HTML University
Science Dept.
Chemistry CHE 100 Notes
R. Acids and Bases
- Arrhenius Definitions
- Acid: yields H1+ in solution
- Base: yields OH1- in solution
- Dissociation
- HCl ----> H1+ + Cl1-
- H2SO4 ----> 2H1+ + SO42-
- HNO3 ----> H1+ + NO31-
- HC2H3O2 ----> H1+ + C2H3O21-
- NaOH ----> Na1+ + OH1-
- NaCl ----> Na1+ + Cl1-
- Bronsted-Lowry Definition
- Acid: Proton Donor
- Base: Proton Acceptor
- Partial Dissociation
- First Dissociation: H2CO3 ---->
H1+ + HCO31-
- Second Dissociation: HCO31- ---->
H1+ + CO32-
- What if: HCO31- +
H1+ ----> H2CO3 ?
- Properties of Acids and Bases
| Property
| ACID
| BASE
|
| Litmus
| blue to red
| red to blue
|
| Taste
| sour
| bitter
|
| Neutralize
| base
| acid
|
| Feel
| N/A
| soapy or slick
|
- Titration
- Neutralization: H1+ + OH1- ----> H2O
- Determine the concentration of an acid
(or base) with a known base (or acid)
- Add a known concentration of an acid to an unknown
concentration of a base
until neutralization or an endpoint is reached
- Indicators
- Methyl Orange: pH 4 Red in acid: yellow in a base
- Methy Red: pH 5 Red in acid; Yellow in a base
- Phenolphthalein: at about pH 8.3 it is colorless: Red in a base
- Calculation
- (ml of known titrant) (N of known titrant) =
(ml of sample) (N of sample)
- Therefore: millimole of known titrant =
millimole of unknown sample
- N.B.: Rearranging equation to solve for unknown:
(ml titrant) x (N titrant)/(ml sample) = (N sample)
- Dissociation of Water
- HOH ----> H1+ + OH1-
- At 25oC the product of the H1+
and OH1- concentrations is 1 x 10-14
= Kw
- Kw at 25oC = [H1+]
[OH1-] = 1 x 10-14
- Values in brackets mean concentration in Molarity =
moles/liter
- pH
- For neutral water [H1+] = [OH1-]
- Therefore the Square Root of 1 x 10-14 =
1 x 10-7 =
[H1+] = [OH1-]
- pH = neg log of the H1+ ion concentration =
-log [H1+]
- pOH = -log [OH1-]
- N.B.: pH + pOH = 14 at 25oC
- Acids
- pH acid: 0-7
- pOH acid: 14-7
- Bases
- pH base: 7-14
- pOH base: 7-0
- Example
- Give the pH of an acid when its concentration is
[H1+] = 1 x 10-6 M
- pH = 6 and pOH = 8
- Example
- Give the pH of an acid when its concentration is
[H1+] = 1 x 10-3 M
- pH = 3; The [OH1-] = 1 x 10-11 AND pOH = 11
- Example
- Give the pH of an acid when its concentration is
[H1+] = 3.00 x 10-2
- Enter scientific notation 3.00 x 10-2
- Press log and it should show -1.52
- Change the sign. Therefore the pH = 1.52 and the pOH = 12.48
- Electrolytes
- Electrolytes: whose aqueous solutions conduct an electric
current either strong or weak
- Nonelectrolytes: whose aqueous solutions do not conduct an electric
current
- Dissociation Equations
- Example: NaCl(aq) ---->
Na1+(aq) + Cl1-(aq)
- Example: NaOH(aq) + HCl(aq) ---->
NaCl(aq) + H2O
- Then: Na1+(aq) + OH1-(aq) +
H1+(aq) + Cl1-(aq) ---->
Na1+(aq) + Cl1-(aq) + H2O
- Finally: H1+(aq) + OH1-(aq) ---->
H2O
- Some Guidelines
- Write and balance equations in regular form
- Then change to ionic
- Molecular form to be kept
- Carefully express polyatomic ions
- Ions appearing on both sides of the equation can be crossed off
- Self Check
- Regular form: AgNO3(aq) + NaCl(aq)---->
AgCl(ppt) + NaNO3(aq)
- Ionic:
- Then: