Question

Calculate the pH of a 0.258 M solution of ethylenediamine (H_2NCH_2CH_2NH_2). The pK_a values for the acidic form of ethylenediamine (^+H_3NCH_2CH_2NH3^+) are 6.848 (pKa_1) and 9.928 (pK_a2). Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H_2NCH_2CH_2NH_2]= [H_2NCH_2CH_2NH_2^+] = [H_3^+ NCH_2CH_2NH_2^+ =

Answer 1

Make the ICE table for the first dissociation of H,NCH,CH,NH 0.258 +X 0.258-x Write the expression to calculate the value of x H.NCH.CH.NH: 11 ОН Substitute the values in the above equation for equilibrium condition 8.47x10-5 0.258-x x2 +8,47x105x-2.185x103-0 x=4.63×10-3 Thus, the concentration [H,NCH,CH NH;]-4.63x10 M, [4.63x10 M and [H,HCH NH0.258-4.63x10 0.253 M Make the ICE table for the second dissociation of H NCH,CH,NH, 4.63×10-3 E 4.63x103 -y Write the expression to calculate the value of x "н,NCH,CH.NH; || ОН- H.NCH,CH,NH b2 Substitute the values in the above equation for equilibrium condition 7.05x10-*- yXy 0.258-y

As the value of Kz2 is very small therefore neglect y from the denominator 7.05x108J 0.258 1.35x104 Thus, the concentration L"HNCH,CH,NHỊ」= 1.35x10"M and LOH-」= 1.35x10 Calculate the total concentration of OH ions -4.63x10-3 +1.35x104 4.765x10-3 M Write the expression to calculate the pH pH-14+log[oHt] Substitute the values in the above equation pH 14+log (4.765x10-3 -14-2.32 =11.68 Hence, the pH of solution is 11.68 As calculated above the concentration of each form of ethylenediamine is, HNCH CHH0.253 M LHNCH CH.NH5] = 4.63×10-M L"HNCH,CHME-1.35x10"M