ch 10 - Acids and Bases Flashcards
Arrhenius acid
dissociates to form an excess of H+ in solution
Arrhenius base
dissociates to form an excess of OH- in solution
Bronsted-Lowry acid
species that donates a hydrogen ion (H+)
Bronsted-Lowry base
species that accepts hydrogen ion (H+)
conjugate acid-base pairs
Bronsted-Lowry acids and bases occur in pairs becaues the definitions require transfer of a proton from the acid to the base
Lewis acid
an electron pair acceptor
Lewis base
an electron pair donor
other terms for Lewis acid-base chemistry
coordinate covalent bond formation; complex ion formation; nucleophile-electrophile interactions
amphoteric
species that reacts like an acid in a base environment and like a base in an acidic environment
amphprotic
a species that can either gain or lose a proton (Bronsted-Lowry)
anion acid nomenclature
acids formed from anions with names that end in -ide have the prefix hydro- and the ending -ic: F- (fluoride) = HF (Hydrofluoric acid); Cl- (Chloride) = HCl (hydrochloric acid); Br- (bromide) = HBr (hydrobromic acid)
naming oxyacids
oxyacids are acids formed from oxyanions; if anion ends in -ite (less oxygen), acid will end with -ous acid; if it ends in -ate (more oxygen), acid will end with -ic acid and prefixes of names are retained
autoionization
water reacted with itself: H20 (l) + H20 (l) ->
water dissociation constant (K sub w)
Kw = [H3O+][OH-] = 10^-14 at 25 degrees C (298 K); at temps above this, Kw will increase as a result of the endothermic nature of the autoionization reaction
p scale
negative logarithm of the number of items: pH and pOH are prototypical examples
pH
-log[H+] = log (1/[H+])
pOH
-log[OH-] = log (1/[OH-]
pH and pOH for aqueous solutions at 298K
pH + pOH = 14; water at equilibrium and 25 degrees C has a concentration of hydroxide ions (10^-7) = to concentration of hydrogen ions: pH of 7 and pOH of 7
how to multiply logs
log (xy) = log x + log y
shortcut to determine p scale values
if the nonlog value is written in proper scientific notation, it will be in the form n x 10^-m where n = number between 1 and 10: -log(n x 10^-m) = -log (n) - log(10^-m) = m - log(n); n will equal number between 1 and 10 which means log n will be a number between 0 and 1 (closer to 1 = closer to 0; closer to 10 = closer to 1) so p value = about m - 0.n where 0.n represents sliding the decimal point of n one position to the left