2. Vitamin C is ascorbic acid - HC6H7O6- (MW=176.12 g mol), which has a K of...

 

Transcribed Image Text:

2. Vitamin C is ascorbic acid - HC6H7O6- (MW=176.12 g mol¹), which has a K₁ of 8.0x105. (a) Calculate the pH of a solution made by dissolving a 500-mg tablet of pure vitamin C in water and diluting to 50.00 mL. (b) Calculate the pH of a solution made by titrating the solution made above with 12.00 mL of 0.1183M NaOH. (c) Calculate the pH of a solution made by titrating the solution made above with 30.00mL of 0.1183M NaOH. a) (0.500 g x 1mol/176.12g) / 0.05000L = 0.00284 mol/0.05000L = 0.0568 M vitamin C 8.0x105 x2/0.0568 x = 0.00213 M = [H3O+] b) 0.01200 L x 0.1183 M = 0.001420 mol OH = ½ way point pH = pKa = 4.10 c) 0.03000 L x 0.1183 M = 0.003549 mol OH = region IV 0.003549 0.00284 = 0.000709 mol OH in excess 0.000709 mol OH / 0.08000 L = 0.00886 M = [OH-] pH = 11.95 pH = 2.67 2. Vitamin C is ascorbic acid - HC6H7O6- (MW=176.12 g mol¹), which has a K₁ of 8.0x105. (a) Calculate the pH of a solution made by dissolving a 500-mg tablet of pure vitamin C in water and diluting to 50.00 mL. (b) Calculate the pH of a solution made by titrating the solution made above with 12.00 mL of 0.1183M NaOH. (c) Calculate the pH of a solution made by titrating the solution made above with 30.00mL of 0.1183M NaOH. a) (0.500 g x 1mol/176.12g) / 0.05000L = 0.00284 mol/0.05000L = 0.0568 M vitamin C 8.0x105 x2/0.0568 x = 0.00213 M = [H3O+] b) 0.01200 L x 0.1183 M = 0.001420 mol OH = ½ way point pH = pKa = 4.10 c) 0.03000 L x 0.1183 M = 0.003549 mol OH = region IV 0.003549 0.00284 = 0.000709 mol OH in excess 0.000709 mol OH / 0.08000 L = 0.00886 M = [OH-] pH = 11.95 pH = 2.67