
  • Both
  • Both
  • Both
  • Both
  • 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
slide4

Bases: produce OH- in solution OR accept H+ from compounds

Mg(OH)2

KOH

CaCO3

NaHCO3

NH3

Any anion

Acids: produce or donate H+ ions in water (forming H3O+)

HNO3

HCl

H2SO4

H3C6H5O7

H3PO4

HClO

slide5

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

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

H2O OH- H2OH2O H3O+ H2O H2O OH- H2O H3O+ H2O H2O H3O+ H2O OH- H2O OH- H2OH2O H3O+ H2O H2O OH- H2O H3O+ H2O H2O H3O+ H2O OH-

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

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

H2O OH- H2OH2O H3O+ H2O H2O H3O+ H2O H3O+ H2O H2O H3O+ H2O H2O OH- H2OH2O H3O+ H2O H2O H3O+ H2O H2OH2O H2O H3O+ H2O

H2O OH- H2OH2O H3O+ H2O H2O H3O+ H2O OH- H2O H2O OH- H2O H2O OH- H2OH2O H3O+ H2O H2OOH- H2O H2OH2O H2O H3O+ H2O

slide6

The pH scale: based on [H+] or [H3O+]

  • Water (pH = 7) is neutral

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

NaOH → Na+ + OH- HCl → H+ + Cl-

- completely dissociate in water

Weak Acids& Bases:(closer to pH 7)

- partially dissociate in water

naming and dissociation practice fill in the blanks
Naming and Dissociation Practice:Fill in the blanks!

Formula

HI

HClO2

HC2H3O2

H2Se

HClO3

Name

hydroiodic acid

chlorous acid

acetic acid

hydroselenic acid

chloric acid

  • Dissociation
  •  H+ + I-
  •  H+ + ClO2-
  • H+ + C2H3O2–
  •  H+ + HSe-
  •  H+ + ClO3-
if the acid and base are both strong the net ionic equation will be the same every time

Neutralization

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

acid + base = salt + water

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)

slide9
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

slide10

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

3

2

6

3

3

hydrocatch
HYDROCATCH

The ball represents a (H+)

conjugates
Conjugates

Write 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+

let s identify the acid and base and their conjugates in the products

lost H+

gained H+

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

HCl + H2O  H3O+ + Cl-

Conjugate

acid

Base

Acid

Conjugate

base

slide14

lost H+

gained H+

NH3 + HOH  NH4+ + OH-

Conjugate

acid

Base

Acid

Conjugate

base

slide15

lost H+

gained H+

H2CO3 + H2O HCO3- + H3O+

Conjugate

acid

Base

Conjugate

base

Acid

calculating ph let s practice some logs

-log 3.000 =

-0.4771

-log(3.9 x 10-4) =

3.4

10-5.6 =

2.5 x 10-6

Calculating pH*let’s practice some logs…

Formulas :

pH = -log[H+]

pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

slide17
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+] is

the 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

slide18
2. Most tap water samples are slightly basic. If the pH of a sample = 7.9, calculate the [H+] in the sample.

pH = -log[H+]

pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

[H+] = 10-pH

[H+] = 10-7.9

[H+] = 1.3 x 10-8 M

Units for concentration are in M, “molar”

3 what is the oh of a solution that has a poh of 3 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

slide20
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

then [OH-] = 10-pOH

[OH-] = 10-6.60

[OH-] = 2.51 x 10-7 M

pH = -log[H+]

pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00

6 what is the h of a solution that has a oh of 9 35 x 10 4 m
6. What is the [H+] of a solution that has a [OH-] of 9.35 x 10-4 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

pH = -log[H+]

pOH = -log[OH-]

[H+] = 10-pH

[OH-] = 10-pOH

pOH + pH = 14.00