Acid-Base Chemistry
Download
1 / 37

Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H + ions - PowerPoint PPT Presentation


  • 105 Views
  • Uploaded on

Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H + ions Arrhenius base: Substance that dissolves in water and provides OH - ions Examples: HCl  H + and Cl - Acid NaOH  Na + + OH - Base.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H + ions' - zhen


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Acid-Base Chemistry

Arrhenius acid: Substance that dissolves in water and provides H+ ions

Arrhenius base: Substance that dissolves in water and provides OH- ions

Examples: HCl  H+ and Cl-Acid

NaOH  Na+ + OH- Base


Bronsted Acid: Substance that donates proton to another substance

Bronsted base: Substance that accepts proton from another substance

Example: HCl + H2O  H3O+ + Cl-

HCl acts as acid; H2O acts as base

In the Reverse Reaction,

H3O+ acts as an acid; Cl- acts as a base

Note: (H3O+ = hydronium ion = H+ = proton)


HCl

  • A diprotic acid contains two acidic protons.

H2SO4

  • A triprotic acid contains three acidic protons.

H3PO4

  • A Brønsted–Lowry acid may be neutral or it may

  • carry a net positive or negative charge.

HCl, H3O+, HSO4−


  • Conjugate acid: Species formed after base accepts a proton

  • Conjugate base: Species remaining after an acid donates its proton

  • Conjugate acid-base pair: an acid and base on opposite sides of the equation that correspond to each other

  • Examples: HNO3 + H2O H3O+ + NO3-

  • acid base acid base

    • Conjugate pairs: HNO3 and NO3-

  • H2O and H3O+


Example: HS- + H2O  H3O+ + S2-

Conjugate pairs: HS- and S2-

H2O and H3O+

Practice: HClO4 + H2O  H3O+ + ClO4-

What are the conjugate pairs?

HClO4 and ClO4-

H2O and H3O+


gain of H+

H2O

H

Br

+

Br−

H3O+

+

acid

acid

base

base

loss of H+

  • HBr and Br− are a conjugate acid–base pair.

  • H2O and H3O+ are a conjugate acid–base pair.

Note: The net charge must be the same on both sides

of the equation.


Water can act as both an acid and a base (amphiprotic)!

HClO4 + H2O  H3O+ + ClO4 (base)

NH3 + H2O OH- + NH4+ (acid)

Strengths of Acids and Bases

Strong acids/bases: dissociate completely when dissolved in solution

Weak acids/bases: dissociate only partially when dissolved in solution


Examples:

Strong Acid: HCl  H+ + Cl- (100% dissociation)

Strong Base: NaOH Na+ + OH (100% dissociation)

Weak Acid: CH3COOH  H+ + CH3COO- (1.3% dissociation)

Weak Base: NH3 + H+ NH4+



Naming Acids

Binary Acids: hydo + root of anion + ic + “acid”

ex. HCl hydrochloric acid, HBr hydrobromic acid

HI

Polyatomic-based Acids: root of polyatomic ion + ic + “acid”

ex. H2SO4 sulfuric acid, H3PO4 phosphoric acid

H2CO3

HNO3

Hydroiodic acid

carbonic acid

nitric acid


The Self-Ionization of Water

H2O + H2O  H3O+ + OH-

Pure water: [H3O+]=[OH- = 10-7 M (at 250C)

Neutral Solution: Any solution in which the concentrations of H3O+ and OH- ions are equal (10-7 M)

Acidic Solution: Solutions having a greater concentration ofH3O+ thanOH- ions ([H3O+] greater than 10-7 M)

Example: A solution with [H3O+] = 10-5 M

Basic Solution: solution having a greater concentration of OH- thanH3O+ions ([H3O+] less than 10-7 M)

Example: A solution with [H3O+] = 10-12 M


  • The pH Scale

    • pH is a measure of acidity

    • Scale ranges from 0-14

  • pH = 7 Neutral

  • pH < 7 Acidic

  • pH > 7 Basic

  • pH represents the concentration of H+ ions in solution

  • Pure water: 1 x 10-7 moles H+ per liter and1 x 10-7 moles OH- per liter


Solutions with equal concentrations of and ions are called Neutral

Solutions with more than 1 x 10-7 moles H+ per liter are Acidic

Solutions with less than 1 x 10-7 moles H+ per liter are Basic

Note: [H+] x [OH-] = 10-14 (always!)


  • pH Scale Summary

    • pH scale refers to amount of H+ ions in solution

    • pH 7 is neutral, less than 7 is acidic, greater than 7 is basic

    • Lower pH = more acidic = more H+ ions

    • Higher pH = more basic = less H+ ions


Each pH unit represents a 10-fold change in H+ ion concentration!

pH 4 has 10 times more H+ ions than pH 5

pH 9 has 10 times fewer H+ ions than pH 8

Mathematical equation for pH:

pH is the negative log of the H3O+ concentration

pH = -log [H3O+]


pH = -log [H3O+]

Any number can be expressed as 10 raised to some exponent: y = 10x

Examples: 100 = 10 2

1000 = 103

0.10 = 10 -1

The log is that exponent!

100 = 10 2; Log of 100 =2

1000 = 103; Log of 1000 = 3

0.10 = 10 –1; Log of 0.10 = -1



  • Calculating pH from [H our calculators for this.3O+]

  • pH = -log [H3O+]

    • Enter [H3O+] into calculator

    • Press the “log” key

    • Change the sign

  • Example: [H3O+] = 1.0 x 10-7 M

  • pH = -log [H3O+]

  • pH = -log [1 x 10-7] = 7


  • Example: our calculators for this. [H3O+] = 1 x 10-11M

  • pH = -log [H3O+]

    • pH = -log [1 x 10-11] = 11

  • Example: [H3O+] = 1 x 10-3 M

  • pH = -log [H3O+]

  • pH = -log [1 x 10-3] = 3


  • Example: our calculators for this. [H3O+] = 4.2 x 10-5

  • pH = -log [H3O+]

    • Enter [H3O+] into calculator (4.2 x 10-5)

    • Press the “log” key (-4.3767507)

    • Change the sign (4.3767507)

      • pH = 4.3767507 = 4.4


  • Example: our calculators for this. [H3O+] = 8.1 x 10-9

  • pH = -log [H3O+]

    • Enter [H3O+]into calculator (8.1 x 10-9)

    • Press the “log” key (-8.091515)

    • Change the sign (8.091515)

      • pH = 8.091515= 8.1


Reactions Between Acids and Bases our calculators for this.

Neutralization: reaction between an acid and a base; always produces salt and water

Example: Write a balanced equation for the reaction of hydrochloric acid with sodium hydroxide.

HCl(aq) + NaOH(aq)

H—OH(l) + NaCl(aq)


Example: our calculators for this.Write a balanced equation for the reaction of hydrochloric acid with magnesium hydroxide.

2

HCl + Mg(OH)2

2

H2O +

MgCl

2


Titration
Titration our calculators for this.

Titration:a technique used to determine the concentration of an acid

or base in a solution

  • If we want to know the concentration of an acid

  • solution, a base of known concentration is

  • added slowly until the acid is neutralized.

  • When the acid is neutralized:

# of moles of acid = # of moles of base

  • This is called the end point of the titration.


Acid base titration
Acid-Base Titration our calculators for this.

  • Titration is a laboratory procedure used to determine the molarity of an acid.

  • In a titration, a base such as NaOH is added to a specific volume of an acid.

Base (NaOH)

Acid

solution




Determining an unknown molarity from titration data flask.

requires three operations:

mole–mole

conversion

factor

[2]

Moles of

base

Moles of

acid

M (mol/L)

conversion

factor

M (mol/L)

conversion

factor

[3]

[1]

Volume of

base solution

Molarity of

acid solution


HOW TO Determine the Molarity of an Acid Solution flask.

from Titration

Example: What is the molarity of an HCl solution if 22.5 mL of a 0.100 M NaOH solution are needed to titrate a 25.0 mL sample of the acid?

volume of base (NaOH)

22.5 mL

volume of acid (HCl)

25.0 mL

conc. of acid (HCl)

?

conc. of base (NaOH)

0.100 M


Determine the number of moles of base flask.

used to neutralize the acid.

Step [1]

M (mol/L)

conversion factor

Volume of

base solution

1 L

1000 mL

0.100 mol NaOH

1 L

x

x

=

22.5 mL NaOH

mL–L

conversion factor

0.00225 mol NaOH


Determine the number of moles of acid that react from the balanced chemical equation.

Step [2]

HCl(aq) + NaOH(aq)

H2O(l) + NaCl(aq)

1 mol HCl

1 mol NaOH

x

0.00225 mol NaOH

0.00225 mol HCl

=

mole–mole

conversion

factor


Determine the molarity of the acid from the number of moles and the known volume.

Step [3]

mol

L

1000 mL

1 L

0.00225 mol HCl

25.0 mL solution

M

=

=

x

mL–L

conversion factor

=

0.0900 M HCl

Answer


Buffers
Buffers and the known volume.

Buffer: a solution whose pH changes very little when acid or base is added.

Most buffers are solutions composed of roughly

equal amounts of

  • a weak acid

  • the salt of its conjugate base

The buffer resists change in pH because

  • added base, −OH, reacts with the weak acid

  • added acid, H3O+, reacts with the conjugate base


  • Buffers contain 2 compounds: and the known volume.

    • Compound with the ability to react with H+ ions

    • Compound with the ability to react with OH- ions

  • Example: HCO3- + H+ H2CO3

  • If acids (H+) are added, react with HCO3-

  • H2CO3 + OH-  HCO3- + H2O

  • If OH- ions are added, react with H2CO3

  • H2CO3 is unstable: H2CO3 H2O + CO2


More examples of a buffer
More Examples of a Buffer and the known volume.

If an acid is added to the following buffer equilibrium,

Adding more

product…

CH3COOH + H2O

H3O+ + CH3COO−

conjugate

base

weak acid

…drives the reaction to the left.

then the excess acid reacts with the conjugate base,

so the overall pH does not change much.


If a and the known volume.base is added to the following buffer equilibrium,

Adding more

reactant…

H2O + CH3COO−

CH3COOH + −OH

conjugate

base

weak acid

…drives the reaction to the right.

then the excess base reacts with the weak acid, so

the overall pH does not change much.


ad