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1. Consider the reaction of dimethylammonium ion ((CH 3 ) 2 NH 2 + ) with water:

ID: 1000980 • Letter: 1

Question

1. Consider the reaction of dimethylammonium ion ((CH3)2NH2+) with water:

1-1. Select the correct chemical equation:
(CH3)2NH2+ (aq) + H2O (l) (CH3)2NH32+ (aq) + H3O+ (aq)
(CH3)2NH2+ (aq) + H2O (l) (CH3)2NH (aq) + OH (aq)
(CH3)2NH2+ (aq) + H2O (l) (CH3)2NH32+ (aq) + OH (aq)
(CH3)2NH2+ (aq) + H2O (l) (CH3)2NH (aq) + H3O+ (aq)

                                                                                                                                                                                              

1-2. The equilibrium expression for this reaction corresponds to:
Ka
Kb

                                                                                                                                                            
1-3. The numerator of the equilibrium expression (from 2) for this reaction (dilute solution) is:

[ (CH3)2NH32+ ][ OH ]
[ (CH3)2NH ]
[ (CH3)2NH ][ H3O+ ]
[ (CH3)2NH32+ ][ H3O+ ]
[ OH ][ (CH3)2NH2+ ]
[ (CH3)2NH32+ ]
[ (CH3)2NH2+ ][ H3O+ ]
[ H3O+ ]
                                                                                                                                                                                                                                                           

1-4. The denominator of the equilibrium expression (from 2) for this reaction (dilute solution) is:

[ (CH3)2NH32+ ]
[ (CH3)2NH2+ ]
[ (CH3)2NH ][ H3O+ ]
[ (CH3)2NH ]
[ (CH3)2NH ][ OH ]
[ (CH3)2NH2+ ][ H2O ]
[ OH ]
[ (CH3)2NH2+ ][ OH ]

                                                                                                                                                                     

2. The pKa of chloroacetic acid is 2.87, determine the Kb of the corresponding conjugate base to two significant figures.

Explanation / Answer

1.1 As dimethylammonium ion ((CH3)2NH2+) is an acid, the conjugate has to be a base (dimethylamine), and the water absorbs the proton left by the acid.

(CH3)2NH2+ (aq) + H2O (l) (CH3)2NH (aq) + H3O+ (aq)

1.2 As is releasing a proton, the correct equilibrium expression is Ka

By definition equilibrium expression is:

K= Products/Reactants

Ka= [(CH3)2NH] [H3O+]/[(CH3)2NH2+ ]

1.3 Numerator= [(CH3)2NH] [H3O+]

1.4 Denominator= [(CH3)2NH2+ ]

2. As pKa + pKb = 14 (remember this comes from Kw= Ka x Kb)

pKb= 14 - pKa

pKb= 14 - 2.87

pKb= 11.13