Question

1.151 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 77.41 mL of water. 21.00 mL of HCl is added to the solution, resulting in a pH of 6.83. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85.

Answer 1

ACES–K is the conjugate base, ACES-H is the corresponding acid in the buffer of ACES–K/ ACES–H.

Use Henderson-hasselbach equation,

pH=pKa+log [ACES–K]/[ ACES–H]……………(1)

concentration of ACES–K=[ ACES–K]=moles of ACES–K/volume of ACES–K

moles of ACES–K=mass/molar mass=1.151g/220.29 g/mol=0.005224 moles

[ACES–K]=0.005224 moles/0.07741 L=0.0675 mol/L

Plug in the values in eqn (1),

6.83=6.85 +log (0.0675 mol/L)/[ ACES–H]

-0.02= log (0.0675 mol/L)/[ ACES–H]

10^(-0.02)= (0.0675 mol/L)/[ ACES–H]

0.0955=(0.0675 mol/L)/[ ACES–H]

[ ACES–H]=0.0675/0.09555=0.7064 mol/L

ACES-K + HCl→ACES-H+ KCl

As ACES-K reacts with HCl in equimolar quantity to give ACES–H

so ,concentration of HCl= conc of ACES-H=0.7064 mol/L