Grade 12th passPhysical Chemistry
pH of 0.01M glycine solution is ( for glycine Ka1 = 4.5 x 10^(-3) and Ka2 = 1.7 x 10^(-10) at 298K
2 Answers
Arun
Approved Tutor Answer8 Years agoDissociation constant, K = k1 × k2 = 4.5 × 10-3 Dissociation constant, K = k1 × k2 = 4.5 × 10-3*1.7 × 10-10 = 7.65 × 10-13
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7 mol L-1
= 7.06
1.7 × 10-10 = 7.65 × 10-13
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7 mol L-1
= 7.06
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7 mol L-1
= 7.06
1.7 × 10-10 = 7.65 × 10-13
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7 mol L-1
= 7.06
Rishi Sharma
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 × 10-13
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7molL-1
= 7.06
1.7 × 10-10= 7.65 × 10-13
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7mol L-1
= 7.06
Thanks and Regards
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 × 10-13
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7molL-1
= 7.06
1.7 × 10-10= 7.65 × 10-13
Hydrogen ion concentration, = K . C = 7 . 65 × 10 - 13 × 0 . 01 = 0 . 87 mol L - 1
pH = -log[H+]
= -log[0.87 × 10-7mol L-1
= 7.06
Thanks and Regards
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