AP Acid Base Chemistry

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Melissa Rathier's picture

drilling types of acid-base solutions and knowing what the process for solving for pH Strong acids, strong bases, weak acids, weak bases, buffers, titrations

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HNO3

Strong Acid
HNO3 --> H+ + NO3-
pH = -log[H+]

HOCN
Ka = 3.5 e-4

weak acid
hydrolyze
ICE

Ca(OH)2

Strong Base
Ca(OH)2 --> Ca2+ + 2OH-
OH- = 2x Ca(OH)2 (ratio)
pOH
pH

HN3
Ka = 1.8 e-5

weak acid
hydrolyze
ICE

H2O2
Ka = 1.8 e-12

Weak acid
hydrolyze
ICE

N2H4
Kb = 9.6 e-7

weak base
hydrolyze
ICE

C5H5N
Kb = 1.5 e-9

weak base
hydrolyze
ICE

KCN
Ka = 6.2 e-10 (HCN)

Salt
KCN --> K+ + CN-
CN- is weak base hydrolyze
KaKb = Kw
ICE

(CH3)2NH
Kb = 9.6 e-4

weak base
hydrolyze
ICE

NaOCl
Ka = 3.0 e-8 (HOCl)

NaOCl -- Na+ + OCl-
hydrolyze OCl- (WB)
ICE
KaKb = Kw
pOH, pH

C5H5NHCl
Kb= 1.5 e-9 (C5H5N)

C5H5NHCl--> C5H5NH+ +Cl-
C5H5NH+ is WA hydrolyze
KaKb = Kw
ICE

CH3NH3Br
Kb = 4.4 e-4 (CH3NH2)

salt
CH3NH3Br --> CH3NH3+ + Br-
CH3NH3 is WA, hydrolyze
KaKb = Kw
ICE

HOCl/ NaOCl
Ka = 3.0 e-8

Buffer
pH = pKa + log (b/a)

HC4H3N2O3/KC4H3N2O3

Ka = 9.8 e-5

Buffer
pH = pKa + log(b/a)

(CH3)3N/ (CH3)3NHBr
Kb = 7.4 e-5

Buffer
KaKb = Kw
pH = pKa + log(b/a)

NH3/ NH4Cl
Kb = 1.8 e-5

Buffer
KaKb = Kw
pH = pKa + log(b/a)

HCl + KOH

Strong acid Strong Base
dilute
neutralize (ICF)
use excess to find pH

Sr(OH)2 + HNO3

Strong acid Strong Base
dilute
neutralize (ICF)
use excess to find pH
(watch ratio!)

[HOCl] > [KOH]

weak acid/ strong base
before equivalence

dilute
neutralize (ICF)
ICE (Buffer!)

[N2H4] > [HCl]

weak base/ strong acid
before equivalence

dilute
neutralize (ICF)
ICE (Buffer!)

[N2H4] = [HCl]

weak base/ strong acid
at equivalence

dilute
neutralize (ICF)
salt of conjugate acid
KaKb = Kw
ICE

1/2[HOCl] = [KOH]

weak acid/ strong base
half-way to equivalence

pH = pKa