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Grade: 12th pass

                        

pH of 0.01M glycine solution is ( for glycine Ka1 = 4.5 x 10^(-3) and Ka2 = 1.7 x 10^(-10) at 298K

3 years ago

Answers : (2)

Arun
24738 Points
							
Dissociation constant, K = k1 × k2 = 4.5 × 10-3 Dissociation constant, K = k1 × k2 = 4.5 × 10-3*1.7 × 10-10 = 7.65 × 10-13
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
3 years ago
Rishi Sharma
askIITians Faculty
614 Points
							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
3 months ago
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