Question

1.074 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 54.37 mL of water. 22.02 mL of HCl is added to the solution, resulting in a pH of 6.78. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85.

Answer 1

let this be a buffer

pH = pKa + log(A-/HA)

A- = ACES

HA = ACES-H (acicid aces)

then

mol of ACES = mass/MW = 1.074/220.29 = 0.004875

V = 54.37 mL

initially

ACES base = 0.004875

after adding

"x" mol of HCl

ACES base = 0.004875 - x

ACES acid = x

substitute in pH equation

6.78 = 6.85 + log(x/(0.004875-x))

10^(6.78 - 6.85) = x/(0.004875-x)

1/0.85113*x = 0.004875-x

(1/0.85113+1)x = 0.004875

x = 0.004875 / (1/0.85113+1) =

x = 0.002241

now... this is mol, change to mmol

mmol = x*10^3 = 2.24 mmol

[HCl] = mmol/mL = 2.24/54.37 = 0.0411 M