1.911 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW

Question

1.911 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 78.59 mL of water. 13.55 mL of HCl is added to the solution, resulting in a pH of 7.00. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85. [HCl]=___M? Hint 1: ACES– is the conjugate base of the weak acid, and common buffer compound, ACES. 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. Develop expressions for the moles of ACES and ACES– in solution after the addition of H. Use the Henderson-Hasselbalch equation to determine the moles of H added to the solution, resulting in a pH = 7.00. Hint 2: 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 x number of moles of HCl. Use the Henderson-Hasselbalch equation to determine the moles of HCl added to the solution.

Explanation / Answer

First let's determine the number of moles of ACES-K:

moles = 1.911 g / 220.29 g/mol = 8.67x10-3 moles

Total volume = 78.59 + 13.55 = 92.14 mL or 0.09214 L

Ka = 10-6.85 = 1.41x10-7 ----> Kb = 1x10-14 / 1.41x10-7 = 7.09x10-8

The HH equation:

pH = pKa + log(A-/HA)

Now, it's time to determine the concentration of A- corresponding to ACES- and then, solve for HA which is HCl so:

HCl + ACES- -------------> ACES + Cl-

And ACES in solution is:

ACES- + H2O ----------> ACES + OH-

i: 8.67x10-3   0 0

e: 8.67x10-3 - x x x

Kb = x2/8.67x10-3    concentration: 8.67x10-3 / 0.09214 = 0.09409 M

7.09x10-8 = x2 / 0.09409

x = (0.09409 * 7.09x10-8 )1/2 = 8.17x10-5 M

now, use the HH equation:

7 = 6.85 + log(8.17x10-5 / HCl)

0.15 = log(8.17x10-5 / HCl)

1.41 = 8.17x10-5 / HCl

[HCl] = 8.17x10-5 / 1.41

[HCl] = 5.78x10-5 M

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