Question

1.521 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (MW=220.29 g/mol) is dissolved in 58.56 mL of water. 22.12 mL of HCl is added to the solution, resulting in a pH of 6.70. Calculate the concentration of the HCl solution. the pKa of ACES is 6.85.

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Answer 1

moles of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt = 1.521/220.29=0.0069

Molarity= 0.0069*1000/58.56=0.1179

ACES+ H3O+ ----> HACES + H2O

let x =moles of HCl supplied ( Supplies H+ ions)

moles of ACES remaining = 0.0069-x mole of HCl added = x*22.12/1000 = 0.02212x

volume after mixing = 22.12+58.56= 80.68ml = 0.08068 L

concentrations of base =

then ACES at equilibrium =0.0069-x and acid = 0.02212x

form PH= Pka+ log [base/acid]

6.7= 6.85+ log {(0.0069-x)/0.02212x}

1.412= log {(0.0069-x)/ 0.02212x}

25.85= 0.0069-x/ 0.02212x

25.85*0.02212x= 0.0069-x

0.572x= 0.0069-x

1.572x= 0.0069

x= 0.0069/1.572=0.0044

Molarity= 0.0044*1000/22.12=0.1989M