Question

A 1.982g sample of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt ( ACES−K+, MW = 220.29 g/mol) is dissolved in 87.51 mL of water.To the solution, 21.98 mL of HCl is added, resulting in a pH of 7.27. Calculate the concentration of the HCl solution. The p?a of ACES is 6.85.

I have calculated this several times and have gotten the same answer, 0.112754 M. However, Sapling says that I am wrong. Then Sapling gives these instructions:

To begin, calculate the original number of moles of ACES− in solution. When a strong acid, such as HCl, is added to a weak base, such as ACES−, the HCl will react completely and ACES− will be converted to ACES. The number of moles of H+ added equals the number of moles of ACES− consumed, which is equal to the corresponding increase in the moles of ACES in solution. Write the reaction between ACES− and HCl and set up a table showing the initial and final number of moles of ACES− and ACES in solution after the addition of ? number of moles of HCl. Use the Henderson-Hasselbalch equation to determine the moles of HCl added to the solution.

I have attempted this and still do not have the correct answer. Please and thank you to whoever knows how to do this.

Answer 1

1 mol No of moles of ACESK+, X = 1.982 g x 5220.29 g = 0.008997 mol pH of the buffer = 7.27 PK, of weak acid ACES, PK, = 6.85 ICE table: + H+ X 0.008997 I(mol): C(mol): E (mol): -x 0.008997- X 0 According to Henderson equation, pH = pK, + log fuxi 0.008997- x 7.27 = 6.85 + logº 0.008997-X -042 log - 0.008997- X -* = 10042 = 2.63 x = 0.002478 mol No of moles of HCI added=(x) mol = 0.002478 mol Volume of HCI added = 21.98 mL = 0.02198 L Concentration of HC1 = 0.02198 L 0.002478 mol = 0.1128 M