Summary of equations for acid and base equilibrium: pure acid, base or salt
pH = –log [H+]
[H+] = 10–pH
pH = 14 – pOH
Kw = [H+][OH–] = 1.0 x 10–14 at 25°C
Weak acid
HA <–––> H+ + A–
pOH = –log [OH–]
Ka = [H+][A–] = x2
Weak Base
B + H2O <–––> BH+ + OH–
[HA] [OH–] = 10–pOH
[HA] Kb = [BH+][OH–] = x2
(notice x = [OH–] with base, Kb )
Diprotic acid
1st step : H2A <–––> H+ + HA–
(2nd step, not important)
HA– <–––> H+ + A–2
[B] [B] Ka1 = [H+][HA–] = x2
[H2A] [H2A]
Ka2 = [H+][A–2] = [A–2]
[HA–] Salt of weak acid: (anion hydrolysis)
NaA –––> Na+ + A–
% ionized = [H+] x 100
[HA]
% ionized = [OH–] x 100
[B]
Use Ka1 to calc. pH
[H+] = [HA–]
Ka2= [A–2]
Kb = [HA][OH–] = Kw = x2
[A–] Ka [A–]
A– + H2O <–––> HA + OH–
“Dibasic salt” Na2A ––> 2 Na+ + A–2
1st step : A–2 + H2O <–––> HA– + OH–
(2nd step, not important)
HA– + H2O <–––> H2A + OH–
Kb1 = [HA–][OH–] = Kw = x2
[A–2] Ka2 [A–2]
Salt of weak base: (cation hydrolysis)
BHCl –––> BH+ + Cl–
Ka = [B][H+] =
BH+ + H2O <–––> B + H3O+
or BH+ <–––> B + H+
Salt of weak acid and weak base:
Both anion and cation hydrolyze
Salt can be acidic or basic!
Hydrogen Salt of di (or tri)protic acid
Salt can be acidic or basic!
[BH+] Ka>Kb : acidic
Kb>Ka : basic
Kw = x 2
Kb
[BH+]
Approximation: [ H + ] =
NaHA –––> Na+ + HA–
acidic:
+][A–2]
Ka2 = [H€
[HA–] (both “1st step,” both important)
HA– <–––> H+ + A–2
Approximation: [H+] = √ Ka1 Ka2
HA–
+ H2O <–––> H2A +
M+3
+ H2O <–––>
basic:
Kb2 = Kw = [H2A][OH–]
Ka1 [HA–] OH–
Metal ion hydrolysis
M(OH)+2
+
H+
Ka = [H+][M(OH)+2] =
[M+3]
KaKw
Kb
x2
[M+3]
(or √Ka2 Ka3 )
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For Buffer and titration pH problems: For mixtures of weak acid + its salt; or
Weak base + its salt, Or Weak acid+ strong base; or Weak base + strong acid
(In other words there is a source of both acid and conj base, or base and conj acid.)
So [H+] ≠ [A–] and [OH–]≠ [BH+], so equations on back don’t work!
Use this Equation: The Henderson-Hasselbalch equation
[H + ]= K a
[HA]
[Acidic form]
_ = K a [Basic form ]
[A ]
pH = pK a + log
[A _ ]
[Basic form]
= pK a + log
[HA]
[Acidic form ]
Weak acid: Acid= Acidic form, Salt = Basic form ; pKa= – log Ka
Weak base: first get Ka = Kw : Base = Basic form, Salt = Acidic form: Then use same equation!
Kb
€
Adding acids, bases to weak acid, base solutions, salts and Buffers
Use the Henderson-Hasselbalch equation:
Adding strong base : adds to the Basic form (A–) and takes away from the Acidic form (HA)
Adding strong acid : adds to the Acidic form (HA) and takes away from the Basic form (A–)
Titration problems : at equivalence point : VaMa(#H) = VbMb(#OH)
Titration curves : Strong acid and strong base (Assume Ma=Mb)
Initial pH : [H+] = Ma (Molarity of strong acid)
Before equivalence point: volume of acid remaining = start volume – vol. base added
total volume = start volume + vol. base added
(vol acid remaining) (Ma) = (total vol.) (M2)
: [H+] = M2
At the equivalence point: [H+] = [OH–]
After equivalence point : volume of excess base = vol. base added – start volume
total volume = start volume + vol. base added
(vol excess base) (Mb) = (total vol.) (M2)
: [OH–] = M2
Titration curves: Weak acid and strong base (Assume Ma=Mb)
Initial pH : [H+] from original weak acid solution (other side!)
Before equivalence point: Use the Henderson Hasselbach equation
(Add strong base to weak acid as described above)
Special case = “Half neutralized”: [H+] = Ka of the weak acid or pH = pKa
At the equivalence point: [OH–] from the diluted salt solution (other side!)
After equivalence point : same as strong acid/strong base titration above:
Calculate [OH–] from diluted excess base.