Acids and Bases Flashcards
Arrhenius base
species that produces OH- in aqueous solution (hydroxide ions)
Arrhenius acid
species that produces H+ (hydrogen ions) in aqueous solution
An Arrhenius base can react with an Arrhenius acid in a/an ___________________ reaction.
neutralization
What is a neutralization reaction?
acid + base –> water + salt
Bronsted-Lowry acid
species that produces H+
True or False: Bronsted-Lowry bases and acids are the same as Arrhenius bases and acids.
FALSE: Bronsted-Lowry ACIDS are the SAME as Arrhenius ACIDS (species that produces H+)…but Bronsted-Lowry BASES are different from Arrhenius BASES.
Bronsted-Lowry Base
species that accepts H+
When a Bronsted-Lowry acid _______ a proton, it forms a conjugate base.
When a Bronsted-Lowry base _______ a proton, it forms a conjugate acid.
acid LOSES proton
base GAINS proton
Lewis acid
electron-pair acceptor
hint: lewis Acid = electron pair Acceptor
Lewis base
electron-pair donor
A Lewis acid is a/an:
a) electrophile
b) nucleophile
a) electrophile
hint: lewis Acid = electron pair Acceptor
A Lewis base is a/an:
a) electrophile
b) nucleophile
b) nucleophile
Which species is fundamental to understanding MCAT OCHEM?
a) Arrhenius
b) Bronsted-Lowry
c) Lewis
c) Lewis
True or False: a given species cannot have multiple classifications (Arrhenius, Bronsted Lowry, Lewis).
FALSE
List the strong acids (7).
HI (hydroiodic acid) HBr (hydrobromic acid) HCl (hydrochloric acid) HNO3 (nitric acid) HClO4 (perchloric acid) HClO3 (chloric acid) H2SO4 (sulfuric acid)
List the strong bases (7).
NaOH (sodium hydroxide) KOH (potassium hydroxide) NH2- (amide ion) H- (hydride ion) Ca(OH)2 (calcium hydroxide) Na2O (sodium oxide) CaO (calcium oxide)
What does the term “pH” refer to?
“potential of the hydrogen ion”
how is pH defined in terms of the proton concentration? (provide equation)
pH = -log[H3O+]
hint: p = -log[whatever]
How is pOH defined in terms of the hydroxide ion concentration? (provide equation)
pOH = -log[OH-]
hint: p = -log[whatever]
pH + pOH = ?
pH + pOH = pKw = 14 (at room temperature)
Calculate the following pH/pOH:
- log[1M] = ?
- log[1 x 10^-1 M] = ?
- log[1 x 10^-2 M] = ?
- log[1 x 10^-13 M] = ?
- log[1 x 10^-14 M] = ?
- log[1M] = 0
- log[1 x 10^-1 M] = 1
- log[1 x 10^-2 M] = 2
- log[1 x 10^-13 M] = 13
- log[1 x 10^-14 M] = 14
Calculate the following:
\+log(1) = ? \+log(3.2) = ? \+log(10) = ?
\+log(1) = 0 \+log(3.2) = 0.5 \+log(10) = 1