A 01276 g sample of an unknown monoprotic acid was dissolved in 250 mL of water

# A 01276 g sample of an unknown monoprotic acid was

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9.A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87. What is the Kaof the unknown acid? (10 points) 3 Copyright © 2016 by Thomas Edison State University. All rights reserved.
0 10. Calculate the pH at the equivalence point for the following titration: 0.20 M HCl versus 0.20 M methylamine (CH3NH3; Kb= 4.4 × 10–4) (5 points)Assume we have 1L container for the reaction:CH3NH2 + HCl ----> CH3NH3++ Cl-Initial 0.2 0.2 0 + x 11. A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL. (25 points) 0 4 Copyright © 2016 by Thomas Edison State University. All rights reserved.