acids & bases

32
Acids & Bases

Upload: ulfah

Post on 23-Jan-2016

27 views

Category:

Documents


0 download

DESCRIPTION

Acids & Bases. 1. Properties of Acids and Bases :. 2. Indicators : Turn 1 color in an acid and another color in a base. A. Litmus Paper: An aci D turns blue litmus paper reD A B ase turns red litmus paper Blue . - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Acids & Bases

Acids & Bases

Page 2: Acids & Bases

1. Properties of Acids and Bases:

Taste Touch Reactions with Metals

Electrical Conductivity

Acid sour

looks like water, burns, stings

Yes-produces

H2 gas

electrolyte in solution

Base bitter

looks like water, feels

slippery

No Reaction

electrolyte in solution

Page 3: Acids & Bases

2. Indicators: Turn 1 color in an acid and

another color in a base.A. Litmus Paper: An aciD turns blue litmus paper reD

A Base turns red litmus paper Blue.

B. Phenolphthalein: colorless in an acid/pink in a base

C. pH paper: range of colors from acid to basic

D. pH meter: measures the concentration of H+ in

solution

Page 4: Acids & Bases

3. Neutralization: A reaction between an acid and base. When an acid and base neutralize, salts form.

Acid + Base → Salt + Water

Ex) HCl + NaOH → NaCl + HOH

Page 5: Acids & Bases

4. Arrhenius Definition:

A. An acid dissociates in water to produce

hydrogen ions, H+.

B. A base dissociates in water to produce hydroxide ions, OH-.

C. Problems with Definition:

• Restricts acids and bases to water solutions.

• Oversimplifies what happens when acids

dissolve in water.

• Does not include certain compounds that have

characteristic properties of acids & bases.

Ex) NH3 (ammonia) doesn’t fit

Page 6: Acids & Bases

5. Bronsted-Lowry Definition:A. An acid is a substance that can donate hydrogen ions. Ex) HCl → H+ + Cl-

B. A base is a substance that can accept

hydrogen ions. Ex) NH3 + H+ → NH4+

C. Advantages of Bronsted-Lowry Definition•Acids and bases are defined independently of how

they behave in water.•Focuses solely on hydrogen ions.

D. Hydrogen ion is the equivalent of a proton. Therefore, acids are often called proton donors and bases are called proton acceptors.

Page 7: Acids & Bases
Page 8: Acids & Bases

6. Hydronium Ion:

A. Hydronium Ion – H3O+ This is a complex that forms in water.

B. To more accurately portray the Bronsted-Lowry, the hydronium ion is used instead of the hydrogen ion.

C. Amphoteric: A substance that can act as either an acid or a base.

Ex.) Water - can gain or lose a H+

Page 9: Acids & Bases

7. Conjugate Acid-Base Pairs:

A pair of compounds that differ by only one hydrogen ion

A. When an acid loses a hydrogen ion, it

becomes its conjugate base.

B. When an base gains a hydrogen ion, it

becomes its conjugate acid.

Page 10: Acids & Bases

Acid (A), Base (B), Conjugate Acid (CA), Conjugate Base (CB)

NH3 + H2O ↔ NH4+ + OH-

HCl + H2O ↔ Cl- + H3O+

• A strong acid will have a weak conjugate base.• A strong base will have a weak conjugate acid.

B

B

A

A

CA CB

CB CA

Page 11: Acids & Bases

STRONG Acid/Base

C. A strong acid or base will completely dissociate (break apart) in water. This is represented by a single () arrow.

HNO3 H+ + NO3-

NaOH Na+ + OH-

Page 12: Acids & Bases

D. A weak acid or base will partially dissociate in water. This is represented by a double (↔) arrow.

CH3COOH ↔ H+ + CH3COO-

NH4OH ↔ NH4+ + OH-

WEAK Acid/Base

Page 13: Acids & Bases

8. Naming Acids Review: A. Binary – H +one anion Prefix “hydro”+ anion name +“ic”acid

Ex) HCl hydrochloric acid

Ex) H3P hydrophosphoric acid

B. Tertiary – H + polyatomic anion no Prefix “hydro”

(oxo) end “ate” = “ic” acidend “ite” = “ous” acid

Ex) H2SO4 sulfuric acid

Ex) H2SO3 sulfurous acid

Page 14: Acids & Bases

15-2 The Self-ionization of Water and pH

1. Water is amphoteric, it acts as both an acid and a base in the same reaction.

Ex) H2O(l) + H2O(l) ↔ H3O+(aq) + OH-

(aq)

2. In pure water at 25C, both hydronium ions and hydroxide ions are found at concentrations of 1 x 10-7 M. Because they are at equilibrium and you do not include liquid water in the equilibrium expression, Keq or Kw (water) can be expressed as follows:

A. Formula for Kw = [H3O+] [OH-]

Page 15: Acids & Bases

A. Formula for Kw = [H3O+] [OH-] 1.0 x 10-14 M = [1.0 x 10-7 M] [1.0x10-7 M]

1.0 x 10-14 M = [H3O+] [OH-]

B. Using Kw in calculations: If the concentration of H3O+ in the blood is 4.0 x 10-8 M, what is the concentration of OH ions in the blood? Is blood acidic, basic or neutral?

Kw = [H3O+] [OH-]

1.0 x 10-14 M = [4.0 x 10-8 M] [OH-] 4.0 x 10-8 M 4.0 x 10-8 M

2.5 x 10-7 M = [OH-]

slightly basic

Page 16: Acids & Bases

3. The pH scale:A. Used to determine if something is an acid or

a base. A way to express H3O+ concentration based on logarithms. pH changes by a factor of 10.

Ex) 10,000 = 104 therefore log 10,000 = 4

0.001 = 10-3 therefore log 0.001 = -3

B. pH 1-6.9: acid

pH 7.1-14: base

pH of 7.0: neutral

Page 17: Acids & Bases
Page 18: Acids & Bases

D. pH = -log [H3O+]

E. [H3O+] [OH-] = 1.0 x 10-14M

F. pH + pOH = 14

H+

OH-

Page 19: Acids & Bases
Page 20: Acids & Bases

H+

OH

-

[H3O+] [H3O+] [OH-] [OH-]

1x10-14 .00000000000001 1x100 1

1x10-13 .0000000000001 1x10-1 .1

1x10-12 .000000000001 1x10-2 .01

1x10-11 .00000000001 1x10-3 .001

1x10-10 .0000000001 1x10-4 .0001

1x10-9 .000000001 1x10-5 .00001

1x10-8 .00000001 1x10-6 .000001

1x10-7 .0000001 1x10-7 .0000001

1x10-6 .000001 1x10-8 .00000001

1x10-5 .00001 1x10-9 .000000001

1x10-4 .0001 1x10-10 .0000000001

1x10-3 .001 1x10-11 .00000000001

1x10-2 .01 1x10-12 .000000000001

1x10-1 .1 1x10-13 .0000000000001

pH1413121110987654321

EQUAL

Page 21: Acids & Bases
Page 22: Acids & Bases

H. Significant Digits Rule

The number of digits AFTER THE DECIMAL POINT in your answer should be equal to the number of significant digits in your original number.

Ex -log[8.7 x 10-4 M] –Calc Answer = 3.0604807474 –Sig Fig pH = 3.06

Page 23: Acids & Bases

Example #1: [H3O+] = 7.3 x 10-5 M What is the pH value? Is it an acid, base or neutral?

Which equation should you use?

Kw = [H3O+] [OH-]OR

pH = -log [H3O+]

pH = -log [7.3 x 10-5M] pH = 4.14 ACID

Page 24: Acids & Bases

Example #2: [OH-] = 5.0 x 10-2 M What is the pH value? Is it an acid, base or neutral?

Which equation should you use?

Kw = [H3O+] [OH-]

ORpH = -log [H3O+]

Kw = [H3O+] [OH-]

1 x 10-14M = [H3O+] [5.0 x 10-2M]

2.0 x 10-13M = [H3O+] pH = -log [2.0 x 10-13M]

pH = 12.70basic

Page 25: Acids & Bases

Ionization of Acids & Bases• H2SO4 2 H+ + SO4

-2 – Sulfuric acid donates 2 H+ ions per mole– This is called a “diprotic” acid

• H3PO3 – Phosphorous acid donates 3 H+ ions per

mole. This is called a “triprotic acid”

• Ca(OH)2 – Calcium hydroxide dissociates into 2 OH-

ions per mole

3 H+ + PO3-3

Ca+2 + 2 OH-1

Page 26: Acids & Bases

15-3 Acid-Base Titration1. An acid-base titration is a carefully controlled

neutralization reaction which can determine the concentration [ ] of an unknown solution.

2. To determine the concentration of an unknown substance, a standard solution is needed. This solution has a known concentration.

3. Titration curve: graph that

shows how pH changes

during a titration.

Page 27: Acids & Bases

4. An indicator, usually phenolphthalein, is used in a titration.

5. The point at which enough standard solution is added to neutralize the unknown solution is called the equivalence point.

6. The point at which the indicator changes color is called the endpoint.

Page 28: Acids & Bases
Page 29: Acids & Bases

7. Therefore: [H+] = [OH-] at the equivalence point (which is usually the endpoint)

Volume (acid) Conc.(acid) = Volume (base) Conc.(base)

VaMa = VbMb

(L) x Moles = (L) x Moles

Liter Liter

moles of H+ = moles of OH-

Page 30: Acids & Bases

VaMa = VbMb Ex #1) Solutions of sodium hydroxide are used to unclog

drains. A 43.0 mL volume of sodium hydroxide was titrated with 32.0 mL of 0.100 M HCl. What is the molarity of the sodium hydroxide solution?

HCl (a) = NaOH (b)

(32.0 mL)(0.100 M HCl) = (Mb)(43.0 mL)

43.0 mL 43.0 mL

0.0744 M = Mb

Page 31: Acids & Bases

Ex #2) A volume of 25.0 mL of 0.120 M sulfuric acid neutralizes 40.0 mL of a sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?

H2SO4 + 2NaOH Na2SO4 + 2H-OH

Note: sulfuric acid has 2 H+ per mole of acid

Therefore you need to multiply the acid side by a factor of 2

VaMa x 2 = VbMb

(25.0 mL) (0.120 M) x 2 = (40.0 mL) x Mb

Mb = 0.150 M

Page 32: Acids & Bases

Ex #3) 24.9 mL of 2.88 M calcium hydroxide completely neutralizes 38.9 mL of a hydrobromic acid solution. What is the molarity of the hyrdobromic acid?

Ca(OH)2 + 2HBr CaBr2 + 2HOHNote: there are 2 hydroxide ions per mole of base

Therefore you need to multiply the base side by a factor of 2

VaMa = VbMb x 2 (38.9 mL) x Ma = (24.9 mL) (2.88 M) x 2

Ma = 3.69 M