The last one Calculate the pH of a 0.293 M solution of ethylenediamine (HNCHCH2N

Question

The last one

Calculate the pH of a 0.293 M solution of ethylenediamine (HNCHCH2NH2). The pK, values for the acia form of ethylenediamine ('H3NCH2CH2NH3) are 6.848 (pKa) and 9.928 (pKa) Number pH-1.7 Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. Number H,NCH, CH,NH 0.288M Number H,NCH,CH2NH0.00498 Incorrect Number The concentration of ethylenediamine in the = 11 M acidic form can be determined by rearranging the K2 expression. H, NCH,CH,NH0.00298 M H,NCH,CH,NH b2

Explanation / Answer

pKb1 = 14 - pKa2 = 14 - 9.928 = 4.072

Kb1 = 8.47 x 10^-5

pKb2 = 14 - 6.848 = 7.152

Kb2 = 7.5 x 10^-8

[+H3NCH2CH2NH3+] is always equal to Kb2

so that

[+H3NCH2CH2NH3+] = Kb2 = 7.5 x 10^-8 M --------------(answer)

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