The equilibrium constant, K, for a neutralization reaction can be symbolized as Kn-Determine K, at 25 °C for the reactionNH, (aq) + HB1O(aq) = NH†(aq) + BrO¯(aq)using Kw, and the K, and K, values from the ionization constant tables.Kn|| AcidFormulaIonization Constant K aBaseFormulaIonization Constant K hAceticCH;COOH1.8×10-5AmmoniaNH31.8×10-5ArsenicH3ASO45.5x10-3|MethylamineCH;NH,5.0×10-4BenzoicCgH;COOH 6.3×10-5|Dimethylamine (CH3)2NH5.4x10-4BoricH;BO35.4x10-10Diethylamine(C,H3)2NH6.9×10–4ButanoicC3H;COOH 1.5×10-5Trimethylamine (CH3)3N6.3×10-5Kai= 4.5x10-7|Triethylamine(C,H3);N5.6x10-4Carbonic|H,CO3Ka2= 4.7×10-11EthylamineCH3CH,NH, 6.3×10-4ChloroaceticCICH,COOH 1.4×10-3Ethylenediamine (CH,NH,), 8.3×10-5ChlorousHCIO,1.1×10-2PyridineC3H;N1.7×10-9Kal= 1.8×10-1Hydroxylamine NH,OH8.7x10-9ChromicH2CrO4Ka2= 3.2x10-7AnilineCgH;NH27.4x10-10|СyanicНCNO| 3.5×10–4HydrazineH2NNH,1.3×10-6FormicНСООН1.8×10-4HydrazoicHN32.5x10-5|HydrocyanicНCN6.2×10-10HydrofluoricHF6.3×10-4Hydrogen peroxide H,O22.4x10-12Ka1= 8.9×10-8HydrosulfuricH,SKą2= 1.0×10-19|НуpobromousHBRO2.8×10-9

The question is based on the concept of chemical equilibrium. we have to calculate equilibrium…

The equilibrium constant, K, for a neutralization reaction can be symbolized as Kn- Determine K, at 25 °C for the reaction NH, (aq) + HBrO(aq) = NH†(aq) + BrO¯(aq) ising Kw, and the K, and K values from the ionization constant tables. Kn

The equilibrium constant, K, for a neutralization reaction can be symbolized as Kn- Determine K, at 25 °C for the reaction NH, (aq) + HBrO(aq) = NH†(aq) + BrO¯(aq) ising Kw, and the K, and K values from the ionization constant tables. Kn

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 56QRT

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Question

The equilibrium constant, K, for a neutralization reaction can be symbolized as Kn-
Determine K, at 25 °C for the reaction
NH, (aq) + HB1O(aq) = NH†(aq) + BrO¯(aq)
using Kw, and the K, and K, values from the ionization constant tables.
Kn
||

Acid
Formula
Ionization Constant K a
Base
Formula
Ionization Constant K h
Acetic
CH;COOH
1.8×10-5
Ammonia
NH3
1.8×10-5
Arsenic
H3ASO4
5.5x10-3
|Methylamine
CH;NH,
5.0×10-4
Benzoic
CgH;COOH 6.3×10-5
|Dimethylamine (CH3)2NH
5.4x10-4
Boric
H;BO3
5.4x10-10
Diethylamine
(C,H3)2NH
6.9×10–4
Butanoic
C3H;COOH 1.5×10-5
Trimethylamine (CH3)3N
6.3×10-5
Kai= 4.5x10-7
|Triethylamine
(C,H3);N
5.6x10-4
Carbonic
|H,CO3
Ka2= 4.7×10-11
Ethylamine
CH3CH,NH, 6.3×10-4
Chloroacetic
CICH,COOH 1.4×10-3
Ethylenediamine (CH,NH,), 8.3×10-5
Chlorous
HCIO,
1.1×10-2
Pyridine
C3H;N
1.7×10-9
Kal= 1.8×10-1
Hydroxylamine NH,OH
8.7x10-9
Chromic
H2CrO4
Ka2= 3.2x10-7
Aniline
CgH;NH2
7.4x10-10
|Сyanic
НCNO
| 3.5×10–4
Hydrazine
H2NNH,
1.3×10-6
Formic
НСООН
1.8×10-4
Hydrazoic
HN3
2.5x10-5
|Hydrocyanic
НCN
6.2×10-10
Hydrofluoric
HF
6.3×10-4
Hydrogen peroxide H,O2
2.4x10-12
Ka1= 8.9×10-8
Hydrosulfuric
H,S
Ką2= 1.0×10-19
|Нуpobromous
HBRO
2.8×10-9

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