Ionization Constants Acids Acid Formula Ionization Constant K a Acetic CH 3 COOH
ID: 949509 • Letter: I
Question
Ionization Constants
Acids
Acid
Formula
Ionization Constant Ka
Acetic
CH3COOH
1.8×10–5
Arsenic
H3AsO4
5.5×10–3
Benzoic
C6H5COOH
6.3×10–5
Boric
H3BO3
5.4×10–10
Butanoic
C3H7COOH
1.5×10–5
Carbonic
H2CO3
Ka1 = 4.5×10–7
Ka2 = 4.7×10–11
Chloroacetic
ClCH2COOH
1.4×10–3
Chlorous
HClO2
1.1×10–2
Chromic
H2CrO4
Ka1 = 1.8×10–1
Ka2 = 3.2×10–7
Cyanic
HCNO
3.5×10–4
Formic
HCOOH
1.8×10–4
Hydrazoic
HN3
2.5×10–5
Hydrocyanic
HCN
6.2×10–10
Hydrofluoric
HF
6.3×10–4
Hydrogen peroxide
H2O2
2.4×10–12
Hydrosulfuric
H2S
Ka1 = 8.9×10–8
Ka2 = 1.0×10–19
Hypobromous
HBrO
2.8×10–9
Hypochlorous
HClO
4.0×10–8
Hypoiodous
HIO
3.2×10–11
Iodic
HIO3
1.7×10–1
Nitrous
HNO2
5.6×10–4
Oxalic
C2H2O4
Ka1 = 5.6×10–2
Ka2 = 1.5×10–4
Paraperiodic
H5IO6
2.8×10–2
Pentanoic
C4H9COOH
1.5×10–5
Periodic
HIO4
7.3×10–2
Phenol
HC6H5O
1.3×10–10
Phosphoric
H3PO4
Ka1 = 6.9×10–3
Ka2 = 6.2×10–8
Ka3 = 4.8×10–13
Phosphorous
H3PO3
Ka1 = 5.0×10–2
Ka2 = 2.0×10–7
Propanoic
C2H5COOH
1.3×10–5
Sulfuric
H2SO4
Ka1 = very large
Ka2 = 1.2×10–2
Sulfurous
H2SO3
Ka1 = 1.4×10–2
Ka2 = 6.3×10–8
Trichloroacetic
Cl3CCOOH
2.2×10–1
Bases
Base
Formula
Ionization Constant Kb
Ammonia
NH3
1.8×10–5
Methylamine
CH3NH2
5.0×10–4
Dimethylamine
(CH3)2NH
5.4×10–4
Diethylamine
(C2H5)2NH
6.9×10–4
Trimethylamine
(CH3)3N
6.3×10–5
Triethylamine
(C2H5)3N
5.6×10–4
Ethylamine
CH3CH2NH2
6.3×10–4
Ethylenediamine
(CH2NH2)2
8.3×10–5
Pyridine
C5H5N
1.7×10–9
Hydroxylamine
NH2OH
8.7×10–9
Aniline
C6H5NH2
7.4×10–10
Hydrazine
H2NNH2
1.3×10–6
Ionization Constants
Acids
Acid
Formula
Ionization Constant Ka
Acetic
CH3COOH
1.8×10–5
Arsenic
H3AsO4
5.5×10–3
Benzoic
C6H5COOH
6.3×10–5
Boric
H3BO3
5.4×10–10
Butanoic
C3H7COOH
1.5×10–5
Carbonic
H2CO3
Ka1 = 4.5×10–7
Ka2 = 4.7×10–11
Chloroacetic
ClCH2COOH
1.4×10–3
Chlorous
HClO2
1.1×10–2
Chromic
H2CrO4
Ka1 = 1.8×10–1
Ka2 = 3.2×10–7
Cyanic
HCNO
3.5×10–4
Formic
HCOOH
1.8×10–4
Hydrazoic
HN3
2.5×10–5
Hydrocyanic
HCN
6.2×10–10
Hydrofluoric
HF
6.3×10–4
Hydrogen peroxide
H2O2
2.4×10–12
Hydrosulfuric
H2S
Ka1 = 8.9×10–8
Ka2 = 1.0×10–19
Hypobromous
HBrO
2.8×10–9
Hypochlorous
HClO
4.0×10–8
Hypoiodous
HIO
3.2×10–11
Iodic
HIO3
1.7×10–1
Nitrous
HNO2
5.6×10–4
Oxalic
C2H2O4
Ka1 = 5.6×10–2
Ka2 = 1.5×10–4
Paraperiodic
H5IO6
2.8×10–2
Pentanoic
C4H9COOH
1.5×10–5
Periodic
HIO4
7.3×10–2
Phenol
HC6H5O
1.3×10–10
Phosphoric
H3PO4
Ka1 = 6.9×10–3
Ka2 = 6.2×10–8
Ka3 = 4.8×10–13
Phosphorous
H3PO3
Ka1 = 5.0×10–2
Ka2 = 2.0×10–7
Propanoic
C2H5COOH
1.3×10–5
Sulfuric
H2SO4
Ka1 = very large
Ka2 = 1.2×10–2
Sulfurous
H2SO3
Ka1 = 1.4×10–2
Ka2 = 6.3×10–8
Trichloroacetic
Cl3CCOOH
2.2×10–1
Bases
Base
Formula
Ionization Constant Kb
Ammonia
NH3
1.8×10–5
Methylamine
CH3NH2
5.0×10–4
Dimethylamine
(CH3)2NH
5.4×10–4
Diethylamine
(C2H5)2NH
6.9×10–4
Trimethylamine
(CH3)3N
6.3×10–5
Triethylamine
(C2H5)3N
5.6×10–4
Ethylamine
CH3CH2NH2
6.3×10–4
Ethylenediamine
(CH2NH2)2
8.3×10–5
Pyridine
C5H5N
1.7×10–9
Hydroxylamine
NH2OH
8.7×10–9
Aniline
C6H5NH2
7.4×10–10
Hydrazine
H2NNH2
1.3×10–6
Acids
Acid
Formula
Ionization Constant Ka
Acetic
CH3COOH
1.8×10–5
Arsenic
H3AsO4
5.5×10–3
Benzoic
C6H5COOH
6.3×10–5
Boric
H3BO3
5.4×10–10
Butanoic
C3H7COOH
1.5×10–5
Carbonic
H2CO3
Ka1 = 4.5×10–7
Ka2 = 4.7×10–11
Chloroacetic
ClCH2COOH
1.4×10–3
Chlorous
HClO2
1.1×10–2
Chromic
H2CrO4
Ka1 = 1.8×10–1
Ka2 = 3.2×10–7
Cyanic
HCNO
3.5×10–4
Formic
HCOOH
1.8×10–4
Hydrazoic
HN3
2.5×10–5
Hydrocyanic
HCN
6.2×10–10
Hydrofluoric
HF
6.3×10–4
Hydrogen peroxide
H2O2
2.4×10–12
Hydrosulfuric
H2S
Ka1 = 8.9×10–8
Ka2 = 1.0×10–19
Hypobromous
HBrO
2.8×10–9
Hypochlorous
HClO
4.0×10–8
Hypoiodous
HIO
3.2×10–11
Iodic
HIO3
1.7×10–1
Nitrous
HNO2
5.6×10–4
Oxalic
C2H2O4
Ka1 = 5.6×10–2
Ka2 = 1.5×10–4
Paraperiodic
H5IO6
2.8×10–2
Pentanoic
C4H9COOH
1.5×10–5
Periodic
HIO4
7.3×10–2
Phenol
HC6H5O
1.3×10–10
Phosphoric
H3PO4
Ka1 = 6.9×10–3
Ka2 = 6.2×10–8
Ka3 = 4.8×10–13
Phosphorous
H3PO3
Ka1 = 5.0×10–2
Ka2 = 2.0×10–7
Propanoic
C2H5COOH
1.3×10–5
Sulfuric
H2SO4
Ka1 = very large
Ka2 = 1.2×10–2
Sulfurous
H2SO3
Ka1 = 1.4×10–2
Ka2 = 6.3×10–8
Trichloroacetic
Cl3CCOOH
2.2×10–1
Calculate the change in pH when 6.00 mL of 0.100 M HCI(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4CI(aq). A list of ionization constants can be found here. Number Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution. NumberExplanation / Answer
Volume of Buffer = 100 mL = 0.1 L
[NH3] = [NH4Cl] = 0.1 M
Moles = Molarity x Volume (L)
=> Moles of NH3 = 0.1 x 0.1 = 0.01
Moles of NH4Cl = 0.1 x 0.1 = 0.01
1)
Moles of HCl added = 0.1 x 0.006 = 0.0006 moles
Kb for NH3 = 1.8 x 10^-5
pKb = - log Kb
=> pKb = 4.745
pOH for the above buffer before adding HCl
pOH = pKb + log (NH4Cl / NH3)
=> pOH = 4.745 + log (0.01 / 0.01) = 4.745
pH = 14 - pOH = 9.255
pOH after adding HCl
pOH = pKb + log (NH4Cl + HCl / NH3 - HCl)
=> pOH = 4.745 + log (0.01 + 0.0006 / 0.01 - 0.0006) = 4.797
pH = 9.203
delta pH = 9.203 - 9.255 = - 0.052
2)
Moles of NaOH added = 0.1 x 0.006 = 0.0006 moles
Kb for NH3 = 1.8 x 10^-5
pKb = - log Kb
=> pKb = 4.745
pOH for the above buffer before adding NaOH
pOH = pKb + log (NH4Cl / NH3)
=> pOH = 4.745 + log (0.01 / 0.01) = 4.745
pH = 14 - pOH = 9.255
pOH after adding NaOH
pOH = pKb + log (NH4Cl - NaOH / NH3 + NaOH)
=> pOH = 4.745 + log (0.01 - 0.0006 / 0.01 + 0.0006) = 4.693
pH = 9.307
delta pH = 9.307 - 9.255 = 0.052
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.