Ionization Constants Acids Acid Formula Ionization Constant K a Acetic CH 3 COOH
ID: 949505 • 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
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 pH for each of the following cases in the titration of 50.0 mL of 0.250 M HCIO(aq) with 0.250 M KOH(aq). The ionization constant for HCIO can be found here. Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH Number (c) after addition of 35.0 mL of KOH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOHExplanation / Answer
pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4
millimoles of HClO= 50 x0.250 = 12.5
a) 0 ml KOH added
pH = 1/2 (pKa- log C)
= 1/2 (7.4 -log (0.250) ) = 4.00
pH= 4.00
(b) after addition of 25.0 mL of KOH
it is half-equivalence point. so
pH = pKa = 7.4
pH = 7.4
(c) after addition of 35.0 mL of KOH
millimoles of KOH = 0.25 x 35 = 8.75
HClO + KOH ------------------------------> KClO + H2O
12.5 8.75 0 0 -----------------------initial
3.75 0 8.75 8.75 -------------------equilibirum
in the solution acid and salt remained so it can form buffer
For acidic buffer
Henderson-Hasselbalch equation
pH = pKa + log[salt/acid]
= 7.4 + log (8.75/3.75)
= 7.77
pH = 7.77
(d) after addition of 50.0 mL of KOH
millimoles of KOH = 0.25 x 50 = 12.5
HClO + KOH ------------------------------> KClO + H2O
12.5 12.5 0 0 -----------------------initial
0 0 12.5 12.5-------------------equilibirum
in the solution salt remained so we have to use salt hydrolysis.
it is the salt of strong base and weak acid so pH should be more than 7
[salt] = salt millimoles /total volume in ml
= 12.5/(50+50)
= 0.125 M
pH = 7 + 1/2[Pka + logC]
= 7 + 1/2 [7.4 + log (0.125)]
= 10.25
pH = 10.25
e) after addition of 60.0 mL of KOH
millimoles of KOH = 0.25 x 60 = 15
HClO + KOH ------------------------------> KClO + H2O
12.5 15 0 0 -----------------------initial
0 2.5 12.5 12.5-------------------equilibirum
in the solution strong base remained
[base ] = 2.5/total volume = 2.5/110 = 0.023
pOH = -log[OH-] = -log(0.023) =1.64
pH + pOH = 14
pH = 12.36
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