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pH of 0.01M glycine solution is ( for glycine Ka1 = 4.5 x 10^(-3) and Ka2 = 1.7 x 10^(-10) at 298K
Dissociation constant, K = k1 × k2 = 4.5 × 10-3 Dissociation constant, K = k1 × k2 = 4.5 × 10-3*1.7 × 10-10 = 7.65 × 10-13Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1pH = -log[H+] = -log[0.87 × 10-7 mol L-1 = 7.061.7 × 10-10 = 7.65 × 10-13Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1pH = -log[H+] = -log[0.87 × 10-7 mol L-1 = 7.06
Dear Student,Please find below the solution to your problem.Dissociation constant, K = k1 × k2= 4.5 × 10-3Dissociation constant, K =k1 × k2= 4.5 × 10-3*1.7 × 10-10= 7.65 × 10-13Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1pH = -log[H+] = -log[0.87 × 10-7molL-1 = 7.061.7 × 10-10= 7.65 × 10-13Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1pH = -log[H+] = -log[0.87 × 10-7mol L-1 = 7.06Thanks and Regards
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