gen Chem page 7 Flashcards
auto ionization of water
Kw= [H+] [OH-] = 10^-14 at 298 K
pkw=
14
pH=
pH=-log [H+]
= log 1/[H+]
[H+]=
10^ -pH
pH + pOH=
14
Ka=
acid diss constant Ka= [H+] [A-]/ [HA]
ka= x^2/M acid initial
Kb=
base diss constant Kb= [OH-] [B+]/[BOH]
Kb= x^2/ M base initial
ka x kb at 298K =
Kw
equivalence point =
NV=NV
henderson hasselbalch equation acid buffer
pH= pKa + log [A-]/[HA]
basic buffer
pOh= pKb + log [B-]/[BOH]
if the ratio the A-: HA or B-: BOH stays the same, but still changes
the numbers don’t change but the buffering capacity does change
strong acid
small pKa, large Ka
strong base
small pKb, large Kb
ph= 7 is neutral
only at 298K
titrant
known
titrand
unknown
the Ka of the acid must be ___ Ka of the indicator
greater (more reactive)
Strong acids
HX (except HF), H2SO4, HClO4, HNO3, H3O+
weak base ex
NH3
arrhenius acid
dissociates into H+
arrhenius base
dissociates into OH-
bronstead lowry acid
donates H+
bronstead lowry base
accepts H+
lewis acid
accepts “lectrons”
lewis base
donates electrons
equivalence point
where acid= base
the part that rises sharply on the curve
half equiv point is the
buffer region where pH=pka because the ration of acid to conjugate base is equal
