1.800 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 89.94 mL of water. 29.91 mL of HCl is added to the solution, resulting in a pH of 7.24. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85.
Finding the equlibrium concentrations
ACES- + H2O <====> ACES + OH-
Moles of ACES- =
Volume of ACES- = 89.94 mL = 0.08994 L
Molarity of ACES- =
pKa = 6.85
pKb = 14 - 6.85 = 7.15
Kb = 10-pKb
Kb = 10-7.15
Kb = 7.08 x 10-8
[ACES] = [OH-] = 8.02 x 10-5 M
[ACES-] = 0.0908 - 8.02x10-5 =0.0906 M
Let 'x' be the moles of H+ ions added
Total volume of solution after addition of HCl = 89.94 + 29.91 = 119.85 mL = 0.11985 L
Moles of ACES- = Molarity x total volume = 0.0906 x 0.11985 =0.01085 mol
Moles of ACES = 8.02x10-5 x 0.11985 =9.61 x 10-6 mol
Using Henderson-Hasselbalch equation
Given pH = 7.24
pOH = 6.76
x = 0.00313 mol
Moles of HCl = 0.00313 mol
Volume of HCl = 29.91 mL = 0.02991 L
Concentration of HCl =
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