Kb Formula NH3 ?6H5NH2 C8H10N4024.1 x 10 C18H2103N8.9 x 10 (C2H5)2NH 6.9 x 104 (

Question

Kb Formula NH3 ?6H5NH2 C8H10N4024.1 x 10 C18H2103N8.9 x 10 (C2H5)2NH 6.9 x 104 (CH3)2NH | 5.9 × 10-4 C2H5NH2 NH20H C9H7N CH3NH2 C17H 1903N | 7.4 × 10 C5H11N C5H5N C9H7N C6H1503N 5.8 x 10 (C2H5)3N 5.2 x 10-4 (CH3)3N N2H4CO Base Ammonia 1.8 x 10 5 Aniline Caffeine Codeine 7.4 × 10-10 -4 7 Diethylamine Dimethylamine Ethylamine Hydroxylamine Isoquinoline Methylamine Morphine Piperidine Pyridine Quinoline Triethanolamine 4.3 × 10 9.1 x 10 2.5 x 10 4.2 x 10-4 -4 9 9 7 1.3 x 10 1.5 x 10 6.3 × 10 9 -10 7 Triethylamine Trimethylamine Urea 6.3 x 10 -14 1.5 × 10

Explanation / Answer

no of moles of (C2H5)3N   = molarity * volume in L

                                           = 0.269*0.0204   = 0.0054876moles

no of moles of HCl              = molarity * volume in L

                                             = 0.249*0.0091   = 0.0022659moles

Pkb   = -logKb

        = log5.2*10^-4

       = 3.2839

                (C2H5)3N(aq)    +    HCl (aq) --------------> (C2H5)3NHCl(aq)

I                0.0054876 0.0022659    0

C -0.0022659 - 0.0022659 0.0022659

E              0.0032217                 0                                  0.002265  

POH    =   Pkb + log[(C2H5)3NHCl]/[(C2H5)3N]

             = 3.2839 + log0.002265/0.0032217

            = 3.2839 -0.153

            = 3.131

PH      = 14-POH

            = 14-3.131 = 10.87 >>>>answer

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